{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0001","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d1ffdccecfe0f54d9fb5433dec7b88999efe3b32969071494c55907e836d3e4","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104001,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,10],[2,10],[7,10],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[0,1,3,99]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/10, 2/10, 7/10, 0/1], aligned values X=[0, 1, 3, 99] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/10], [1, [2, 3], 7/10]], Yes, 23/10, 121/100)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/10], [1, [2, 3], 7/10]], Yes, 23/10, 121/100)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0002","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1556b27f9e412c0a52cd4d945082ec6c5e6381dae59d0cc712e04bfd2df0908","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104002,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,11],[2,11],[8,11],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[1,2,4,100]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/11, 2/11, 8/11, 0/1], aligned values X=[1, 2, 4, 100] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/11], [1, [2, 3], 8/11]], Yes, 37/11, 138/121)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/11], [1, [2, 3], 8/11]], Yes, 37/11, 138/121)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0003","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"356ca45b9777a265e48ef394006bfc0d71f5726382df71bef29adb01e90482b6","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104003,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,12],[2,12],[9,12],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[2,3,5,101]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/12, 2/12, 9/12, 0/1], aligned values X=[2, 3, 5, 101] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 1/4], [1, [2, 3], 3/4]], Yes, 53/12, 155/144)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 1/4], [1, [2, 3], 3/4]], Yes, 53/12, 155/144)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0004","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e578037ed7ae50323c286dd9874457374a3684b70e45fa8a0aaa874520a02af","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104004,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,13],[2,13],[10,13],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[3,4,6,102]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/13, 2/13, 10/13, 0/1], aligned values X=[3, 4, 6, 102] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/13], [1, [2, 3], 10/13]], Yes, 71/13, 172/169)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/13], [1, [2, 3], 10/13]], Yes, 71/13, 172/169)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0005","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00783071392ea969f855bfcf85ef269467e14fbcfbfec89d97fd75760f5ef22b","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104005,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,14],[2,14],[11,14],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[4,5,7,103]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/14, 2/14, 11/14, 0/1], aligned values X=[4, 5, 7, 103] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/14], [1, [2, 3], 11/14]], Yes, 13/2, 27/28)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/14], [1, [2, 3], 11/14]], Yes, 13/2, 27/28)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0006","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"366ab5150c7b9dee0b36987dc29cf4a077cef4592dbafbfa7cd4fbafb47a1c3c","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104006,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,15],[2,15],[12,15],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[5,6,8,104]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/15, 2/15, 12/15, 0/1], aligned values X=[5, 6, 8, 104] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 1/5], [1, [2, 3], 4/5]], Yes, 113/15, 206/225)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 1/5], [1, [2, 3], 4/5]], Yes, 113/15, 206/225)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0007","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f5ee3065693a3af3310f3b5467bb5cf1ff1660cd48945aa663d5b63bff343402","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104007,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,16],[2,16],[13,16],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[6,7,9,105]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/16, 2/16, 13/16, 0/1], aligned values X=[6, 7, 9, 105] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/16], [1, [2, 3], 13/16]], Yes, 137/16, 223/256)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/16], [1, [2, 3], 13/16]], Yes, 137/16, 223/256)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0008","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5bd12d0bfa011d7b85cdf1aeab38b4ecb82338842dc16728fa3db933d0d20960","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104008,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,17],[2,17],[14,17],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[7,8,10,106]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/17, 2/17, 14/17, 0/1], aligned values X=[7, 8, 10, 106] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/17], [1, [2, 3], 14/17]], Yes, 163/17, 240/289)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/17], [1, [2, 3], 14/17]], Yes, 163/17, 240/289)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0009","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d3626dfabe1b8ee0f2eaeb56a4a8587749c2edd3b43f68ff1622f2c7b7473d9","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104009,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,18],[2,18],[15,18],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[8,9,11,107]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/18, 2/18, 15/18, 0/1], aligned values X=[8, 9, 11, 107] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 1/6], [1, [2, 3], 5/6]], Yes, 191/18, 257/324)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 1/6], [1, [2, 3], 5/6]], Yes, 191/18, 257/324)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0010","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99c1fba111dc41dcb2a09b1b6a8dcba840b55860a22b901a89674324b69986c5","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104010,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,19],[2,19],[16,19],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[9,10,12,108]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/19, 2/19, 16/19, 0/1], aligned values X=[9, 10, 12, 108] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/19], [1, [2, 3], 16/19]], Yes, 221/19, 274/361)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/19], [1, [2, 3], 16/19]], Yes, 221/19, 274/361)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0011","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"460f926f211dfdbc388022b577d1f3445a0aa0591e7d138556da53ddb7d578d5","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104011,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,20],[2,20],[17,20],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[10,11,13,109]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/20, 2/20, 17/20, 0/1], aligned values X=[10, 11, 13, 109] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/20], [1, [2, 3], 17/20]], Yes, 253/20, 291/400)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/20], [1, [2, 3], 17/20]], Yes, 253/20, 291/400)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0012","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"12009d1271357c898a46f931ff3c5520d378fa2cdd3ba7f286d1f1e8904a32d8","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104012,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,21],[2,21],[18,21],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[11,12,14,110]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/21, 2/21, 18/21, 0/1], aligned values X=[11, 12, 14, 110] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 1/7], [1, [2, 3], 6/7]], Yes, 41/3, 44/63)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 1/7], [1, [2, 3], 6/7]], Yes, 41/3, 44/63)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0013","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b46890ae48c3523210537ee4fa5df2072d5ff3548331d4dcc25ed12c72eb342","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104013,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,22],[2,22],[19,22],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[12,13,15,111]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/22, 2/22, 19/22, 0/1], aligned values X=[12, 13, 15, 111] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/22], [1, [2, 3], 19/22]], Yes, 323/22, 325/484)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/22], [1, [2, 3], 19/22]], Yes, 323/22, 