{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261108-r04-chebyshev-inequality-0001","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78aec56a3da2176bbaf24cdf06af9627e9778b696cedb4070d83c31b41974d7f","family_id":"qf-mod39-cheb-qualification","seed":202611084001,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,6],[2,6],[3,6]],\"values\":[0,1,4]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/6, 2/6, 3/6] and aligned values X=[0, 1, 4]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/6, 1/3, 1/2], [0, 1, 4], 7/3, 26/9, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/6, 1/3, 1/2], [0, 1, 4], 7/3, 26/9, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0002","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a3cbc1ba09365927e9757e70e97b128b2e205733bada14fad73ce01afbf392ba","family_id":"qf-mod39-cheb-qualification","seed":202611084002,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,7],[2,7],[4,7]],\"values\":[0,2,5]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/7, 2/7, 4/7] and aligned values X=[0, 2, 5]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/7, 2/7, 4/7], [0, 2, 5], 24/7, 180/49, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/7, 2/7, 4/7], [0, 2, 5], 24/7, 180/49, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0003","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"79ea4fe1eed2c30952f91ee01d7bf1aeb9e7025366df526350b361786778b7fe","family_id":"qf-mod39-cheb-qualification","seed":202611084003,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,8],[2,8],[5,8]],\"values\":[0,3,6]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/8, 2/8, 5/8] and aligned values X=[0, 3, 6]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/8, 1/4, 5/8], [0, 3, 6], 9/2, 9/2, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/8, 1/4, 5/8], [0, 3, 6], 9/2, 9/2, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0004","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a411189b6bfc8988a64f51f014fe028022d8a22d2a0975f77715d2604f0393b","family_id":"qf-mod39-cheb-qualification","seed":202611084004,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,9],[2,9],[6,9]],\"values\":[0,4,7]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/9, 2/9, 6/9] and aligned values X=[0, 4, 7]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/9, 2/9, 2/3], [0, 4, 7], 50/9, 434/81, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/9, 2/9, 2/3], [0, 4, 7], 50/9, 434/81, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0005","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9eec74cc9ab5d2d56c1725f6a35af84bca447723eb3eb39d3858d4d60a0ef473","family_id":"qf-mod39-cheb-qualification","seed":202611084005,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,10],[2,10],[7,10]],\"values\":[0,5,8]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/10, 2/10, 7/10] and aligned values X=[0, 5, 8]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/10, 1/5, 7/10], [0, 5, 8], 33/5, 156/25, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/10, 1/5, 7/10], [0, 5, 8], 33/5, 156/25, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0006","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17b5853186093d82d81676edf5d89280c395702fb554035a95b2d006899e1c47","family_id":"qf-mod39-cheb-qualification","seed":202611084006,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,11],[2,11],[8,11]],\"values\":[0,6,9]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/11, 2/11, 8/11] and aligned values X=[0, 6, 9]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/11, 2/11, 8/11], [0, 6, 9], 84/11, 864/121, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/11, 2/11, 8/11], [0, 6, 9], 84/11, 864/121, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0007","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26521956e5cebfdd2a3ec8ccdadc6977617379ee02620c3ae257383e268d50bf","family_id":"qf-mod39-cheb-qualification","seed":202611084007,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,12],[2,12],[9,12]],\"values\":[0,7,10]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/12, 2/12, 9/12] and aligned values X=[0, 7, 10]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/12, 1/6, 3/4], [0, 7, 10], 26/3, 145/18, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/12, 1/6, 3/4], [0, 7, 10], 26/3, 145/18, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0008","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"615dcb1449c7c4a29ec91c0e8036b2df43ac097c51bbbcf2b176bb2b44a27fa0","family_id":"qf-mod39-cheb-qualification","seed":202611084008,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,13],[2,13],[10,13]],\"values\":[0,8,11]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/13, 2/13, 10/13] and aligned values X=[0, 8, 11]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/13, 2/13, 10/13], [0, 8, 11], 126/13, 1518/169, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/13, 2/13, 10/13], [0, 8, 11], 126/13, 1518/169, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0009","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"beee60fe63e36504b50df91b3a262e386b6238f852cc057629dc2c8d00d4f79e","family_id":"qf-mod39-cheb-qualification","seed":202611084009,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,14],[2,14],[11,14]],\"values\":[0,9,12]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/14, 2/14, 11/14] and aligned values X=[0, 9, 12]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/14, 1/7, 11/14], [0, 9, 12], 75/7, 486/49, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/14, 1/7, 11/14], [0, 9, 12], 75/7, 486/49, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0010","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5cb5659aea8e69523b5f2e157326d278117da30d93ddbafefffa01419782889f","family_id":"qf-mod39-cheb-qualification","seed":202611084010,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,15],[2,15],[12,15]],\"values\":[0,10,13]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/15, 2/15, 12/15] and aligned values X=[0, 10, 13]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/15, 2/15, 4/5], [0, 10, 13], 176/15, 2444/225, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/15, 2/15, 4/5], [0, 10, 13], 176/15, 2444/225, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0011","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"421c761ce3e593f4e3198fdd068af3ae3cc21a9734bd1baf3e1097f4bf42a809","family_id":"qf-mod39-cheb-qualification","seed":202611084011,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,16],[2,16],[13,16]],\"values\":[0,11,14]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/16, 2/16, 13/16] and aligned values X=[0, 11, 14]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/16, 1/8, 13/16], [0, 11, 14], 51/4, 189/16, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/16, 1/8, 13/16], [0, 11, 14], 51/4, 189/16, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0012","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b003b568935e785641d36b7492a76d50625ff846d39341bdb9befdc2b89d87d","family_id":"qf-mod39-cheb-qualification","seed":202611084012,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,17],[2,17],[14,17]],\"values\":[0,12,15]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/17, 2/17, 14/17] and aligned values X=[0, 12, 15]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/17, 2/17, 14/17], [0, 12, 15], 234/17, 3690/289, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/17, 2/17, 14/17], [0, 12, 15], 234/17, 3690/289, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0013","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a4eb6150b92b721c009ef0832d7fdad71448329bd35ffaba84cf62de461dd7b","family_id":"qf-mod39-cheb-qualification","seed":202611084013,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,18],[2,18],[15,18]],\"values\":[0,13,16]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/18, 2/18, 15/18] and aligned values X=[0, 13, 16]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/18, 1/9, 5/6], [0, 13, 16], 133/9, 1112/81, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/18, 1/9, 5/6], [0, 13, 