{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0001","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0ed63d486d5f83374980c63157e3e9c7e45e1d134fed6c9a47c796842e7c530","family_id":"qf-variance-weighted","seed":202610057001,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/1\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2}, with event family F=𝒫(S). The total function X:S→R has table {(0, 5), (1, 5), (2, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(3), 2/(3), 0/(3)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"0/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 0/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0002","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ecd07a6af6450cf90549ff00adaca6221f6380da99d7a46c6665f4942e86497","family_id":"qf-variance-weighted","seed":202610057002,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"3\",\"1/2\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"9/4","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 9/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0003","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"48e8b2c0d56d97e5a70938322d0c4e0ef204cc0380cc7808d46a8d3da0d24477","family_id":"qf-variance-weighted","seed":202610057003,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"4\",\"1/2\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"4/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 4/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0004","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d1980a6d64ed1cfba5680f559acddcab964b512a90d6a1acfd1538ba0caa5cec","family_id":"qf-variance-weighted","seed":202610057004,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"5\",\"1/2\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"25/4","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 25/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0005","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"529a550595de6b17935caba8be7117b9a52820cc4cf3be07a65685eb9c4fbed5","family_id":"qf-variance-weighted","seed":202610057005,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"7/10\"],[\"6\",\"3/10\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"189/25","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 189/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0006","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e91944ede98b7fa6dc9b92a925bd42265cb6bd57fb00311591a97e2724316e13","family_id":"qf-variance-weighted","seed":202610057006,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"2/3\"],[\"7\",\"1/3\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"98/9","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 98/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0007","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86236d4f73365f7f634261528a13022483922667ef1ee876358229fbe232f844","family_id":"qf-variance-weighted","seed":202610057007,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"9/14\"],[\"8\",\"5/14\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"720/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 720/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0008","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5434326df0dba3e3a9701adfe65071119937816acc0c15078124336d61ae0da","family_id":"qf-variance-weighted","seed":202610057008,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/8\"],[\"9\",\"3/8\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"1215/64","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 1215/64. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0009","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f2352f4f208741e59027f9b565d81deed51ae44e0b45643443128efbfa0e5d8","family_id":"qf-variance-weighted","seed":202610057009,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"11/14\"],[\"10\",\"3/14\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"825/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 825/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0010","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a564dcf844c7359afd78e06834a5f28f0843c69e4ffa94894b87f429a548bed0","family_id":"qf-variance-weighted","seed":202610057010,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"3/4\"],[\"11\",\"1/4\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"363/16","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 363/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0011","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9c63c6343506028913e4d90c8def10238d64f3b9dc789fa67682601977f4a96","family_id":"qf-variance-weighted","seed":202610057011,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"13/18\"],[\"12\",\"5/18\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"260/9","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 260/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0012","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90ca39628839cf6f887c889140ff933a7f361a3e65aae57499c69ea25f9ef0d8","family_id":"qf-variance-weighted","seed":202610057012,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"7/10\"],[\"13\",\"3/10\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"3549/100","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 3549/100. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0013","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"adc3a5901326de7ba4f1f9f1865dce54c20ed29a732422bf4c42249d3a383b48","family_id":"qf-variance-weighted","seed":202610057013,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"14\",\"1/6\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"245/9","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 245/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0014","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a6a0c8132d4821f887dff48a853a5d8edcbbaf8ea8dce9eca16e9017fd32f1f","family_id":"qf-variance-weighted","seed":202610057014,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"4/5\"],[\"15\",\"1/5\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"36/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 36/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0015","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8d4f177e97c60fd9bddc2bad0d391b0d519536b736a24326adabde4a713a4d26","family_id":"qf-variance-weighted","seed":202610057015,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"17/22\"],[\"16\",\"5/22\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"5440/121","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 5440/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0016","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"529c51e6ba84b752731392a659ef757c2acd9fd30dd258565606c7837755af30","family_id":"qf-variance-weighted","seed":202610057016,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"3/4\"],[\"17\",\"1/4\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"867/16","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 