325/484)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0014","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"473c80a1737df68d59bf1065ce4df0154d98ac7e3bfcb924a78bc485baded8b1","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104014,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,23],[2,23],[20,23],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[13,14,16,112]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/23, 2/23, 20/23, 0/1], aligned values X=[13, 14, 16, 112] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/23], [1, [2, 3], 20/23]], Yes, 361/23, 342/529)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/23], [1, [2, 3], 20/23]], Yes, 361/23, 342/529)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0015","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93a575cf7c03fb6619b16a387bef62a898edcae327c8b261e6c3df1cb3fffbe1","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104015,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,24],[2,24],[21,24],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[14,15,17,113]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/24, 2/24, 21/24, 0/1], aligned values X=[14, 15, 17, 113] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 1/8], [1, [2, 3], 7/8]], Yes, 401/24, 359/576)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 1/8], [1, [2, 3], 7/8]], Yes, 401/24, 359/576)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0016","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"188ad63cd3cd039215533eb50861267e4b7fad2f0278bacd896addd477d2e4ec","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104016,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,25],[2,25],[22,25],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[15,16,18,114]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/25, 2/25, 22/25, 0/1], aligned values X=[15, 16, 18, 114] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/25], [1, [2, 3], 22/25]], Yes, 443/25, 376/625)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/25], [1, [2, 3], 22/25]], Yes, 443/25, 376/625)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0017","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c6b2bab55110d5cc908d289f91870eb58d5bc6d7156ec1216aa0946f77f36a9b","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104017,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,26],[2,26],[23,26],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[16,17,19,115]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/26, 2/26, 23/26, 0/1], aligned values X=[16, 17, 19, 115] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/26], [1, [2, 3], 23/26]], Yes, 487/26, 393/676)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/26], [1, [2, 3], 23/26]], Yes, 487/26, 393/676)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0018","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf7fcc8e157647a623c60e80b085d322c598a7179bcc327b9dfcbcca5da12af6","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104018,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,27],[2,27],[24,27],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[17,18,20,116]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/27, 2/27, 24/27, 0/1], aligned values X=[17, 18, 20, 116] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 1/9], [1, [2, 3], 8/9]], Yes, 533/27, 410/729)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 1/9], [1, [2, 3], 8/9]], Yes, 533/27, 410/729)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0019","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a9c5849a01528ba11057cec3fc1938c270dd8f1e64ed0fd65d5f34966f6ecc0","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104019,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,28],[2,28],[25,28],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[18,19,21,117]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/28, 2/28, 25/28, 0/1], aligned values X=[18, 19, 21, 117] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/28], [1, [2, 3], 25/28]], Yes, 83/4, 61/112)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/28], [1, [2, 3], 25/28]], Yes, 83/4, 61/112)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0020","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a18fc65a9b1975ed721bb661406e2560d51492929742f46fc3cd0f84ea43cb87","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104020,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,29],[2,29],[26,29],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[19,20,22,118]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/29, 2/29, 26/29, 0/1], aligned values X=[19, 20, 22, 118] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/29], [1, [2, 3], 26/29]], Yes, 631/29, 444/841)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/29], [1, [2, 3], 26/29]], Yes, 631/29, 444/841)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0021","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db6072b3666933158c2dcaf39126fd5903dd76da412703f836c3bcf3debd3f70","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104021,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,30],[2,30],[27,30],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[20,21,23,119]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/30, 2/30, 27/30, 0/1], aligned values X=[20, 21, 23, 119] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 1/10], [1, [2, 3], 9/10]], Yes, 683/30, 461/900)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 1/10], [1, [2, 3], 9/10]], Yes, 683/30, 461/900)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0022","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a391c329f0049d993f9e6377c245ccf21f9986d2bd877840595070f655f7d4f","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104022,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,31],[2,31],[28,31],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[21,22,24,120]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/31, 2/31, 28/31, 0/1], aligned values X=[21, 22, 24, 120] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/31], [1, [2, 3], 28/31]], Yes, 737/31, 478/961)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/31], [1, [2, 3], 28/31]], Yes, 737/31, 478/961)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0023","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5040807800d89ad338491e7f0e9b7cb28729450412aec9880e69c6f0c86e8303","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104023,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,32],[2,32],[29,32],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[22,23,25,121]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/32, 2/32, 29/32, 0/1], aligned values X=[22, 23, 25, 121] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/32], [1, [2, 3], 29/32]], Yes, 793/32, 495/1024)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/32], [1, [2, 3], 29/32]], Yes, 793/32, 495/1024)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0024","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0cdba871ff0573b42410fce54233f621878ecc3c02dfd399a57e31bdb1e01b81","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104024,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,33],[2,33],[30,33],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[23,24,26,122]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/33, 2/33, 30/33, 0/1], aligned values X=[23, 24, 26, 122] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 1/11], [1, [2, 3], 10/11]], Yes, 851/33, 512/1089)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 1/11], [1, [2, 3], 10/11]], Yes, 851/33, 512/1089)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0025","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2802c07bbe7b06f240c3a0aff059dd25d63d5b59271aba74e32c7ee262369f15","family_id":"qf-mod41-totalvar-partition-qualification","seed":202611104025,"semantic_template":"qf-mod41-totalvar-partition-qualification/{\"masses\":[[1,34],[2,34],[31,34],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[24,25,27,123]}"},"fields":{"title":"Nonempty Cells May Have Zero Mass without Becoming Empty Events","stem":"Use the same complete normalized finite law masses=[1/34, 2/34, 31/34, 0/1], aligned values X=[24, 25, 27, 123] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (partition, cell_mass_rows, complete_disjoint_certificate, mean, variance).","answer":{"kind":"expression","value":"([[0, 1], [2, 3]], [[0, [0, 1], 3/34], [1, [2, 3], 31/34]], Yes, 911/34, 529/1156)","explanation":"Report cells and [cell index,cell atoms,cell mass], then the complete disjoint certificate, global mean and global variance. Zero-mass atoms remain part of the carrier and their containing nonempty cell. The exact result, retaining the requested order, is ([[0, 1], [2, 3]], [[0, [0, 1], 3/34], [1, [2, 3], 31/34]], Yes, 911/34, 529/1156)."