16], 133/9, 1112/81, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0014","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"804bf2793bdd05bc95208295758902d74fd4130176d3f4a452d6a1a8646fd678","family_id":"qf-mod39-cheb-qualification","seed":202611084014,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,19],[2,19],[16,19]],\"values\":[0,14,17]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/19, 2/19, 16/19] and aligned values X=[0, 14, 17]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/19, 2/19, 16/19], [0, 14, 17], 300/19, 5304/361, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/19, 2/19, 16/19], [0, 14, 17], 300/19, 5304/361, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0015","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7aa91d03c542b50ee9d3e50a18b17468f3db84fa31fd7a0fad0982c729f4d91","family_id":"qf-mod39-cheb-qualification","seed":202611084015,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,20],[2,20],[17,20]],\"values\":[0,15,18]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/20, 2/20, 17/20] and aligned values X=[0, 15, 18]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/20, 1/10, 17/20], [0, 15, 18], 84/5, 783/50, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/20, 1/10, 17/20], [0, 15, 18], 84/5, 783/50, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0016","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"101896838f3aa8e1c89390ea2df66a947398685523d4deb11b18d5841493f461","family_id":"qf-mod39-cheb-qualification","seed":202611084016,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,21],[2,21],[18,21]],\"values\":[0,16,19]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/21, 2/21, 18/21] and aligned values X=[0, 16, 19]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/21, 2/21, 6/7], [0, 16, 19], 374/21, 7334/441, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/21, 2/21, 6/7], [0, 16, 19], 374/21, 7334/441, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0017","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68f946ed54742e7920a93fb89874b45cb282f3287a02c4c4e2c934104a7dacfb","family_id":"qf-mod39-cheb-qualification","seed":202611084017,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,22],[2,22],[19,22]],\"values\":[0,17,20]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/22, 2/22, 19/22] and aligned values X=[0, 17, 20]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/22, 1/11, 19/22], [0, 17, 20], 207/11, 2130/121, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/22, 1/11, 19/22], [0, 17, 20], 207/11, 2130/121, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0018","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b49b931b9f9d1f83c56093ccfcd998361196bae75ec6b5049851521ced5d6ce7","family_id":"qf-mod39-cheb-qualification","seed":202611084018,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,23],[2,23],[20,23]],\"values\":[0,18,21]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/23, 2/23, 20/23] and aligned values X=[0, 18, 21]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/23, 2/23, 20/23], [0, 18, 21], 456/23, 9828/529, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/23, 2/23, 20/23], [0, 18, 21], 456/23, 9828/529, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0019","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4406e2ffa2cdeabd72bc7b9596a57ed6cc9250bab237684f3d5b9f8ecb39b3b3","family_id":"qf-mod39-cheb-qualification","seed":202611084019,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,24],[2,24],[21,24]],\"values\":[0,19,22]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/24, 2/24, 21/24] and aligned values X=[0, 19, 22]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/24, 1/12, 7/8], [0, 19, 22], 125/6, 176/9, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/24, 1/12, 7/8], [0, 19, 22], 125/6, 176/9, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0020","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c0279fc7e4343cba4932afeaafba0de090f10e69e7ff8742db62707c3e573c45","family_id":"qf-mod39-cheb-qualification","seed":202611084020,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,25],[2,25],[22,25]],\"values\":[0,20,23]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/25, 2/25, 22/25] and aligned values X=[0, 20, 23]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/25, 2/25, 22/25], [0, 20, 23], 546/25, 12834/625, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/25, 2/25, 22/25], [0, 20, 23], 546/25, 12834/625, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0021","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"01c4b399aad978ef79e16da31969fc30ec68bdef3e5b46c3fbde9d2484f077bb","family_id":"qf-mod39-cheb-qualification","seed":202611084021,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,26],[2,26],[23,26]],\"values\":[0,21,24]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/26, 2/26, 23/26] and aligned values X=[0, 21, 24]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/26, 1/13, 23/26], [0, 21, 24], 297/13, 3636/169, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/26, 1/13, 23/26], [0, 21, 24], 297/13, 3636/169, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0022","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d6b05dd0ddaaceba7b1de1acc098432bec1945bb0bde40b09a5b1408302ae77","family_id":"qf-mod39-cheb-qualification","seed":202611084022,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,27],[2,27],[24,27]],\"values\":[0,22,25]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/27, 2/27, 24/27] and aligned values X=[0, 22, 25]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/27, 2/27, 8/9], [0, 22, 25], 644/27, 16400/729, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/27, 2/27, 8/9], [0, 22, 25], 644/27, 16400/729, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0023","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"108c084bc878bed910adcd01d6c8561456684dc73b57a9012068d1b7d65006a2","family_id":"qf-mod39-cheb-qualification","seed":202611084023,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,28],[2,28],[25,28]],\"values\":[0,23,26]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/28, 2/28, 25/28] and aligned values X=[0, 23, 26]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/28, 1/14, 25/28], [0, 23, 26], 174/7, 2301/98, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/28, 1/14, 25/28], [0, 23, 26], 174/7, 2301/98, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0024","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05c1139448e916ad41075b18fb77c70c699e3eae5a543cc3fabfe817a5fa7e7c","family_id":"qf-mod39-cheb-qualification","seed":202611084024,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,29],[2,29],[26,29]],\"values\":[0,24,27]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/29, 2/29, 26/29] and aligned values X=[0, 24, 27]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/29, 2/29, 26/29], [0, 24, 27], 750/29, 20574/841, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/29, 2/29, 26/29], [0, 24, 27], 750/29, 20574/841, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0025","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"509a2237cc77a6875e9f01b0d180ad951eaace6ba8dad1a958ed40b496e0ba53","family_id":"qf-mod39-cheb-qualification","seed":202611084025,"semantic_template":"qf-mod39-cheb-qualification/{\"masses\":[[1,30],[2,30],[27,30]],\"values\":[0,25,28]}"},"fields":{"title":"Finite Mean and Variance Do Not Need a Nonnegative Original Variable","stem":"Use complete normalized finite law masses=[1/30, 2/30, 27/30] and aligned values X=[0, 25, 28]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Give (masses, values, mean, variance, finite_moment_certificate, certificate).","answer":{"kind":"expression","value":"([1/30, 1/15, 9/10], [0, 25, 28], 403/15, 5726/225, Yes, finite_mean_and_variance)","explanation":"Compute exact normalized masses, values, mean and variance. Signed X is allowed; the nonnegative variable used in the derivation is the squared centered deviation. These finite moments are not assumptions of independence. The exact result, retaining the requested order, is ([1/30, 1/15, 9/10], [0, 25, 28], 403/15, 5726/225, Yes, finite_mean_and_variance)."