867/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0017","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58b61a33ca9c6bc9bff0015c08f1daa2ee272fcff88413a0ab9ee9c27f52b562","family_id":"qf-variance-weighted","seed":202610057017,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"19/22\"],[\"18\",\"3/22\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"4617/121","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 4617/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0018","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cac7c8d86223bae4564457cb305dc80f60d3c69c15c4a137cb6a35d38903a402","family_id":"qf-variance-weighted","seed":202610057018,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"19\",\"1/6\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"1805/36","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 1805/36. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0019","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a4c4f4312c52ca3b8092741ec53e66960dad0e75e8e07f1aeddc919a7e8db5f4","family_id":"qf-variance-weighted","seed":202610057019,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"21/26\"],[\"20\",\"5/26\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"10500/169","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 10500/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0020","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8bf7773b57ae14e3b9767b48bf69955481df54f227c3139a6195923071d4139d","family_id":"qf-variance-weighted","seed":202610057020,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"11/14\"],[\"21\",\"3/14\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"297/4","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 297/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0021","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6acaabdc1fc046b3a59fe5e8391e7a54ea9b2a7feb13dba9543c2ed251bdb3cd","family_id":"qf-variance-weighted","seed":202610057021,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"23/26\"],[\"22\",\"3/26\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"8349/169","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 8349/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0022","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39db17bddd27454d20f65ad795db3f94a12433168a612e9b5e97de75d3640ba6","family_id":"qf-variance-weighted","seed":202610057022,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"6/7\"],[\"23\",\"1/7\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"3174/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 3174/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0023","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2b018bee39bc0e55f32076d7e6a2295ed88afc8754adab8cc4dc8f536626757","family_id":"qf-variance-weighted","seed":202610057023,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"24\",\"1/6\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"80/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 80/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0024","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c5ff74ff334707f73ec188bb74e499d031bb2774f86c127cd13df75c471efa11","family_id":"qf-variance-weighted","seed":202610057024,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"13/16\"],[\"25\",\"3/16\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"24375/256","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 24375/256. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0025","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7e1fe815d98db3de498993c4a9de1fbe724ab30e4c5da9e6e273278705614f9a","family_id":"qf-variance-weighted","seed":202610057025,"semantic_template":"qf-variance-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"9/10\"],[\"26\",\"1/10\"]],\"operation\":\"variance\"}"},"fields":{"title":"Exact Finite Probability Variance","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Put Z=X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"1521/25","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 1521/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Exact Finite Probability Variance"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r07-variance-0026","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a43e59ab75c5ca46c08a2896a83e5880be2e7576ba68dd1f25b84e0ca13a69ab","family_id":"qf-variance-shift","seed":202610057026,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/1\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2}, with event family F=𝒫(S). The total function X:S→R has table {(0, 5), (1, 5), (2, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(3), 2/(3), 0/(3)], summing exactly to one. Put Z=X+(-12). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"0/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 0/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0027","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bda4e063a4b16ac72f020db14f2711c0afe4f3ce604167f8ae88482397db2972","family_id":"qf-variance-shift","seed":202610057027,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"3\",\"1/2\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Put Z=X+(-11). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"9/4","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 9/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0028","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b41b7e84b6f38dab76c227cbcdf80d64d038ba5ee4d9c82379f3b7437940c1ea","family_id":"qf-variance-shift","seed":202610057028,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"4\",\"1/2\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Put Z=X+(-10). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"4/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 4/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0029","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aebd5fd2f4bbc30f44f5d55ceb7fd08b8f8d4a8065089ca342c2877dd84fc88a","family_id":"qf-variance-shift","seed":202610057029,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"5\",\"1/2\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Put Z=X+(-9). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"25/4","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 25/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0030","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1fe34f9485518ff9d9a24a25a2b91ad8ed14d374564a1844a7eb3015ec34fef4","family_id":"qf-variance-shift","seed":202610057030,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"7/10\"],[\"6\",\"3/10\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Put Z=X+(-8). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"189/25","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 189/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0031","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf8d0706225df7a93e3d004bb49655d414c83e963e499275a4344c8b2c1d58fa","family_id":"qf-variance-shift","seed":202610057031,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"2/3\"],[\"7\",\"1/3\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Put Z=X+(-7). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"98/9","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 98/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0032","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f786702984a35535d1550873b089fc91ca43b4db79d7998ccc58498727e30150","family_id":"qf-variance-shift","seed":202610057032,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"9/14\"],[\"8\",\"5/14\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Put Z=X+(-6). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"720/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 720/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0033","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6475cd3730a959db47336d25a00deccb401391f521a231ce7ca820b936e43d62","family_id":"qf-variance-shift","seed":202610057033,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/8\"],[\"9\",\"3/8\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Put Z=X+(-5). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"1215/64","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 1215/64. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0034","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cb5c0fbf39c6573be1db3b559c8a8fe91023d586aba7e7494c7427bc1b5ea90","family_id":"qf-variance-shift","seed":202610057034,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"11/14\"],[\"10\",\"3/14\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Put Z=X+(-4). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"825/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 825/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0035","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35d654b82aa512c6b94ba4140a7cd2dafc4101ea470bef792625d4a26888a7c2","family_id":"qf-variance-shift","seed":202610057035,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"3/4\"],[\"11\",\"1/4\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Put Z=X+(-3). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"363/16","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 363/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0036","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd058399c0d4a397aa21b32298910b41a208d6089ecdf2f1ad7feb6c7add1c48","family_id":"qf-variance-shift","seed":202610057036,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"13/18\"],[\"12\",\"5/18\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Put Z=X+(-2). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"260/9","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 260/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0037","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc75b6ecfcd3be801fcba7a6e93fefefc304b85d68574a9c73c5fd4aa3b5c2f5","family_id":"qf-variance-shift","seed":202610057037,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"7/10\"],[\"13\",\"3/10\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Put Z=X+(-1). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"3549/100","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 3549/100. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0038","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1af4b76e5cc7406d668dcb94fc49624eddfae8aea8c4b741782b387103eb44c2","family_id":"qf-variance-shift","seed":202610057038,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"14\",\"1/6\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Put Z=X+(0). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"245/9","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 245/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0039","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"857f5d2de44afb081b4dda11968650b27a376c18cfc4e6354c12e8f3fd160ffc","family_id":"qf-variance-shift","seed":202610057039,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"4/5\"],[\"15\",\"1/5\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Put Z=X+(1). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"36/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 36/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0040","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f52dcc7949711bfeca020458bc2d53bfc7b6965b671fc8517caa5f301f553a6c","family_id":"qf-variance-shift","seed":202610057040,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"17/22\"],[\"16\",\"5/22\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Put Z=X+(2). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"5440/121","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 5440/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0041","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66ecb0806870e8dfb619b7a915d8f91a1dfe65bcdd25a76043b3eb93af2187c0","family_id":"qf-variance-shift","seed":202610057041,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"3/4\"],[\"17\",\"1/4\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Put Z=X+(3). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"867/16","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 867/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0042","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"440a1f9dc57180460b35ee43435f73ec90a3d24a9ea887685d127161c2bc7d92","family_id":"qf-variance-shift","seed":202610057042,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"19/22\"],[\"18\",\"3/22\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Put Z=X+(4). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"4617/121","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 4617/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0043","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba857290ba31405aed80604a4a59ff5b7ca147aea091c6418b5552255141df83","family_id":"qf-variance-shift","seed":202610057043,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"19\",\"1/6\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Put Z=X+(5). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"1805/36","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 1805/36. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0044","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5330b7a21025b4ebe8c22218e9316a61876947ac1d2294a538c22e86a65fddd2","family_id":"qf-variance-shift","seed":202610057044,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"21/26\"],[\"20\",\"5/26\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Put Z=X+(6). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"10500/169","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 10500/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0045","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0da1771392b61a2091313cb194ba628b396ddb80d9a9073ad6ad793ec0917f87","family_id":"qf-variance-shift","seed":202610057045,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"11/14\"],[\"21\",\"3/14\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Put Z=X+(7). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"297/4","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 297/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0046","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c69078f7f3653114db090c95a99229bb8c0ae023897f8e790d31c8a10cf93ed","family_id":"qf-variance-shift","seed":202610057046,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"23/26\"],[\"22\",\"3/26\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Put Z=X+(8). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"8349/169","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 8349/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0047","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6bb311717a711e65977ecea8645c30c344a706987623efa1e4f6affe40207484","family_id":"qf-variance-shift","seed":202610057047,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"6/7\"],[\"23\",\"1/7\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Put Z=X+(9). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"3174/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 3174/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0048","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dbaa541582d52c093dbf69f087c3687338ffc60cba91ff89d5bbad5336bdd986","family_id":"qf-variance-shift","seed":202610057048,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"24\",\"1/6\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Put Z=X+(10). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"80/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 80/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0049","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80c8ed0a2521b7cbbfba36ee7b6c3a32fd80dde2f854d46b90aefc263f581647","family_id":"qf-variance-shift","seed":202610057049,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"13/16\"],[\"25\",\"3/16\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Put Z=X+(11). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"24375/256","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 24375/256. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0050","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb0bcd91a13bbd9be0b609ace5e898fa43e229124d5be58896c1fe6a15d5a174","family_id":"qf-variance-shift","seed":202610057050,"semantic_template":"qf-variance-shift/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"9/10\"],[\"26\",\"1/10\"]],\"operation\":\"shift_variance\"}"},"fields":{"title":"Probability Variance after Translation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Put Z=X+(12). Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"1521/25","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 1521/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Translation"],"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-20261004-advance-20261005-r07-variance-0051","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ed5fa570a41e4a046e86fa80f12c287882650ce5f83852e47531153bb9156f6","family_id":"qf-variance-scale","seed":202610057051,"semantic_template":"qf-variance-scale/{\"k\":0,\"normalized_model\":true,\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2}, with event family F=𝒫(S). The total function X:S→R has table {(0, 5), (1, 5), (2, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(3), 2/(3), 0/(3)], summing exactly to one. Put Z=(0)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"0/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 0/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0052","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a557ef56bc9a301d1f2216750aa60c0b7d4f2dba393b43671bb22bbb3ced23cf","family_id":"qf-variance-scale","seed":202610057052,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":2,\"law\":[[\"0\",\"1/2\"],[\"3\",\"1/2\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Put Z=(-2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"9/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 9/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0053","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4719b98ca2fbc64eeb043cb14321a9da2649210555993cde44faecccb1784dd4","family_id":"qf-variance-scale","seed":202610057053,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":3,\"law\":[[\"0\",\"1/2\"],[\"4\",\"1/2\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Put Z=(3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"36/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 36/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0054","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"061237e2908c7c4fdacc29c06508c49af3f09f1185970627b029d04daaf94673","family_id":"qf-variance-scale","seed":202610057054,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":4,\"law\":[[\"0\",\"1/2\"],[\"5\",\"1/2\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Put Z=(-4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"100/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 100/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0055","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f21ed111d4487e1a08ff3306cdd92f36353765200b9b0dd445cab7ed94f8e82","family_id":"qf-variance-scale","seed":202610057055,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":5,\"law\":[[\"0\",\"7/10\"],[\"6\",\"3/10\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Put Z=(5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"189/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 189/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0056","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b56b6a9f2636aa189188cc09eba7b8108a374b5deadde84861000ead55ac1905","family_id":"qf-variance-scale","seed":202610057056,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":1,\"law\":[[\"0\",\"2/3\"],[\"7\",\"1/3\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Put Z=(-1)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"98/9","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 98/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0057","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca242f0cdd9e5a183eeafeb68d0701593508f836b7bd4ffdc7faa15b714b286b","family_id":"qf-variance-scale","seed":202610057057,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":2,\"law\":[[\"0\",\"9/14\"],[\"8\",\"5/14\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Put Z=(2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"2880/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 2880/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0058","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16292cd7d216cef4dcc0aaac11d17a3e770bef57a10f9b548654c9468b4f12c0","family_id":"qf-variance-scale","seed":202610057058,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":3,\"law\":[[\"0\",\"5/8\"],[\"9\",\"3/8\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Put Z=(-3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"10935/64","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 10935/64. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0059","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9d1747014ac47d3f373f80c23028a1e7ab51a33f2b58cf88aeb102d65f50bb6","family_id":"qf-variance-scale","seed":202610057059,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":4,\"law\":[[\"0\",\"11/14\"],[\"10\",\"3/14\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Put Z=(4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"13200/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 13200/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0060","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5220a92a25a400ee10ab843cf6db7f92d465b9caf8242d30537e38869b9f440e","family_id":"qf-variance-scale","seed":202610057060,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":5,\"law\":[[\"0\",\"3/4\"],[\"11\",\"1/4\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Put Z=(-5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"9075/16","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 9075/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0061","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd538cd1bba05afcbbdf7353b5504495aa7ca31e52aad07ad15cd2c74d81e29f","family_id":"qf-variance-scale","seed":202610057061,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":1,\"law\":[[\"0\",\"13/18\"],[\"12\",\"5/18\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Put Z=(1)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"260/9","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 260/9. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0062","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c46a30573e997ec71cd7df32b78e216b53f2ace4802d165403a2c728ec78f8f8","family_id":"qf-variance-scale","seed":202610057062,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":2,\"law\":[[\"0\",\"7/10\"],[\"13\",\"3/10\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Put Z=(-2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"3549/25","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 3549/25. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0063","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b35cf13622f99615ea7f6147cc1a735b2d6e53dc74db12529c6ad98c33dae0b","family_id":"qf-variance-scale","seed":202610057063,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":3,\"law\":[[\"0\",\"5/6\"],[\"14\",\"1/6\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Put Z=(3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"245/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 245/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0064","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"845672aa5e530db9f6d4bb7888fc7c77b8a178b99c62db9a46826654b6b8dd9a","family_id":"qf-variance-scale","seed":202610057064,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":4,\"law\":[[\"0\",\"4/5\"],[\"15\",\"1/5\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Put Z=(-4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"576/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 576/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0065","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"78be1c4362390485316d4745c31352075fef92235d5746bcac743b3b176ede3c","family_id":"qf-variance-scale","seed":202610057065,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":5,\"law\":[[\"0\",\"17/22\"],[\"16\",\"5/22\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Put Z=(5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"136000/121","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 136000/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0066","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"686f58a7466720ef6120f68c78dc6f85c7043b428d7cc9c7ddd2e3f4b33539bb","family_id":"qf-variance-scale","seed":202610057066,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":1,\"law\":[[\"0\",\"3/4\"],[\"17\",\"1/4\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Put Z=(-1)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"867/16","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 867/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0067","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7622acbc361acd806dca2f231fa363907bea90521d6d4e2554558168170397bc","family_id":"qf-variance-scale","seed":202610057067,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":2,\"law\":[[\"0\",\"19/22\"],[\"18\",\"3/22\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Put Z=(2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"18468/121","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 18468/121. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0068","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b5393cf2f4b1044261cd7e7d337f8fc098a73ae4def878b4b93fc064eda736b4","family_id":"qf-variance-scale","seed":202610057068,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":3,\"law\":[[\"0\",\"5/6\"],[\"19\",\"1/6\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Put Z=(-3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"1805/4","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 1805/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0069","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c7391d21f97bebc4ba79d91507cea46fde64ce62e22e0b9da268ea168912b59f","family_id":"qf-variance-scale","seed":202610057069,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":4,\"law\":[[\"0\",\"21/26\"],[\"20\",\"5/26\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Put Z=(4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"168000/169","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 168000/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0070","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97cf4e19b375f44e0f90d9e1500b731b2efbade471d89717a58ff5d1e76d7ef8","family_id":"qf-variance-scale","seed":202610057070,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":5,\"law\":[[\"0\",\"11/14\"],[\"21\",\"3/14\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Put Z=(-5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"7425/4","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 7425/4. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0071","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"043967cbaf1d8921459dc50ac34545099277ce0839ea05d153f5f3a20773d8d3","family_id":"qf-variance-scale","seed":202610057071,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":1,\"law\":[[\"0\",\"23/26\"],[\"22\",\"3/26\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Put Z=(1)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"8349/169","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 8349/169. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0072","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87e4740cebf0b3c3bab223513c4f870f21b6e689ec3a79d0ff79d44df9fd6d94","family_id":"qf-variance-scale","seed":202610057072,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":2,\"law\":[[\"0\",\"6/7\"],[\"23\",\"1/7\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Put Z=(-2)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"12696/49","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 12696/49. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0073","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b654e09e88d534eab6340ee410ace9239f1cb2725fc920a4ec8f6bcd1d12988a","family_id":"qf-variance-scale","seed":202610057073,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":3,\"law\":[[\"0\",\"5/6\"],[\"24\",\"1/6\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Put Z=(3)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"720/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 720/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0074","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db0f6b7e43d0867b9d8a58b4bf7dcaa601784fc61c66a01e77babc01b48828af","family_id":"qf-variance-scale","seed":202610057074,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":4,\"law\":[[\"0\",\"13/16\"],[\"25\",\"3/16\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Put Z=(-4)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"24375/16","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 24375/16. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0075","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02232537c43c663522f1ea9a90d7816a587b9141dd5bcb281b4b055c9c55afb6","family_id":"qf-variance-scale","seed":202610057075,"semantic_template":"qf-variance-scale/{\"F\":\"powerset(S)\",\"k\":5,\"law\":[[\"0\",\"9/10\"],[\"26\",\"1/10\"]],\"operation\":\"scale_variance\"}"},"fields":{"title":"Probability Variance after Scaling","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Put Z=(5)X. Find Var(Z)=E[(Z−E[Z])²] exactly as a reduced positive-denominator fraction. This is probability-distribution variance, not sample variance with an n−1 denominator.","answer":{"kind":"expression","value":"1521/1","explanation":"Compute the mean, then sum weighted squared deviations exactly to obtain 1521/1. Translation preserves deviations; scaling multiplies variance by k², including zero or negative k. Zero-weight outcomes contribute nothing. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Probability Variance after Scaling"],"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-20261004-advance-20261005-r07-variance-0076","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b0a9acfe39579ebc972b3368879810f540fa5c73c28c46f83e08ed3fbbf6b20e","family_id":"qf-variance-zero-support","seed":202610057076,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/1\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2}, with event family F=𝒫(S). The total function X:S→R has table {(0, 5), (1, 5), (2, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(3), 2/(3), 0/(3)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"Yes","explanation":"The attained values on positive-weight support are {5}. Constancy there gives the answer Yes. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0077","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83b08cec000ab0d4c697d346eb048ca6425e80d85b0cdf012bb17320d611e921","family_id":"qf-variance-zero-support","seed":202610057077,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"3\",\"1/2\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-7, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0078","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea8f0cd06b4834a611f6875ec704f29c4f999fda1a2a70505d950747cbb7a204","family_id":"qf-variance-zero-support","seed":202610057078,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"4\",\"1/2\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-8, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0079","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"da02e92c53f3576d05d05e2ea405781797e11bf64acad87ed788d176df5fbe0a","family_id":"qf-variance-zero-support","seed":202610057079,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"1/2\"],[\"5\",\"1/2\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-9, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0080","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2999537c78c9fa6e67d87aea225fc383b0a56ae66ad171e802cb1cf9b615e40","family_id":"qf-variance-zero-support","seed":202610057080,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"7/10\"],[\"6\",\"3/10\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-6, 0}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0081","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f24dfbeff8d00a2264669f275966b93253b58fb7a398cbb01d4d9aba9654fea6","family_id":"qf-variance-zero-support","seed":202610057081,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"2/3\"],[\"7\",\"1/3\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-11, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0082","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15a7e7b840342f1e562340ed4c5bd18270fdc4438fbc375c934ec2f2eb81e0fa","family_id":"qf-variance-zero-support","seed":202610057082,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"9/14\"],[\"8\",\"5/14\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-4, 4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0083","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5906079908c002ed36e4848c03d5fbbbfd909f0d55ed294d4d5a918610a7346","family_id":"qf-variance-zero-support","seed":202610057083,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/8\"],[\"9\",\"3/8\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-13, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0084","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"010cbb84d439a5d3b794196ac7fd0e6ae6b26989b1c6aa0bd41f80d04fc4cf1e","family_id":"qf-variance-zero-support","seed":202610057084,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"11/14\"],[\"10\",\"3/14\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-2, 