},"skills":["Nonempty Cells May Have Zero Mass without Becoming Empty Events"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0026","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b9038b9875b469424fc74b72edb0fd047a4534d3730e91428236d04e1203bf0","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104026,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,10],[2,10],[7,10],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[0,1,3,99]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/10, 2/10, 7/10, 0/1], aligned values X=[0, 1, 3, 99] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/10, 2/3, 2/9], [1, [2, 3], 7/10, 3, 0]], 1/15, 343/300, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/10, 2/3, 2/9], [1, [2, 3], 7/10, 3, 0]], 1/15, 343/300, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0027","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f64080ff7240a4c19cf1a23f0a2e317755caee68dd5fb3a2437184479df9a81","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104027,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,11],[2,11],[8,11],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[1,2,4,100]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/11, 2/11, 8/11, 0/1], aligned values X=[1, 2, 4, 100] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/11, 5/3, 2/9], [1, [2, 3], 8/11, 4, 0]], 2/33, 392/363, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/11, 5/3, 2/9], [1, [2, 3], 8/11, 4, 0]], 2/33, 392/363, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0028","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dad7fb7b5cdb97419ddcd0261b87fdde80d49ee79eb34e36dfbdf25fdc7e07c7","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104028,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,12],[2,12],[9,12],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[2,3,5,101]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/12, 2/12, 9/12, 0/1], aligned values X=[2, 3, 5, 101] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 1/4, 8/3, 2/9], [1, [2, 3], 3/4, 5, 0]], 1/18, 49/48, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 1/4, 8/3, 2/9], [1, [2, 3], 3/4, 5, 0]], 1/18, 49/48, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0029","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d3dff880eac89fec7e222e711f04b8412dd39568b5166a3ac3af8295cfe07ce","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104029,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,13],[2,13],[10,13],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[3,4,6,102]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/13, 2/13, 10/13, 0/1], aligned values X=[3, 4, 6, 102] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/13, 11/3, 2/9], [1, [2, 3], 10/13, 6, 0]], 2/39, 490/507, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/13, 11/3, 2/9], [1, [2, 3], 10/13, 6, 0]], 2/39, 490/507, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0030","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e64e1c8332d5e157e16579759245e43503282cfff3ed50aca569675abd664862","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104030,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,14],[2,14],[11,14],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[4,5,7,103]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/14, 2/14, 11/14, 0/1], aligned values X=[4, 5, 7, 103] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/14, 14/3, 2/9], [1, [2, 3], 11/14, 7, 0]], 1/21, 11/12, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/14, 14/3, 2/9], [1, [2, 3], 11/14, 7, 0]], 1/21, 11/12, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0031","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c87e3e3aacc6702d1eb60795ae310e5ab4ab1c5ceb68d3b13b8c6c38018f0ed3","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104031,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,15],[2,15],[12,15],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[5,6,8,104]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/15, 2/15, 12/15, 0/1], aligned values X=[5, 6, 8, 104] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 1/5, 17/3, 2/9], [1, [2, 3], 4/5, 8, 0]], 2/45, 196/225, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 1/5, 17/3, 2/9], [1, [2, 3], 4/5, 8, 0]], 2/45, 196/225, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0032","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6750072c770b577f9556951ae090b23d0e89d13975d93252716f484e8536c755","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104032,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,16],[2,16],[13,16],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[6,7,9,105]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/16, 2/16, 13/16, 0/1], aligned values X=[6, 7, 9, 105] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/16, 20/3, 2/9], [1, [2, 3], 13/16, 9, 0]], 1/24, 637/768, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/16, 20/3, 2/9], [1, [2, 3], 13/16, 9, 0]], 1/24, 637/768, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0033","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f6ce5d1d074e2cc300ac8f4c33dcdb6dc26eee59bfc0eefb1c97881dcc7085a","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104033,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,17],[2,17],[14,17],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[7,8,10,106]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/17, 2/17, 14/17, 0/1], aligned values X=[7, 8, 10, 106] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/17, 23/3, 2/9], [1, [2, 3], 14/17, 10, 0]], 2/51, 686/867, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/17, 23/3, 2/9], [1, [2, 3], 14/17, 10, 0]], 2/51, 686/867, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0034","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea6b266abda640992ac1ca07f14787f29d850f856d5f5b45b1c4b6c4838da4d1","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104034,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,18],[2,18],[15,18],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[8,9,11,107]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/18, 2/18, 15/18, 0/1], aligned values X=[8, 9, 11, 107] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 1/6, 26/3, 2/9], [1, [2, 3], 5/6, 11, 0]], 1/27, 245/324, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 1/6, 26/3, 2/9], [1, [2, 3], 5/6, 11, 0]], 1/27, 245/324, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0035","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6595dfc5d8a25179ec71b98e37102509f7a75e78869ae780b3489f65c8c91081","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104035,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,19],[2,19],[16,19],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[9,10,12,108]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/19, 2/19, 16/19, 0/1], aligned values X=[9, 10, 12, 108] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/19, 29/3, 2/9], [1, [2, 3], 16/19, 12, 0]], 2/57, 784/1083, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/19, 29/3, 2/9], [1, [2, 3], 16/19, 12, 0]], 2/57, 784/1083, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0036","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"447125eed738be1c115bbf0829559f04eeafc971a78a189043ff437575c2c50b","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104036,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,20],[2,20],[17,20],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[10,11,13,109]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/20, 2/20, 17/20, 0/1], aligned values X=[10, 11, 13, 109] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/20, 32/3, 2/9], [1, [2, 3], 17/20, 13, 0]], 1/30, 833/1200, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/20, 32/3, 2/9], [1, [2, 3], 17/20, 13, 0]], 1/30, 833/1200, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0037","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3e8a96e9921b57761ed5de0685b607be706e86fc1308bd4d193c740d6997d9e","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104037,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,21],[2,21],[18,21],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[11,12,14,110]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/21, 2/21, 18/21, 0/1], aligned values X=[11, 12, 14, 110] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 1/7, 35/3, 2/9], [1, [2, 3], 6/7, 14, 0]], 2/63, 2/3, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 1/7, 35/3, 2/9], [1, [2, 3], 6/7, 14, 0]], 2/63, 2/3, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0038","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34aa4c48fa078d764ea3dfb4d56e166b493c489a081f9013257b9ed85384db0a","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104038,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,22],[2,22],[19,22],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[12,13,15,111]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/22, 2/22, 19/22, 0/1], aligned values X=[12, 13, 15, 111] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/22, 38/3, 2/9], [1, [2, 3], 19/22, 15, 0]], 1/33, 931/1452, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/22, 38/3, 2/9], [1, [2, 3], 19/22, 15, 0]], 1/33, 931/1452, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0039","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0696f7f385b7bbd86fe1415c0697f24b480e7912db0c9b5404419e02b5cb1551","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104039,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,23],[2,23],[20,23],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[13,14,16,112]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/23, 