},"skills":["Finite Mean and Variance Do Not Need a Nonnegative Original Variable"],"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-20261108-r04-chebyshev-inequality-0026","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"100f473f53137d8c616c730eb81dab44aefdfdaba299d3d4a26a8e0d38bbf824","family_id":"qf-mod39-cheb-tail-probability","seed":202611084026,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,6],[2,6],[3,6]],\"values\":[0,1,4]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/6, 2/6, 3/6] and aligned values X=[0, 1, 4]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(7/3, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (7/3, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0027","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c060bd95246d043c338d4ae618005878a22509a243aff098b5a596e2fe9013be","family_id":"qf-mod39-cheb-tail-probability","seed":202611084027,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[2,1],\"masses\":[[1,7],[2,7],[4,7]],\"values\":[0,2,5]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/7, 2/7, 4/7] and aligned values X=[0, 2, 5]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(24/7, 2, 1/7, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (24/7, 2, 1/7, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0028","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e08c0818911b0e3ee89384eb61872783c948ee87ba8daa84011b39150ee672c1","family_id":"qf-mod39-cheb-tail-probability","seed":202611084028,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[3,1],\"masses\":[[1,8],[2,8],[5,8]],\"values\":[0,3,6]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/8, 2/8, 5/8] and aligned values X=[0, 3, 6]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(9/2, 3, 1/8, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (9/2, 3, 1/8, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0029","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a958233d8fcb81e67da31a413c40241c8350efe1c4e4e38c1877e15b2da8c96","family_id":"qf-mod39-cheb-tail-probability","seed":202611084029,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,9],[2,9],[6,9]],\"values\":[0,4,7]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/9, 2/9, 6/9] and aligned values X=[0, 4, 7]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(50/9, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (50/9, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0030","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b2c24f965fa17b2b384d32404fa2677852a46f53ec41d62658473c1915804c5","family_id":"qf-mod39-cheb-tail-probability","seed":202611084030,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[2,1],\"masses\":[[1,10],[2,10],[7,10]],\"values\":[0,5,8]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/10, 2/10, 7/10] and aligned values X=[0, 5, 8]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(33/5, 2, 1/10, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (33/5, 2, 1/10, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0031","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e087ad252f740934fd9e97bcea0bdfdd6927f5cc159c53af2a7a94173dfd53e5","family_id":"qf-mod39-cheb-tail-probability","seed":202611084031,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[3,1],\"masses\":[[1,11],[2,11],[8,11]],\"values\":[0,6,9]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/11, 2/11, 8/11] and aligned values X=[0, 6, 9]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(84/11, 3, 1/11, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (84/11, 3, 1/11, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0032","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f87a534daba94e067c7771b6d4df298b3adb24d05d58bf3e2302fb5bc013d030","family_id":"qf-mod39-cheb-tail-probability","seed":202611084032,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,12],[2,12],[9,12]],\"values\":[0,7,10]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/12, 2/12, 9/12] and aligned values X=[0, 7, 10]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(26/3, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (26/3, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0033","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64bfb1d657410e6b3eb8fe43677308471cf549c249a47b2bb66f88a6f8a254f3","family_id":"qf-mod39-cheb-tail-probability","seed":202611084033,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[2,1],\"masses\":[[1,13],[2,13],[10,13]],\"values\":[0,8,11]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/13, 2/13, 10/13] and aligned values X=[0, 8, 11]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(126/13, 2, 1/13, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (126/13, 2, 1/13, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0034","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fdfefdad53894b287b3a2d9133b399810dca92764ca9fbe80b9dd0c796428c20","family_id":"qf-mod39-cheb-tail-probability","seed":202611084034,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[3,1],\"masses\":[[1,14],[2,14],[11,14]],\"values\":[0,9,12]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/14, 2/14, 11/14] and aligned values X=[0, 9, 12]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(75/7, 3, 1/14, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (75/7, 3, 1/14, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0035","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b90782aa2e087231af9f92a7afda7e1d8184a952b4b45693160d7664e83570c","family_id":"qf-mod39-cheb-tail-probability","seed":202611084035,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,15],[2,15],[12,15]],\"values\":[0,10,13]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/15, 2/15, 12/15] and aligned values X=[0, 10, 13]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(176/15, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (176/15, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0036","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"819ff30ecfdabca3c0fa6b25cf751d364bc63530f3fdfa589d5e50d1503793b8","family_id":"qf-mod39-cheb-tail-probability","seed":202611084036,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[2,1],\"masses\":[[1,16],[2,16],[13,16]],\"values\":[0,11,14]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/16, 2/16, 13/16] and aligned values X=[0, 11, 14]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(51/4, 2, 1/16, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (51/4, 2, 1/16, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0037","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52ca3624bac8612a4657d6444aba2fd81ae2e0a6e74ea990ab37357c934deae6","family_id":"qf-mod39-cheb-tail-probability","seed":202611084037,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[3,1],\"masses\":[[1,17],[2,17],[14,17]],\"values\":[0,12,15]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/17, 2/17, 14/17] and aligned values X=[0, 12, 15]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(234/17, 3, 1/17, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (234/17, 3, 1/17, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0038","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52a3fdc2a39f69d0f67479c96cbfefad1efed04e4f3ab8c81375108a5a38163b","family_id":"qf-mod39-cheb-tail-probability","seed":202611084038,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,18],[2,18],[15,18]],\"values\":[0,13,16]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/18, 2/18, 15/18] and aligned values X=[0, 13, 16]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(133/9, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (133/9, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0039","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5ceb3d3da4b70574c59066eee8ca49b59cd42176c8f30539930c75190f62fd9","family_id":"qf-mod39-cheb-tail-probability","seed":202611084039,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[2,1],\"masses\":[[1,19],[2,19],[16,19]],\"values\":[0,14,17]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/19, 