8}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0085","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"252b0c99d6cc3902c777f5318cd5f6e57d6833e1e3e0771dd9f91abc824a4086","family_id":"qf-variance-zero-support","seed":202610057085,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"3/4\"],[\"11\",\"1/4\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-15, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0086","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d685ad39087566c89d721568d5824b9e2b4ef703e5ba65926caaf2b974d2dcc0","family_id":"qf-variance-zero-support","seed":202610057086,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"13/18\"],[\"12\",\"5/18\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {0, 12}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0087","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c91598325333e75885b68dac0c80d9d8968bad82a84bb7d007f51bed8f60c62","family_id":"qf-variance-zero-support","seed":202610057087,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"7/10\"],[\"13\",\"3/10\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-17, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0088","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eec7a16cac99b9fc3a174a3cc8a2a13fecc36f99dba52a18370954915d90e594","family_id":"qf-variance-zero-support","seed":202610057088,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"14\",\"1/6\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {2, 16}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0089","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"796162f975aff10df099c13508110610feb9db39438fe85fe0567114ccf6e57f","family_id":"qf-variance-zero-support","seed":202610057089,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"4/5\"],[\"15\",\"1/5\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-19, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0090","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca23bd566cd2eb78002a711be0fe1a9a12e54179e92ec528b99094fdac4b95b7","family_id":"qf-variance-zero-support","seed":202610057090,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"17/22\"],[\"16\",\"5/22\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {4, 20}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0091","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f2facf3848b98c07dae017532a4f616843d47bf288e1a537a4b0db08269c0f6","family_id":"qf-variance-zero-support","seed":202610057091,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"3/4\"],[\"17\",\"1/4\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-21, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0092","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f97286ef3cf43ebac49c601cb882bb59755ca0bf8230e43235dd403c4a5e3f0","family_id":"qf-variance-zero-support","seed":202610057092,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"19/22\"],[\"18\",\"3/22\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {6, 24}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0093","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a664289ab850e78a111faf0cd479bd0907008f93eb93e648e91aa699d74c2034","family_id":"qf-variance-zero-support","seed":202610057093,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"19\",\"1/6\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-23, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0094","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a214aa6c6213ab2c69536bed0af2ce833bdc1accc2143c6aa614914c7e312ae9","family_id":"qf-variance-zero-support","seed":202610057094,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"21/26\"],[\"20\",\"5/26\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {8, 28}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0095","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6ec8dff44e991586b55cd441ed5c51b2c55a016680bc1404fbbc35b00af984d","family_id":"qf-variance-zero-support","seed":202610057095,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"11/14\"],[\"21\",\"3/14\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-25, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0096","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe6df90ae2e3052a40e1059ad5d07ad3e66d31f8c65dc2dfc51b14f4e04d7876","family_id":"qf-variance-zero-support","seed":202610057096,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"23/26\"],[\"22\",\"3/26\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {10, 32}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0097","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"555e2fd42c97397cc7ed1e4af61f2161d004fac45544c7cc8ff070f4c75d6695","family_id":"qf-variance-zero-support","seed":202610057097,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"6/7\"],[\"23\",\"1/7\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-27, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0098","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dada6392a4f08745429ec617f441b8dbb0fab78dcd7946473a17de2ff42d6241","family_id":"qf-variance-zero-support","seed":202610057098,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"5/6\"],[\"24\",\"1/6\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {12, 36}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0099","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"85fabc989d7b201d8464cc7dcb3e2f1ef5855cb0927f52e2129e90fe07beab73","family_id":"qf-variance-zero-support","seed":202610057099,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"13/16\"],[\"25\",\"3/16\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {-29, -4}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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-20261004-advance-20261005-r07-variance-0100","batch_id":"batch-20261004-advance-20261005-r07-variance-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68e5ae1e951ebf761105a54e28ef9c705c2cc338597fc755a7fc0a1f28bbec82","family_id":"qf-variance-zero-support","seed":202610057100,"semantic_template":"qf-variance-zero-support/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"9/10\"],[\"26\",\"1/10\"]],\"operation\":\"zero_variance\"}"},"fields":{"title":"Zero Variance on Positive-Probability Support","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Is Var(X)=0 under this probability law? Constancy is required only at positive-weight outcomes.","answer":{"kind":"short_text","value":"No","explanation":"The attained values on positive-weight support are {14, 40}. Constancy there gives the answer No. Zero-weight outcomes may have different values without affecting variance; constancy on all of S is not required. This is probability-distribution variance, not sample variance with an n−1 denominator."},"skills":["Zero Variance on Positive-Probability Support"],"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."}}