2/23, 20/23, 0/1], aligned values X=[13, 14, 16, 112] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/23, 41/3, 2/9], [1, [2, 3], 20/23, 16, 0]], 2/69, 980/1587, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/23, 41/3, 2/9], [1, [2, 3], 20/23, 16, 0]], 2/69, 980/1587, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0040","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68be97d947754d02b24ba7177e9c800ddeff6f48e26761c34d50375b7180b342","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104040,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,24],[2,24],[21,24],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[14,15,17,113]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/24, 2/24, 21/24, 0/1], aligned values X=[14, 15, 17, 113] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 1/8, 44/3, 2/9], [1, [2, 3], 7/8, 17, 0]], 1/36, 343/576, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 1/8, 44/3, 2/9], [1, [2, 3], 7/8, 17, 0]], 1/36, 343/576, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0041","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e51c975012b040dec40a56fc4a8201c56cb6dcb9b620b7b6946f73a61a6e3f2f","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104041,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,25],[2,25],[22,25],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[15,16,18,114]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/25, 2/25, 22/25, 0/1], aligned values X=[15, 16, 18, 114] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/25, 47/3, 2/9], [1, [2, 3], 22/25, 18, 0]], 2/75, 1078/1875, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/25, 47/3, 2/9], [1, [2, 3], 22/25, 18, 0]], 2/75, 1078/1875, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0042","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b89b3075084f6beed773f70b2adbf708cd2a92fc6454d4144440db282df30972","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104042,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,26],[2,26],[23,26],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[16,17,19,115]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/26, 2/26, 23/26, 0/1], aligned values X=[16, 17, 19, 115] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/26, 50/3, 2/9], [1, [2, 3], 23/26, 19, 0]], 1/39, 1127/2028, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/26, 50/3, 2/9], [1, [2, 3], 23/26, 19, 0]], 1/39, 1127/2028, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0043","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30aa8ada54cf51f965589cf123b87ee0152b94a894115967839bb7294edf3fbe","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104043,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,27],[2,27],[24,27],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[17,18,20,116]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/27, 2/27, 24/27, 0/1], aligned values X=[17, 18, 20, 116] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 1/9, 53/3, 2/9], [1, [2, 3], 8/9, 20, 0]], 2/81, 392/729, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 1/9, 53/3, 2/9], [1, [2, 3], 8/9, 20, 0]], 2/81, 392/729, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0044","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26d3e1f756677096e1a86239a8bcb27bab192467907827e14fb7bb566d3f175a","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104044,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,28],[2,28],[25,28],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[18,19,21,117]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/28, 2/28, 25/28, 0/1], aligned values X=[18, 19, 21, 117] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/28, 56/3, 2/9], [1, [2, 3], 25/28, 21, 0]], 1/42, 25/48, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/28, 56/3, 2/9], [1, [2, 3], 25/28, 21, 0]], 1/42, 25/48, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0045","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a282892c0d27ec2f7d06d2a3a7d10b3d41791ba04d6620db2b52c1c4e496ab12","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104045,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,29],[2,29],[26,29],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[19,20,22,118]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/29, 2/29, 26/29, 0/1], aligned values X=[19, 20, 22, 118] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/29, 59/3, 2/9], [1, [2, 3], 26/29, 22, 0]], 2/87, 1274/2523, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/29, 59/3, 2/9], [1, [2, 3], 26/29, 22, 0]], 2/87, 1274/2523, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0046","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bfff60abce19eee30b5518e9758edd80bbffc732374084d1cb138647754c2f5","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104046,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,30],[2,30],[27,30],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[20,21,23,119]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/30, 2/30, 27/30, 0/1], aligned values X=[20, 21, 23, 119] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 1/10, 62/3, 2/9], [1, [2, 3], 9/10, 23, 0]], 1/45, 49/100, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 1/10, 62/3, 2/9], [1, [2, 3], 9/10, 23, 0]], 1/45, 49/100, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0047","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5eec49a7bcf9bc64bb63a38ec422a4b6b69d2f19b23d887cc6b9ab2921dc3ea","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104047,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,31],[2,31],[28,31],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[21,22,24,120]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/31, 2/31, 28/31, 0/1], aligned values X=[21, 22, 24, 120] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/31, 65/3, 2/9], [1, [2, 3], 28/31, 24, 0]], 2/93, 1372/2883, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/31, 65/3, 2/9], [1, [2, 3], 28/31, 24, 0]], 2/93, 1372/2883, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0048","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51041586c139cea72b7783f7bcefcc81269967b2587ad25a098c6d51fd573e63","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104048,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,32],[2,32],[29,32],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[22,23,25,121]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/32, 2/32, 29/32, 0/1], aligned values X=[22, 23, 25, 121] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/32, 68/3, 2/9], [1, [2, 3], 29/32, 25, 0]], 1/48, 1421/3072, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/32, 68/3, 2/9], [1, [2, 3], 29/32, 25, 0]], 1/48, 1421/3072, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0049","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b11628a893bb740ffe7275d5903b2fb2b0a22d245498759b4f4eaea50bb1a91","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104049,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,33],[2,33],[30,33],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[23,24,26,122]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/33, 2/33, 30/33, 0/1], aligned values X=[23, 24, 26, 122] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 1/11, 71/3, 2/9], [1, [2, 3], 10/11, 26, 0]], 2/99, 490/1089, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 1/11, 71/3, 2/9], [1, [2, 3], 10/11, 26, 0]], 2/99, 490/1089, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0050","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16c2f5c2b502ed2309ace236b9490e9ce22482f52ed15fb6aa47431b597eff2b","family_id":"qf-mod41-totalvar-within-between-components","seed":202611104050,"semantic_template":"qf-mod41-totalvar-within-between-components/{\"masses\":[[1,34],[2,34],[31,34],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[24,25,27,123]}"},"fields":{"title":"Average Cell Variances and Vary Cell Means under the Same Law","stem":"Use the same complete normalized finite law masses=[1/34, 2/34, 31/34, 0/1], aligned values X=[24, 25, 27, 123] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, certificate).","answer":{"kind":"expression","value":"([[0, [0, 1], 3/34, 74/3, 2/9], [1, [2, 3], 31/34, 27, 0]], 1/51, 1519/3468, same_law_same_partition_within_and_between_terms)","explanation":"Each row is [index,cell atoms,mass,conditional mean or null,conditional variance or null]. The within term is the mass-weighted cell variance; the between term is the mass-weighted squared deviation of cell means from the global mean. Use identical law and partition for both; null cells have zero outer weight. The exact result, retaining the requested order, is ([[0, [0, 1], 3/34, 74/3, 2/9], [1, [2, 3], 31/34, 27, 0]], 1/51, 1519/3468, same_law_same_partition_within_and_between_terms)."