2/19, 16/19] and aligned values X=[0, 14, 17]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(300/19, 2, 1/19, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (300/19, 2, 1/19, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0040","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4857e91177667834cad3ba0498b696d57cddbf2f982ee0910ea423c57d8ea82e","family_id":"qf-mod39-cheb-tail-probability","seed":202611084040,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[3,1],\"masses\":[[1,20],[2,20],[17,20]],\"values\":[0,15,18]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/20, 2/20, 17/20] and aligned values X=[0, 15, 18]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(84/5, 3, 1/20, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (84/5, 3, 1/20, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0041","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2fc10d6390d26f91dd0f8df9ab3a78b00837ba483373bdb3693a560085609c3f","family_id":"qf-mod39-cheb-tail-probability","seed":202611084041,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,21],[2,21],[18,21]],\"values\":[0,16,19]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/21, 2/21, 18/21] and aligned values X=[0, 16, 19]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(374/21, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (374/21, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0042","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfdc2747fdbe9ccaac8b6c313af7a5e4c7b046933b8e39a9236d03eed77645a5","family_id":"qf-mod39-cheb-tail-probability","seed":202611084042,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[2,1],\"masses\":[[1,22],[2,22],[19,22]],\"values\":[0,17,20]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/22, 2/22, 19/22] and aligned values X=[0, 17, 20]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(207/11, 2, 1/22, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (207/11, 2, 1/22, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0043","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7daf83fd80ea54eaa8e14d0dc6db29735b11a37752ca7e57a1b6df87e21102b9","family_id":"qf-mod39-cheb-tail-probability","seed":202611084043,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[3,1],\"masses\":[[1,23],[2,23],[20,23]],\"values\":[0,18,21]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/23, 2/23, 20/23] and aligned values X=[0, 18, 21]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(456/23, 3, 1/23, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (456/23, 3, 1/23, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0044","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"865f883fb8198d26a57a1ace637a8f752be77dd0b21118617694d435f5b9b92a","family_id":"qf-mod39-cheb-tail-probability","seed":202611084044,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,24],[2,24],[21,24]],\"values\":[0,19,22]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/24, 2/24, 21/24] and aligned values X=[0, 19, 22]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(125/6, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (125/6, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0045","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"67620ff1689bad62c94092012c70aba1f8fe43d48d782066523cb01eb3215754","family_id":"qf-mod39-cheb-tail-probability","seed":202611084045,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[2,1],\"masses\":[[1,25],[2,25],[22,25]],\"values\":[0,20,23]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/25, 2/25, 22/25] and aligned values X=[0, 20, 23]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(546/25, 2, 1/25, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (546/25, 2, 1/25, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0046","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e5bd30dd9de2d08c76d526f588c28c0bb6252d6276a5cf28e23109e624c614f","family_id":"qf-mod39-cheb-tail-probability","seed":202611084046,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[3,1],\"masses\":[[1,26],[2,26],[23,26]],\"values\":[0,21,24]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/26, 2/26, 23/26] and aligned values X=[0, 21, 24]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(297/13, 3, 1/26, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (297/13, 3, 1/26, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0047","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d2685a70b80cfbc8e5a1d17a316c3dc051b794d8cda798d73fa19934bca1d45","family_id":"qf-mod39-cheb-tail-probability","seed":202611084047,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,27],[2,27],[24,27]],\"values\":[0,22,25]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/27, 2/27, 24/27] and aligned values X=[0, 22, 25]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(644/27, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (644/27, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0048","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a17a0cc4e7fe438fc45abf391e0895705f0bf02cad5492e8974e5a0c68634ea5","family_id":"qf-mod39-cheb-tail-probability","seed":202611084048,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[2,1],\"masses\":[[1,28],[2,28],[25,28]],\"values\":[0,23,26]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/28, 2/28, 25/28] and aligned values X=[0, 23, 26]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(174/7, 2, 1/28, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (174/7, 2, 1/28, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0049","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"242733c0403a971b426d8e313abc057673c6e7e87e0d9c6b6d31a5e299681d9e","family_id":"qf-mod39-cheb-tail-probability","seed":202611084049,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[3,1],\"masses\":[[1,29],[2,29],[26,29]],\"values\":[0,24,27]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/29, 2/29, 26/29] and aligned values X=[0, 24, 27]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(750/29, 3, 1/29, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (750/29, 3, 1/29, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0050","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b73691584d58e8fbbafdd717c23d8b16eb52ba79b21a471cd0542d9a5b93acb6","family_id":"qf-mod39-cheb-tail-probability","seed":202611084050,"semantic_template":"qf-mod39-cheb-tail-probability/{\"epsilon\":[1,1],\"masses\":[[1,30],[2,30],[27,30]],\"values\":[0,25,28]}"},"fields":{"title":"The Absolute-Deviation Event Includes Equality at Epsilon","stem":"Use complete normalized finite law masses=[1/30, 2/30, 27/30] and aligned values X=[0, 25, 28]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (mean, epsilon, tail_probability, certificate).","answer":{"kind":"expression","value":"(403/15, 1, 1, exact_absolute_deviation_tail)","explanation":"Sum the complete law over |X−mean|≥epsilon. Retain the exact mean, epsilon and tail; the threshold event is non-strict even though epsilon itself must be positive. The exact result, retaining the requested order, is (403/15, 1, 1, exact_absolute_deviation_tail)."},"skills":["The Absolute-Deviation Event Includes Equality at Epsilon"],"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-20261108-r04-chebyshev-inequality-0051","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca327acb3258edddc84811aef59c2a704af74318670448e9593ed9eac253327c","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084051,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,6],[2,6],[3,6]],\"values\":[0,1,4]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/6, 2/6, 3/6] and aligned values X=[0, 1, 4]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 26/9, 26/9, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 26/9, 26/9, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0052","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b9946f4944ed732fc420140d8f3a5dff84972421093ba6958d36ab3e0c00a5c1","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084052,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[2,1],\"masses\":[[1,7],[2,7],[4,7]],\"values\":[0,2,5]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/7, 2/7, 4/7] and aligned values X=[0, 2, 5]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/7, 180/49, 45/49, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/7, 180/49, 45/49, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0053","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c40705e3bd2b2980f71d843c16312fb957e640f6ff5ec2604fa03339fe53689f","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084053,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[3,1],\"masses\":[[1,8],[2,8],[5,8]],\"values\":[0,3,6]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/8, 2/8, 5/8] and aligned values X=[0, 3, 6]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/8, 9/2, 1/2, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/8, 9/2, 1/2, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0054","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a15bbbda7f48736463a8f20fa470b59812bda99f42873e1b8142efcb74e44669","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084054,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,9],[2,9],[6,9]],\"values\":[0,4,7]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/9, 2/9, 6/9] and aligned values X=[0, 4, 7]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 434/81, 434/81, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 434/81, 434/81, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0055","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d49ee5e1999373bc7352901b784af5ca2f83057e500f4d1289f6708d4d37ae9","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084055,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[2,1],\"masses\":[[1,10],[2,10],[7,10]],\"values\":[0,5,8]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/10, 2/10, 7/10] and aligned values X=[0, 5, 8]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/10, 156/25, 39/25, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/10, 156/25, 39/25, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0056","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7384a7359dcd7395cf8fb4810bfdd9ce0afbb720b40a7b7f6d2d05a92bd940ab","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084056,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[3,1],\"masses\":[[1,11],[2,11],[8,11]],\"values\":[0,6,9]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/11, 2/11, 8/11] and aligned values X=[0, 6, 9]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/11, 864/121, 96/121, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/11, 864/121, 96/121, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0057","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"479b26aa17195076d9c5828e3847c19dea6474dc83c143d8f7c7fd355f47f2ce","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084057,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,12],[2,12],[9,12]],\"values\":[0,7,10]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/12, 2/12, 9/12] and aligned values X=[0, 7, 10]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 145/18, 145/18, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 145/18, 145/18, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0058","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0e1054fb8b0628233ae91cdfbcc57f5a64c29caec1eed50d81bb205fde04affd","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084058,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[2,1],\"masses\":[[1,13],[2,13],[10,13]],\"values\":[0,8,11]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/13, 2/13, 10/13] and aligned values X=[0, 8, 11]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/13, 1518/169, 759/338, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/13, 1518/169, 759/338, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0059","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9af10ba98fa307216a74e367278811aff369f191ac74ae8a6f6a340ce7eed253","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084059,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[3,1],\"masses\":[[1,14],[2,14],[11,14]],\"values\":[0,9,12]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/14, 2/14, 11/14] and aligned values X=[0, 9, 12]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/14, 486/49, 54/49, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/14, 486/49, 54/49, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0060","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c03065850b6eb8c760543cb562c0e20fd486505762b5c19b2958c14e11d8d76","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084060,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,15],[2,15],[12,15]],\"values\":[0,10,13]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/15, 2/15, 12/15] and aligned values X=[0, 10, 13]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 2444/225, 2444/225, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 2444/225, 2444/225, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0061","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c74673c810500188e7ba43c46e8a0fd9a2b00a8e01a94d25c7ce7f282929109a","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084061,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[2,1],\"masses\":[[1,16],[2,16],[13,16]],\"values\":[0,11,14]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/16, 2/16, 13/16] and aligned values X=[0, 11, 14]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/16, 189/16, 189/64, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/16, 189/16, 189/64, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0062","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d2649bb6340d3a9007896a8cb50413980049d4283d99fd71000ca5d62a079f8","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084062,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[3,1],\"masses\":[[1,17],[2,17],[14,17]],\"values\":[0,12,15]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/17, 2/17, 14/17] and aligned values X=[0, 12, 15]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/17, 3690/289, 410/289, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/17, 3690/289, 410/289, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0063","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f016f1e079349b359e4f1b1a54dcb6a31fca287c7ea9d51803f12ebe5d7a4097","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084063,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,18],[2,18],[15,18]],\"values\":[0,13,16]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/18, 2/18, 15/18] and aligned values X=[0, 13, 16]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 1112/81, 1112/81, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 1112/81, 1112/81, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0064","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b7d290fcdb45143dc56fcce7fb46eb3dd1bee6e784930646ca698f4e89c3c86","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084064,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[2,1],\"masses\":[[1,19],[2,19],[16,19]],\"values\":[0,14,17]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/19, 2/19, 16/19] and aligned values X=[0, 14, 17]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/19, 5304/361, 1326/361, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/19, 5304/361, 1326/361, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0065","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"023bb3cc0d18e248b020db59bd18fb7848a97fd1d1083c4fbe8d2bd25d35a012","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084065,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[3,1],\"masses\":[[1,20],[2,20],[17,20]],\"values\":[0,15,18]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/20, 2/20, 17/20] and aligned values X=[0, 15, 18]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/20, 783/50, 87/50, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/20, 783/50, 