},"skills":["Average Cell Variances and Vary Cell Means under the Same Law"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0051","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c6dce5c9697849bfe2c650a85efd461592362749214b9a9a509d0dbe7b679d0","family_id":"qf-mod41-totalvar-full-identity","seed":202611104051,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,10],[2,10],[7,10],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[0,1,3,99]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/10, 2/10, 7/10, 0/1], aligned values X=[0, 1, 3, 99] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(121/100, 1/15, 343/300, 121/100, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (121/100, 1/15, 343/300, 121/100, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0052","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4201384d97ad5157ab5a760ccb0cde0844d7e9a3893adc0348d56eee191df9b","family_id":"qf-mod41-totalvar-full-identity","seed":202611104052,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,11],[2,11],[8,11],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[1,2,4,100]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/11, 2/11, 8/11, 0/1], aligned values X=[1, 2, 4, 100] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(138/121, 2/33, 392/363, 138/121, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (138/121, 2/33, 392/363, 138/121, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0053","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ceca8a06f724159a6a714e3f7966044d917de29b7df94feb5f92d0165b823591","family_id":"qf-mod41-totalvar-full-identity","seed":202611104053,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,12],[2,12],[9,12],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[2,3,5,101]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/12, 2/12, 9/12, 0/1], aligned values X=[2, 3, 5, 101] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(155/144, 1/18, 49/48, 155/144, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (155/144, 1/18, 49/48, 155/144, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0054","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1c153e9ffb531156f1d9c47153ac780c0b701486dba538e7d87577820b92cc6","family_id":"qf-mod41-totalvar-full-identity","seed":202611104054,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,13],[2,13],[10,13],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[3,4,6,102]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/13, 2/13, 10/13, 0/1], aligned values X=[3, 4, 6, 102] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(172/169, 2/39, 490/507, 172/169, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (172/169, 2/39, 490/507, 172/169, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0055","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92454fdc22e8a40a3334de1ef327ab47015956e37ed9bccd7babaf5b20554ce1","family_id":"qf-mod41-totalvar-full-identity","seed":202611104055,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,14],[2,14],[11,14],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[4,5,7,103]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/14, 2/14, 11/14, 0/1], aligned values X=[4, 5, 7, 103] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(27/28, 1/21, 11/12, 27/28, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (27/28, 1/21, 11/12, 27/28, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0056","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c4cd9d8d01b6d7fcc9d11383b12e8bc6c0cc9b40d610bc87325401728850d8a","family_id":"qf-mod41-totalvar-full-identity","seed":202611104056,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,15],[2,15],[12,15],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[5,6,8,104]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/15, 2/15, 12/15, 0/1], aligned values X=[5, 6, 8, 104] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(206/225, 2/45, 196/225, 206/225, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (206/225, 2/45, 196/225, 206/225, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0057","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa81b00a397c29eabd53bc4f2d49864acc1e8c94f1ff04125d543f53ae9e7ee2","family_id":"qf-mod41-totalvar-full-identity","seed":202611104057,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,16],[2,16],[13,16],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[6,7,9,105]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/16, 2/16, 13/16, 0/1], aligned values X=[6, 7, 9, 105] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(223/256, 1/24, 637/768, 223/256, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (223/256, 1/24, 637/768, 223/256, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0058","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05d441310d91146e794c4ad3505c4f8c1361ad0d0839343d92b65958721b006b","family_id":"qf-mod41-totalvar-full-identity","seed":202611104058,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,17],[2,17],[14,17],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[7,8,10,106]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/17, 2/17, 14/17, 0/1], aligned values X=[7, 8, 10, 106] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(240/289, 2/51, 686/867, 240/289, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (240/289, 2/51, 686/867, 240/289, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0059","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"08291a6af2447550e59758df6be99d3773fd7186c3fda5c6f538d9c18bac0f27","family_id":"qf-mod41-totalvar-full-identity","seed":202611104059,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,18],[2,18],[15,18],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[8,9,11,107]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/18, 2/18, 15/18, 0/1], aligned values X=[8, 9, 11, 107] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(257/324, 1/27, 245/324, 257/324, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (257/324, 1/27, 245/324, 257/324, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0060","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"077eacb0c202433cdd6459a0dfec87e0f38e571e0baadda22cd1ff5dc3e7c87c","family_id":"qf-mod41-totalvar-full-identity","seed":202611104060,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,19],[2,19],[16,19],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[9,10,12,108]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/19, 2/19, 16/19, 0/1], aligned values X=[9, 10, 12, 108] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(274/361, 2/57, 784/1083, 274/361, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (274/361, 2/57, 784/1083, 274/361, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0061","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7accca45e4c48b69b7c51bfbab1ba09b6b48afbe61bdbdf90f456dcca309c624","family_id":"qf-mod41-totalvar-full-identity","seed":202611104061,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,20],[2,20],[17,20],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[10,11,13,109]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/20, 2/20, 17/20, 0/1], aligned values X=[10, 11, 13, 109] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(291/400, 1/30, 833/1200, 291/400, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (291/400, 1/30, 833/1200, 291/400, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0062","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ae7ad0641c4d8666fa4d9bf1bf0c545512ccd50600601e7a75256826ee97934","family_id":"qf-mod41-totalvar-full-identity","seed":202611104062,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,21],[2,21],[18,21],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[11,12,14,110]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/21, 2/21, 18/21, 0/1], aligned values X=[11, 12, 14, 110] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(44/63, 2/63, 2/3, 44/63, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (44/63, 2/63, 2/3, 44/63, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0063","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16d4150f252ef9b11de781de12d3d7d7e2cb9e0890188cb3aa364b14ec4da5cf","family_id":"qf-mod41-totalvar-full-identity","seed":202611104063,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,22],[2,22],[19,22],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[12,13,15,111]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/22, 2/22, 19/22, 0/1], aligned values X=[12, 13, 15, 111] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(325/484, 1/33, 931/1452, 325/484, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (325/484, 1/33, 931/1452, 325/484, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0064","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"934195648170cd4e45e711b3bb43a0590eef1ea6cbe0efb96871dbeea68f0a6b","family_id":"qf-mod41-totalvar-full-identity","seed":202611104064,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,23],[2,23],[20,23],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[13,14,16,112]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/23, 2/23, 20/23, 0/1], aligned