87/50, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0066","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7e7f18ab2369a0aa862dbe6c384ffc31016d8def3d6f0ccb185fec1b333198a7","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084066,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,21],[2,21],[18,21]],\"values\":[0,16,19]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/21, 2/21, 18/21] and aligned values X=[0, 16, 19]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 7334/441, 7334/441, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 7334/441, 7334/441, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0067","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e2d027fc940d84a806b184ce3d56f61e71e01bc344861854fb441b3ddfd27b37","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084067,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[2,1],\"masses\":[[1,22],[2,22],[19,22]],\"values\":[0,17,20]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/22, 2/22, 19/22] and aligned values X=[0, 17, 20]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/22, 2130/121, 1065/242, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/22, 2130/121, 1065/242, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0068","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ada1e4cdf86cfbcc7d3d8b4fc04142579e757904e98c577c101e70303d8abfe","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084068,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[3,1],\"masses\":[[1,23],[2,23],[20,23]],\"values\":[0,18,21]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/23, 2/23, 20/23] and aligned values X=[0, 18, 21]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/23, 9828/529, 1092/529, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/23, 9828/529, 1092/529, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0069","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb79dd5eb878ca38c24ce70ad93fb059e3abc80acc2021ab4b86d1239bf5fb7c","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084069,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,24],[2,24],[21,24]],\"values\":[0,19,22]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/24, 2/24, 21/24] and aligned values X=[0, 19, 22]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 176/9, 176/9, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 176/9, 176/9, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0070","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4eba2f01108c019db3367d2b5d5c43d51ce1040ad3277e37e4b669f5fdbb1aab","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084070,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[2,1],\"masses\":[[1,25],[2,25],[22,25]],\"values\":[0,20,23]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/25, 2/25, 22/25] and aligned values X=[0, 20, 23]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/25, 12834/625, 6417/1250, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/25, 12834/625, 6417/1250, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0071","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8859b47db6ef34251caeb69dea51a330d1e7a5c5690eb560a77e3cd588da0655","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084071,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[3,1],\"masses\":[[1,26],[2,26],[23,26]],\"values\":[0,21,24]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/26, 2/26, 23/26] and aligned values X=[0, 21, 24]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/26, 3636/169, 404/169, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/26, 3636/169, 404/169, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0072","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbda7cd7bf1a6a0a4720f74dd90231bd457e625653d8593b75bd18a5877a427a","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084072,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,27],[2,27],[24,27]],\"values\":[0,22,25]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/27, 2/27, 24/27] and aligned values X=[0, 22, 25]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 16400/729, 16400/729, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 16400/729, 16400/729, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0073","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9fe9b1747ad09572bc42e6da07a81978fe373b620b0dd1418e0e5c21784848c5","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084073,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[2,1],\"masses\":[[1,28],[2,28],[25,28]],\"values\":[0,23,26]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/28, 2/28, 25/28] and aligned values X=[0, 23, 26]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/28, 2301/98, 2301/392, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/28, 2301/98, 2301/392, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0074","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f7ba898fa5a0cd4666647925dad2c3c6dc0ac12fa91422fe503f0cdf672c04d3","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084074,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[3,1],\"masses\":[[1,29],[2,29],[26,29]],\"values\":[0,24,27]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/29, 2/29, 26/29] and aligned values X=[0, 24, 27]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/29, 20574/841, 2286/841, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1/29, 20574/841, 2286/841, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0075","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d53bd7f9547371e05dbd392befab0d155ea3b705f404693ef0ee0bdc1130bb3a","family_id":"qf-mod39-cheb-bound-comparison","seed":202611084075,"semantic_template":"qf-mod39-cheb-bound-comparison/{\"epsilon\":[1,1],\"masses\":[[1,30],[2,30],[27,30]],\"values\":[0,25,28]}"},"fields":{"title":"Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence","stem":"Use complete normalized finite law masses=[1/30, 2/30, 27/30] and aligned values X=[0, 25, 28]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Retain the raw bound without clipping it to1. Give (tail_probability, variance, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1, 5726/225, 5726/225, Yes, upper_bound_not_tail_equality_and_no_independence_needed)","explanation":"Compute the actual tail, variance and variance/epsilon^2 separately. The upper bound may exceed1. With variance0, values are constant almost surely under this law, not necessarily at zero-mass table entries. The exact result, retaining the requested order, is (1, 5726/225, 5726/225, Yes, upper_bound_not_tail_equality_and_no_independence_needed)."},"skills":["Chebyshev Gives a Raw Upper Bound, Not Tail Equality or Independence"],"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-20261108-r04-chebyshev-inequality-0076","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0ba297547b651c41d87bf53821f13f0327ec0244769b1d8f6cd82273611c389","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084076,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,6],[2,6],[3,6]],\"values\":[0,1,4]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/6, 2/6, 3/6] and aligned values X=[0, 1, 4]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([49/9, 16/9, 25/9], 1, 1, 1, Yes, 26/9, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([49/9, 16/9, 25/9], 1, 1, 1, Yes, 26/9, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0077","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11f9ac6daeda048394d7e09ea14b4a226c6c1ee5e9ce4144ecb7ec786398a1ea","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084077,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[2,1],\"masses\":[[1,7],[2,7],[4,7]],\"values\":[0,2,5]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/7, 2/7, 4/7] and aligned values X=[0, 2, 5]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([576/49, 100/49, 121/49], 4, 1/7, 1/7, Yes, 45/49, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([576/49, 100/49, 121/49], 4, 1/7, 1/7, Yes, 45/49, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0078","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"525de9e0b09cfed4d79183933d64c57da8f24ec3342ed15116b3aebf278a4a92","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084078,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[3,1],\"masses\":[[1,8],[2,8],[5,8]],\"values\":[0,3,6]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/8, 2/8, 5/8] and aligned values X=[0, 3, 6]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([81/4, 9/4, 9/4], 9, 1/8, 1/8, Yes, 1/2, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([81/4, 9/4, 9/4], 9, 1/8, 1/8, Yes, 1/2, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0079","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"054ad6ee364d31744cd99fb518a595b444ee556c0e0ab80c76c45fcafa2e21c7","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084079,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,9],[2,9],[6,9]],\"values\":[0,4,7]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/9, 2/9, 6/9] and aligned values X=[0, 4, 7]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([2500/81, 196/81, 169/81], 1, 1, 1, Yes, 434/81, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([2500/81, 196/81, 169/81], 1, 1, 1, Yes, 434/81, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0080","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"217e7d84f9961383d0f9d4ded51dabf8ef5a9f3df49152c80e71b3f9fc9ce8be","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084080,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[2,1],\"masses\":[[1,10],[2,10],[7,10]],\"values\":[0,5,8]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/10, 2/10, 7/10] and aligned values X=[0, 5, 8]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([1089/25, 64/25, 49/25], 4, 1/10, 1/10, Yes, 39/25, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([1089/25, 64/25, 49/25], 4, 1/10, 1/10, Yes, 39/25, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0081","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb4eecba326c34526a04caebb8e826c34c9b0cfa733803017504e2c90be0a10d","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084081,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[3,1],\"masses\":[[1,11],[2,11],[8,11]],\"values\":[0,6,9]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/11, 2/11, 8/11] and aligned values X=[0, 6, 9]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([7056/121, 324/121, 225/121], 9, 1/11, 1/11, Yes, 96/121, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([7056/121, 324/121, 225/121], 9, 1/11, 1/11, Yes, 96/121, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0082","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff3a089621361d88bb70839f4355129c2ea217a711225c582e78bcb7dc817b2b","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084082,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,12],[2,12],[9,12]],\"values\":[0,7,10]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/12, 2/12, 9/12] and aligned values X=[0, 7, 10]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([676/9, 25/9, 16/9], 1, 1, 1, Yes, 145/18, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([676/9, 25/9, 16/9], 1, 1, 1, Yes, 145/18, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0083","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2847dec967d083173358bf084f0d20e22f8994a9849d2a6810f24db398b94bb1","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084083,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[2,1],\"masses\":[[1,13],[2,13],[10,13]],\"values\":[0,8,11]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/13, 2/13, 10/13] and aligned values X=[0, 8, 11]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([15876/169, 484/169, 289/169], 4, 1/13, 1/13, Yes, 759/338, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([15876/169, 484/169, 289/169], 4, 1/13, 1/13, Yes, 759/338, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0084","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92a90df2a66a9b04a56d47de7746ae9fede3e14128d7da32c51aea61f37bcdb6","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084084,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[3,1],\"masses\":[[1,14],[2,14],[11,14]],\"values\":[0,9,12]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/14, 2/14, 11/14] and aligned values X=[0, 9, 12]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([5625/49, 144/49, 81/49], 9, 1/14, 1/14, Yes, 54/49, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([5625/49, 144/49, 81/49], 9, 1/14, 1/14, Yes, 54/49, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0085","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d6235792ed892b764d0768593865107d212d7bced7181ba8417681a0067efd11","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084085,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,15],[2,15],[12,15]],\"values\":[0,10,13]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/15, 2/15, 12/15] and aligned values X=[0, 10, 13]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([30976/225, 676/225, 361/225], 1, 1, 1, Yes, 2444/225, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([30976/225, 676/225, 361/225], 1, 1, 1, Yes, 2444/225, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0086","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ab40e73d67596c64a22d685e8b0be6fa50d9468deb1eb948abade0574c8e1acf","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084086,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[2,1],\"masses\":[[1,16],[2,16],[13,16]],\"values\":[0,11,14]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/16, 2/16, 13/16] and aligned values X=[0, 11, 14]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([2601/16, 49/16, 25/16], 4, 1/16, 1/16, Yes, 189/64, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([2601/16, 49/16, 25/16], 4, 1/16, 1/16, Yes, 189/64, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0087","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e06543f2850ff829590e81c0898e73e3c710081bbf91ab7bbfc3b8812aabbd1e","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084087,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[3,1],\"masses\":[[1,17],[2,17],[14,17]],\"values\":[0,12,15]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/17, 2/17, 14/17] and aligned values X=[0, 12, 15]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([54756/289, 900/289, 441/289], 9, 1/17, 1/17, Yes, 410/289, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([54756/289, 900/289, 441/289], 9, 1/17, 1/17, Yes, 410/289, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0088","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0cd2f450cb7cbd03c525652cf90e139c5dbaa53c4173422a2a5e7802465a8e58","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084088,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,18],[2,18],[15,18]],\"values\":[0,13,16]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/18, 2/18, 15/18] and aligned values X=[0, 13, 16]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([17689/81, 256/81, 121/81], 1, 1, 1, Yes, 1112/81, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([17689/81, 256/81, 121/81], 1, 1, 1, Yes, 1112/81, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0089","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3434447eff695d2191e389136ab52722d9556cebc6766254af5d7d96e3ea06b1","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084089,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[2,1],\"masses\":[[1,19],[2,19],[16,19]],\"values\":[0,14,17]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/19, 2/19, 16/19] and aligned values X=[0, 14, 17]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([90000/361, 1156/361, 529/361], 4, 1/19, 1/19, Yes, 1326/361, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([90000/361, 1156/361, 529/361], 4, 1/19, 1/19, Yes, 1326/361, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0090","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e8d7249aa8ed9d02a198a2e022089ae98567b2dfbe58516ce642927e8fd1c42","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084090,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[3,1],\"masses\":[[1,20],[2,20],[17,20]],\"values\":[0,15,18]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/20, 2/20, 17/20] and aligned values X=[0, 15, 18]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([7056/25, 81/25, 