values X=[13, 14, 16, 112] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(342/529, 2/69, 980/1587, 342/529, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (342/529, 2/69, 980/1587, 342/529, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0065","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6747d5de78059c5b1184be9332f5d2f1a5ba34b02cd4cbd99a0b870a0c0a6a93","family_id":"qf-mod41-totalvar-full-identity","seed":202611104065,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,24],[2,24],[21,24],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[14,15,17,113]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/24, 2/24, 21/24, 0/1], aligned values X=[14, 15, 17, 113] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(359/576, 1/36, 343/576, 359/576, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (359/576, 1/36, 343/576, 359/576, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0066","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df7cd464281a99f4d84b3b8de3fb21b77c35be298d45c74a30f15cd2a0dd1aeb","family_id":"qf-mod41-totalvar-full-identity","seed":202611104066,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,25],[2,25],[22,25],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[15,16,18,114]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/25, 2/25, 22/25, 0/1], aligned values X=[15, 16, 18, 114] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(376/625, 2/75, 1078/1875, 376/625, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (376/625, 2/75, 1078/1875, 376/625, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0067","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb09ddd137d7443b03c551da88a7083005915de535fd2591837741d799489689","family_id":"qf-mod41-totalvar-full-identity","seed":202611104067,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,26],[2,26],[23,26],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[16,17,19,115]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/26, 2/26, 23/26, 0/1], aligned values X=[16, 17, 19, 115] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(393/676, 1/39, 1127/2028, 393/676, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (393/676, 1/39, 1127/2028, 393/676, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0068","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c49ec7fd184413e63d10a3ecde63ac8f85a123379b4a85d81e43ca89d546079","family_id":"qf-mod41-totalvar-full-identity","seed":202611104068,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,27],[2,27],[24,27],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[17,18,20,116]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/27, 2/27, 24/27, 0/1], aligned values X=[17, 18, 20, 116] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(410/729, 2/81, 392/729, 410/729, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (410/729, 2/81, 392/729, 410/729, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0069","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1a92a3c9957571fa66a10d1c4596b4c01cbfcee2e45b08bbdbe940a51c9de22","family_id":"qf-mod41-totalvar-full-identity","seed":202611104069,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,28],[2,28],[25,28],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[18,19,21,117]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/28, 2/28, 25/28, 0/1], aligned values X=[18, 19, 21, 117] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(61/112, 1/42, 25/48, 61/112, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (61/112, 1/42, 25/48, 61/112, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0070","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8d383d0f6d994f55c6c1aac545bf64397843d62b6839a68d025b16c0ab0d4d6","family_id":"qf-mod41-totalvar-full-identity","seed":202611104070,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,29],[2,29],[26,29],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[19,20,22,118]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/29, 2/29, 26/29, 0/1], aligned values X=[19, 20, 22, 118] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(444/841, 2/87, 1274/2523, 444/841, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (444/841, 2/87, 1274/2523, 444/841, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0071","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21c923212b1737547da8be793efe71a8fa71a394fdeca69f4f60bcddcaa72232","family_id":"qf-mod41-totalvar-full-identity","seed":202611104071,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,30],[2,30],[27,30],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[20,21,23,119]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/30, 2/30, 27/30, 0/1], aligned values X=[20, 21, 23, 119] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(461/900, 1/45, 49/100, 461/900, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (461/900, 1/45, 49/100, 461/900, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0072","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b63c16ff192d7f22b37153b433396286ddabd4d3548a3cbabe2350bd4d398237","family_id":"qf-mod41-totalvar-full-identity","seed":202611104072,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,31],[2,31],[28,31],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[21,22,24,120]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/31, 2/31, 28/31, 0/1], aligned values X=[21, 22, 24, 120] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(478/961, 2/93, 1372/2883, 478/961, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (478/961, 2/93, 1372/2883, 478/961, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0073","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee82c113bfe6ccb539722b116e0267674738c3d719d2ee903ef9b9ce625dfa30","family_id":"qf-mod41-totalvar-full-identity","seed":202611104073,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,32],[2,32],[29,32],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[22,23,25,121]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/32, 2/32, 29/32, 0/1], aligned values X=[22, 23, 25, 121] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(495/1024, 1/48, 1421/3072, 495/1024, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (495/1024, 1/48, 1421/3072, 495/1024, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0074","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dbf785380f06302cdd425e74eaef4fdeb51e65b3ccfb7d194c0ef2e5f31f273","family_id":"qf-mod41-totalvar-full-identity","seed":202611104074,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,33],[2,33],[30,33],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[23,24,26,122]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/33, 2/33, 30/33, 0/1], aligned values X=[23, 24, 26, 122] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(512/1089, 2/99, 490/1089, 512/1089, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (512/1089, 2/99, 490/1089, 512/1089, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0075","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"054872100a424579988e33a97aac04e53cd98bc8eb426fda19d738d365f96a4c","family_id":"qf-mod41-totalvar-full-identity","seed":202611104075,"semantic_template":"qf-mod41-totalvar-full-identity/{\"masses\":[[1,34],[2,34],[31,34],[0,1]],\"partition\":[[0,1],[2,3]],\"values\":[24,25,27,123]}"},"fields":{"title":"Both Within and Between Terms Are Needed for Total Variance","stem":"Use the same complete normalized finite law masses=[1/34, 2/34, 31/34, 0/1], aligned values X=[24, 25, 27, 123] and complete disjoint partition=[[0, 1], [2, 3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (variance, within, between, sum, equal).","answer":{"kind":"expression","value":"(529/1156, 1/51, 1519/3468, 529/1156, Yes)","explanation":"Compare Var(X), E[Var(X|G)], Var(E[X|G]) and their sum with exact rational arithmetic. Do not omit a term or interchange their meanings. No independence of partition events is assumed. The exact result, retaining the requested order, is (529/1156, 1/51, 1519/3468, 529/1156, Yes)."},"skills":["Both Within and Between Terms Are Needed for Total Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0076","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6aa683f074f4f2032d40c998fd66077414ad2a7ac0f78646ef33e3f316832130","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104076,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,10],[2,10],[7,10],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[0,1,3,99]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/10, 2/10, 7/10, 0/1], aligned values X=[0, 1, 3, 99] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 23/10, 121/100], [1, [3], 0, null, null]], 121/100, 0, 121/100, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 23/10, 121/100], [1, [3], 0, null, null]], 121/100, 0, 121/100, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0077","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d0c9e7f7506f2ec3944230d98dca9a24c1eaf0327355cd864b933a8ce19584c","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104077,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,11],[2,11],[8,11],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[1,2,4,100]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/11, 2/11, 8/11, 0/1], aligned values X=[1, 2, 4, 100] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 37/11, 138/121], [1, [3], 0, null, null]], 138/121, 0, 138/121, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 37/11, 138/121], [1, [3], 0, null, null]], 138/121, 0, 138/121, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0078","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b11a5edc45b24ea206e5738aad8aa78cd6ca4c0544f371b48f742f3108a6203f","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104078,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,12],[2,12],[9,12],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[2,3,5,101]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/12, 2/12, 9/12, 0/1], aligned values X=[2, 3, 5, 101] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 53/12, 155/144], [1, [3], 0, null, null]], 155/144, 0, 155/144, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 53/12, 155/144], [1, [3], 0, null, null]], 155/144, 0, 155/144, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0079","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"84d236ca108f8a7a1924fb64f2955e6f8526461cc8472ee057b427fce7d9e8df","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104079,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,13],[2,13],[10,13],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[3,4,6,102]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/13, 2/13, 10/13, 0/1], aligned values X=[3, 4, 6, 102] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 71/13, 172/169], [1, [3], 0, null, null]], 172/169, 0, 172/169, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 71/13, 172/169], [1, [3], 0, null, null]], 172/169, 0, 172/169, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0080","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"24486247605bfae55456a17849435ca05253005e1c5390f89509b7edb90abb07","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104080,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,14],[2,14],[11,14],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[4,5,7,103]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/14, 2/14, 11/14, 0/1], aligned values X=[4, 5, 7, 103] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 13/2, 27/28], [1, [3], 0, null, null]], 27/28, 0, 27/28, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 13/2, 27/28], [1, [3], 0, null, null]], 27/28, 0, 27/28, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0081","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"111ea542199dc521f9e2df2833099afb469be1b94533a47c9beaa0f5691e083f","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104081,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,15],[2,15],[12,15],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[5,6,8,104]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/15, 2/15, 12/15, 0/1], aligned values X=[5, 6, 8, 104] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 113/15, 206/225], [1, [3], 0, null, null]], 206/225, 0, 206/225, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 113/15, 206/225], [1, [3], 0, null, null]], 206/225, 0, 206/225, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0082","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7699ceeb8449cbde39f6d7f6c65716535b7fd8954c9d25b0a59f9b98a6899e7","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104082,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,16],[2,16],[13,16],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[6,7,9,105]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/16, 2/16, 13/16, 0/1], aligned values X=[6, 7, 9, 105] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 137/16, 223/256], [1, [3], 0, null, null]], 223/256, 0, 223/256, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 137/16, 223/256], [1, [3], 0, null, null]], 223/256, 0, 223/256, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0083","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ce76a42908c7b0f5c6f50affaeace7d3654fe3f5e2f5a8935ba25009067b8ca","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104083,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,17],[2,17],[14,17],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[7,8,10,106]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/17, 2/17, 14/17, 0/1], aligned values X=[7, 8, 10, 106] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 163/17, 240/289], [1, [3], 0, null, null]], 240/289, 0, 240/289, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 163/17, 240/289], [1, [3], 0, null, null]], 240/289, 0, 240/289, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0084","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6928aa34aae457ed524b35b72bee9b0ba1406693f4c381a7134a53c50fd7bc1f","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104084,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,18],[2,18],[15,18],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[8,9,11,107]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/18, 2/18, 15/18, 0/1], aligned values X=[8, 9, 11, 107] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 191/18, 257/324], [1, [3], 0, null, null]], 257/324, 0, 257/324, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 191/18, 257/324], [1, [3], 0, null, null]], 257/324, 0, 257/324, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0085","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"323bc851e4027dcb28386cfbe014fe723006499c2f76382241a33d161d19db74","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104085,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,19],[2,19],[16,19],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[9,10,12,108]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/19, 2/19, 16/19, 0/1], aligned values X=[9, 10, 12, 108] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 221/19, 274/361], [1, [3], 0, null, null]], 274/361, 0, 274/361, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 221/19, 274/361], [1, [3], 0, null, null]], 274/361, 0, 274/361, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0086","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef42b57de3edbfa505bbd13f676aa7e4eb4217381489b19e14a8ab13084c5b90","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104086,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,20],[2,20],[17,20],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[10,11,13,109]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/20, 2/20, 17/20, 0/1], aligned values X=[10, 11, 13, 109] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 253/20, 291/400], [1, [3], 0, null, null]], 291/400, 0, 291/400, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 253/20, 291/400], [1, [3], 0, null, null]], 291/400, 0, 291/400, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0087","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff30917cb929928d7f771180cf7131345375bfce49a6ebd95b7c5e2cfc9c2012","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104087,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,21],[2,21],[18,21],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[11,12,14,110]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/21, 2/21, 18/21, 0/1], aligned values X=[11, 12, 14, 110] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 41/3, 44/63], [1, [3], 0, null, null]], 44/63, 0, 44/63, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 41/3, 44/63], [1, [3], 0, null, null]], 44/63, 0, 44/63, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0088","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6779ac7534825ec933c0df14c33a576aa4f25f4b557730a4454b8047f4061c8a","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104088,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,22],[2,22],[19,22],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[12,13,15,111]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/22, 2/22, 19/22, 0/1], aligned values X=[12, 13, 15, 111] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 323/22, 325/484], [1, [3], 0, null, null]], 325/484, 0, 325/484, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 323/22, 325/484], [1, [3], 0, null, null]], 325/484, 0, 325/484, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0089","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f6fccd444b7c7a70b1d74c16ce45ef5e277a880a871dab9eb2634f238df2aa0","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104089,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,23],[2,23],[20,23],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[13,14,16,112]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/23, 2/23, 20/23, 0/1], aligned values X=[13, 14, 16, 112] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 361/23, 342/529], [1, [3], 0, null, null]], 342/529, 0, 342/529, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 361/23, 342/529], [1, [3], 0, null, null]], 342/529, 0, 