36/25], 9, 1/20, 1/20, Yes, 87/50, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([7056/25, 81/25, 36/25], 9, 1/20, 1/20, Yes, 87/50, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0091","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"202143341e66f3af3710bd77521c104a9b31ad83ec5799895317ad903c3d679d","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084091,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,21],[2,21],[18,21]],\"values\":[0,16,19]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/21, 2/21, 18/21] and aligned values X=[0, 16, 19]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([139876/441, 1444/441, 625/441], 1, 1, 1, Yes, 7334/441, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([139876/441, 1444/441, 625/441], 1, 1, 1, Yes, 7334/441, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0092","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e34c85c72913419c6081253e88a3c3d3ca0cf8a01b7cc9a5da2d37771c53ef2","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084092,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[2,1],\"masses\":[[1,22],[2,22],[19,22]],\"values\":[0,17,20]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/22, 2/22, 19/22] and aligned values X=[0, 17, 20]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([42849/121, 400/121, 169/121], 4, 1/22, 1/22, Yes, 1065/242, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([42849/121, 400/121, 169/121], 4, 1/22, 1/22, Yes, 1065/242, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0093","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d61e8d92b277b9b71363907b67dab592842538d2684613e1223eff8e66af3054","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084093,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[3,1],\"masses\":[[1,23],[2,23],[20,23]],\"values\":[0,18,21]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/23, 2/23, 20/23] and aligned values X=[0, 18, 21]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([207936/529, 1764/529, 729/529], 9, 1/23, 1/23, Yes, 1092/529, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([207936/529, 1764/529, 729/529], 9, 1/23, 1/23, Yes, 1092/529, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0094","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c34081723ce237a8c3b547628bca7d5f9623d92d3bd51c098a5c620e7a9067f0","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084094,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,24],[2,24],[21,24]],\"values\":[0,19,22]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/24, 2/24, 21/24] and aligned values X=[0, 19, 22]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([15625/36, 121/36, 49/36], 1, 1, 1, Yes, 176/9, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([15625/36, 121/36, 49/36], 1, 1, 1, Yes, 176/9, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0095","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e6367c4b9a39cefedce5d9276fd9c13c863cd4617f3d236de0f849cb43104ab","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084095,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[2,1],\"masses\":[[1,25],[2,25],[22,25]],\"values\":[0,20,23]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/25, 2/25, 22/25] and aligned values X=[0, 20, 23]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([298116/625, 2116/625, 841/625], 4, 1/25, 1/25, Yes, 6417/1250, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([298116/625, 2116/625, 841/625], 4, 1/25, 1/25, Yes, 6417/1250, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0096","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e12f855738c25adaeeae34553dad1a2a4bf891360ca96277335c87d9f6ba1e2","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084096,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[3,1],\"masses\":[[1,26],[2,26],[23,26]],\"values\":[0,21,24]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/26, 2/26, 23/26] and aligned values X=[0, 21, 24]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([88209/169, 576/169, 225/169], 9, 1/26, 1/26, Yes, 404/169, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([88209/169, 576/169, 225/169], 9, 1/26, 1/26, Yes, 404/169, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0097","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"598d43b6d10c41d3df4fa759ef8393b38ee0eac0b4b8ce99d82849db126ca627","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084097,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,27],[2,27],[24,27]],\"values\":[0,22,25]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/27, 2/27, 24/27] and aligned values X=[0, 22, 25]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([414736/729, 2500/729, 961/729], 1, 1, 1, Yes, 16400/729, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([414736/729, 2500/729, 961/729], 1, 1, 1, Yes, 16400/729, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0098","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5841b851f4275fe5a39f1d001297e3a53285974f711e495a5838f589474a325e","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084098,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[2,1],\"masses\":[[1,28],[2,28],[25,28]],\"values\":[0,23,26]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/28, 2/28, 25/28] and aligned values X=[0, 23, 26]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=2/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([30276/49, 169/49, 64/49], 4, 1/28, 1/28, Yes, 2301/392, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([30276/49, 169/49, 64/49], 4, 1/28, 1/28, Yes, 2301/392, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0099","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2199ace50d8aace7097e82228988f0d1f2c79e1f5d57ca2d517b3d04ab55ff2","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084099,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[3,1],\"masses\":[[1,29],[2,29],[26,29]],\"values\":[0,24,27]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/29, 2/29, 26/29] and aligned values X=[0, 24, 27]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=3/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([562500/841, 2916/841, 1089/841], 9, 1/29, 1/29, Yes, 2286/841, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([562500/841, 2916/841, 1089/841], 9, 1/29, 1/29, Yes, 2286/841, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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-20261108-r04-chebyshev-inequality-0100","batch_id":"batch-20261014-advance-20261108-r04-chebyshev-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6813ba8fbab9fb1ca55f7cc0f255e010c11802ef23b0c981bd7cb21e5818aff2","family_id":"qf-mod39-cheb-markov-square-derivation","seed":202611084100,"semantic_template":"qf-mod39-cheb-markov-square-derivation/{\"epsilon\":[1,1],\"masses\":[[1,30],[2,30],[27,30]],\"values\":[0,25,28]}"},"fields":{"title":"Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared","stem":"Use complete normalized finite law masses=[1/30, 2/30, 27/30] and aligned values X=[0, 25, 28]. Masses are nonnegative and sum to1; values may be signed. Mean and variance use the same law. No independence premise is required for this one-variable Chebyshev bound; zero-mass values remain in the table but contribute no event mass or moment. Positive epsilon=1/1. Give (squared_deviations, epsilon_squared, squared_tail, absolute_tail, same_event, raw_upper_bound, certificate).","answer":{"kind":"expression","value":"([162409/225, 784/225, 289/225], 1, 1, 1, Yes, 5726/225, Markov_applied_to_nonnegative_squared_deviation)","explanation":"Compute every (X−mean)^2, use threshold epsilon^2, and compare the squared-event probability with the absolute-deviation tail. They are the same event. The final raw bound is variance/epsilon^2, not an event-probability equality claim. The exact result, retaining the requested order, is ([162409/225, 784/225, 289/225], 1, 1, 1, Yes, 5726/225, Markov_applied_to_nonnegative_squared_deviation)."},"skills":["Apply Markov to the Nonnegative Squared Deviation at Epsilon Squared"],"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."}}