342/529, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0090","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"917720299823c189782315fbd43fd8880d73ed8a011d490196e5a40928c031ba","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104090,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,24],[2,24],[21,24],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[14,15,17,113]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/24, 2/24, 21/24, 0/1], aligned values X=[14, 15, 17, 113] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 401/24, 359/576], [1, [3], 0, null, null]], 359/576, 0, 359/576, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 401/24, 359/576], [1, [3], 0, null, null]], 359/576, 0, 359/576, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0091","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"966942061900faaa24287a26f15c76c2cc392b0555c085b3edea4df79d82a89d","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104091,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,25],[2,25],[22,25],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[15,16,18,114]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/25, 2/25, 22/25, 0/1], aligned values X=[15, 16, 18, 114] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 443/25, 376/625], [1, [3], 0, null, null]], 376/625, 0, 376/625, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 443/25, 376/625], [1, [3], 0, null, null]], 376/625, 0, 376/625, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0092","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4bf87b811b279c1aaf2b07ea1440cb59424a3d964869f6170ac4cfd0553fea8c","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104092,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,26],[2,26],[23,26],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[16,17,19,115]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/26, 2/26, 23/26, 0/1], aligned values X=[16, 17, 19, 115] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 487/26, 393/676], [1, [3], 0, null, null]], 393/676, 0, 393/676, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 487/26, 393/676], [1, [3], 0, null, null]], 393/676, 0, 393/676, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0093","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e881ea2c721ae640b58870625a2e54305cff10c82b99fbd08703bb071cb083b","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104093,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,27],[2,27],[24,27],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[17,18,20,116]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/27, 2/27, 24/27, 0/1], aligned values X=[17, 18, 20, 116] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 533/27, 410/729], [1, [3], 0, null, null]], 410/729, 0, 410/729, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 533/27, 410/729], [1, [3], 0, null, null]], 410/729, 0, 410/729, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0094","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87bb0bda636118c60ca12c1067a67a2e58fd6b876a12951cba188c5bcf624867","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104094,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,28],[2,28],[25,28],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[18,19,21,117]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/28, 2/28, 25/28, 0/1], aligned values X=[18, 19, 21, 117] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 83/4, 61/112], [1, [3], 0, null, null]], 61/112, 0, 61/112, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 83/4, 61/112], [1, [3], 0, null, null]], 61/112, 0, 61/112, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0095","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a46d2b2fb47856e2dafdbe9f1a59e3f5d5dad641910f50ea8994236af23edcf","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104095,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,29],[2,29],[26,29],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[19,20,22,118]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/29, 2/29, 26/29, 0/1], aligned values X=[19, 20, 22, 118] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 631/29, 444/841], [1, [3], 0, null, null]], 444/841, 0, 444/841, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 631/29, 444/841], [1, [3], 0, null, null]], 444/841, 0, 444/841, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0096","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a31af9e690d3d4935f65ef3717c933cb16b1a14917c56560728a84d712dd1b8","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104096,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,30],[2,30],[27,30],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[20,21,23,119]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/30, 2/30, 27/30, 0/1], aligned values X=[20, 21, 23, 119] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 683/30, 461/900], [1, [3], 0, null, null]], 461/900, 0, 461/900, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 683/30, 461/900], [1, [3], 0, null, null]], 461/900, 0, 461/900, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0097","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a3e097865cba89943db590fd49461525fb12ec9d32471c1a0e8450f411937d3","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104097,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,31],[2,31],[28,31],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[21,22,24,120]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/31, 2/31, 28/31, 0/1], aligned values X=[21, 22, 24, 120] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 737/31, 478/961], [1, [3], 0, null, null]], 478/961, 0, 478/961, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 737/31, 478/961], [1, [3], 0, null, null]], 478/961, 0, 478/961, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0098","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8064b7331fcc2e685858f5608a342d03ce59e0c8d37e667ddd360a9e8009b281","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104098,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,32],[2,32],[29,32],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[22,23,25,121]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/32, 2/32, 29/32, 0/1], aligned values X=[22, 23, 25, 121] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 793/32, 495/1024], [1, [3], 0, null, null]], 495/1024, 0, 495/1024, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 793/32, 495/1024], [1, [3], 0, null, null]], 495/1024, 0, 495/1024, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0099","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6af5b5b08a0faf224278cdcbea7e9b5d6af78bb511d523766fbb80a1ff98be2","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104099,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,33],[2,33],[30,33],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[23,24,26,122]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/33, 2/33, 30/33, 0/1], aligned values X=[23, 24, 26, 122] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 851/33, 512/1089], [1, [3], 0, null, null]], 512/1089, 0, 512/1089, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 851/33, 512/1089], [1, [3], 0, null, null]], 512/1089, 0, 512/1089, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261110-r04-law-of-total-variance-0100","batch_id":"batch-20261014-advance-20261110-r04-law-of-total-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ac914a68ebc9a2097412b0e8780c62f2dbd15bbe45fcfb850867280126cb7b03","family_id":"qf-mod41-totalvar-zero-mass-boundary","seed":202611104100,"semantic_template":"qf-mod41-totalvar-zero-mass-boundary/{\"masses\":[[1,34],[2,34],[31,34],[0,1]],\"partition\":[[0,1,2],[3]],\"values\":[24,25,27,123]}"},"fields":{"title":"Undefined Null-Cell Mean and Variance Have Zero Outer Contribution","stem":"Use the same complete normalized finite law masses=[1/34, 2/34, 31/34, 0/1], aligned values X=[24, 25, 27, 123] and complete disjoint partition=[[0, 1, 2], [3]]. Every cell is nonempty even if all its atoms have zero mass; cells cover the whole carrier. G is generated by this partition. Conditional division requires positive cell mass. No independence premise is required. Give (cell_rows, within, between, variance, equal, null_cell_division_claim, boundary).","answer":{"kind":"expression","value":"([[0, [0, 1, 2], 1, 911/34, 529/1156], [1, [3], 0, null, null]], 529/1156, 0, 529/1156, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)","explanation":"Require a zero-mass nonempty cell. Keep null for both conditional quotients, never divide by zero, and retain both complete within and between terms. The fixed No withholds null-cell division, not the total-variance identity. The exact result, retaining the requested order, is ([[0, [0, 1, 2], 1, 911/34, 529/1156], [1, [3], 0, null, null]], 529/1156, 0, 529/1156, Yes, No, zero_mass_cell_not_divided_and_outer_weighted_contribution_zero)."},"skills":["Undefined Null-Cell Mean and Variance Have Zero Outer Contribution"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}