{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0001","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c945653d1f368bc9632c386f9d325126d467cc73f92eccb87c859fd67b1d575","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144001,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[1,2]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=1/2. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(1/2, (2*pi)^(-1/2)*exp(-1/8), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (1/2, (2*pi)^(-1/2)*exp(-1/8), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0002","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a04bb96ea8a297633baccb78e9e336ab8112c38285e5440972fe74cac3d9e17","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144002,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-2,3]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-2/3. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-2/3, (2*pi)^(-1/2)*exp(-2/9), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-2/3, (2*pi)^(-1/2)*exp(-2/9), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0003","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95dfb843d3e44d0b221c2c9f44e7b7e46662b7786c275adaf90089a8457045c0","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144003,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[3,4]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=3/4. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(3/4, (2*pi)^(-1/2)*exp(-9/32), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (3/4, (2*pi)^(-1/2)*exp(-9/32), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0004","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2320b48efbe23f4955e78bcaa2e37eeb105d218708745b77158af46e1db19ab6","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144004,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-4,5]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-4/5. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-4/5, (2*pi)^(-1/2)*exp(-8/25), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-4/5, (2*pi)^(-1/2)*exp(-8/25), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0005","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c637519addbc50032272b43dd02e440d8142c27fabd49eda181a54f10ca8d9ca","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144005,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[5,6]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=5/6. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(5/6, (2*pi)^(-1/2)*exp(-25/72), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (5/6, (2*pi)^(-1/2)*exp(-25/72), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0006","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"966c5a179a58c8d4b59a6061f9e0baa84b9a0dcb38f5eefb5965a7b778e17438","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144006,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-6,7]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-6/7. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-6/7, (2*pi)^(-1/2)*exp(-18/49), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-6/7, (2*pi)^(-1/2)*exp(-18/49), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0007","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"667557adb242288f13fa9862b7a4cb4626b496bfc4b461b66f65aeb3e444d91b","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144007,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[7,8]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=7/8. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(7/8, (2*pi)^(-1/2)*exp(-49/128), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (7/8, (2*pi)^(-1/2)*exp(-49/128), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0008","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d0dd22ff68140bfa8a24a7fa2ed3639e0ab7ca9b7115c946c35de3f386fde1e","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144008,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-8,9]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-8/9. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-8/9, (2*pi)^(-1/2)*exp(-32/81), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-8/9, (2*pi)^(-1/2)*exp(-32/81), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0009","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c47826a810c2ed7a6691dc226bcc77188968a41c0257a994010806ec27a126c6","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144009,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[9,10]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=9/10. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(9/10, (2*pi)^(-1/2)*exp(-81/200), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (9/10, (2*pi)^(-1/2)*exp(-81/200), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0010","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7077b1a4b3bc80c07812f815e1265feb1c293f989d95d0991657bf05b06d27f7","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144010,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-10,11]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-10/11. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-10/11, (2*pi)^(-1/2)*exp(-50/121), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-10/11, (2*pi)^(-1/2)*exp(-50/121), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0011","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14dbcd71966cec8ab4d311b2f6ba832b0a8e2a9fb580d91c77de534d0b6288b3","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144011,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[11,12]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=11/12. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(11/12, (2*pi)^(-1/2)*exp(-121/288), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (11/12, (2*pi)^(-1/2)*exp(-121/288), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0012","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"202560d7c3223fc2736be4696395dbb68bdc90da40c315dcfa5943b25f49412a","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144012,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-12,13]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-12/13. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-12/13, (2*pi)^(-1/2)*exp(-72/169), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-12/13, (2*pi)^(-1/2)*exp(-72/169), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0013","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31b0b498b6608aacb9fd0db052605cfa40e450f03e863264552e675d80182e1d","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144013,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[13,14]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=13/14. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(13/14, (2*pi)^(-1/2)*exp(-169/392), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (13/14, (2*pi)^(-1/2)*exp(-169/392), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0014","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9ebcf3f5afdd2ea21a9183d26d1b8640be8c7e29216869669f0882a08d0d5fec","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144014,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-14,15]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-14/15. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-14/15, (2*pi)^(-1/2)*exp(-98/225), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-14/15, (2*pi)^(-1/2)*exp(-98/225), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0015","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"006b448ad1d037c1f7433854601e61077ee8c59c5c0197154bbec68ecf0c604a","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144015,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[15,16]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=15/16. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(15/16, (2*pi)^(-1/2)*exp(-225/512), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (15/16, (2*pi)^(-1/2)*exp(-225/512), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0016","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5fb922bf2aeed6383490232dc02d646df7fe76288dd58c0fe523842a9e97269c","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144016,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-16,17]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-16/17. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-16/17, (2*pi)^(-1/2)*exp(-128/289), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-16/17, (2*pi)^(-1/2)*exp(-128/289), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0017","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91c7d059683beabf3cac22ed4f47f84b8b54a8834e3919cb985651999674f77b","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144017,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[17,18]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=17/18. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(17/18, (2*pi)^(-1/2)*exp(-289/648), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (17/18, (2*pi)^(-1/2)*exp(-289/648), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0018","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb5ad9b1f1b9a849a96f985b377f2fd6965e7bb1556e80348a853b1275cc78ca","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144018,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-18,19]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-18/19. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-18/19, (2*pi)^(-1/2)*exp(-162/361), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-18/19, (2*pi)^(-1/2)*exp(-162/361), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0019","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0dde3435e5874a64e17dae76585fb306dc25cd6a80bb85fb675ab67c534ea83d","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144019,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[19,20]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=19/20. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(19/20, (2*pi)^(-1/2)*exp(-361/800), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (19/20, (2*pi)^(-1/2)*exp(-361/800), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0020","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"178e1351dfdc297683af79fd13568d286f62fb6fa13d0e69be9efca7ca5b63ab","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144020,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-20,21]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-20/21. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-20/21, (2*pi)^(-1/2)*exp(-200/441), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-20/21, (2*pi)^(-1/2)*exp(-200/441), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0021","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8bbf93b1a0ceed1588bdabd959826b9d8a56eb8c0048ac293f9633da9761aba","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144021,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[21,22]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=21/22. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(21/22, (2*pi)^(-1/2)*exp(-441/968), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (21/22, (2*pi)^(-1/2)*exp(-441/968), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0022","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cfb245e84cb8302baf622098227dc84378ad699b862b63dd5c4f83e4579c627","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144022,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-22,23]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-22/23. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-22/23, (2*pi)^(-1/2)*exp(-242/529), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-22/23, (2*pi)^(-1/2)*exp(-242/529), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0023","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37c397d9e3d6e8f7fdbed7c98cf6c6b7cac577771700345d4025b37c5898ca6d","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144023,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[23,24]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=23/24. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(23/24, (2*pi)^(-1/2)*exp(-529/1152), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (23/24, (2*pi)^(-1/2)*exp(-529/1152), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0024","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cdfbfeb7209962f646506d7dd73ee8b1eb7255b64bf63b1b07a0e63c6cbdc8e","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144024,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[-24,25]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=-24/25. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(-24/25, (2*pi)^(-1/2)*exp(-288/625), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (-24/25, (2*pi)^(-1/2)*exp(-288/625), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0025","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63c614fc471a3093efebee1a3cff175e3cfbcf9fe49d90c1188ccac9f6fa0d1d","family_id":"qf-mod45-normaldist-symbolic-density","seed":202611144025,"semantic_template":"qf-mod45-normaldist-symbolic-density/{\"x\":[25,26]}"},"fields":{"title":"The Standard-Normal Density Value Remains an Exact Symbolic Expression","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use exact x=25/26. Give (x, symbolic_density, density_law, normalization_scope).","answer":{"kind":"expression","value":"(25/26, (2*pi)^(-1/2)*exp(-625/1352), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)","explanation":"Compute the rational exponent −x^2/2 and retain the symbolic normalization factor and exponential. A symbolic finite argument is not a proof of Gaussian normalization, a point mass or an elementary CDF evaluation. The exact result, retaining the requested order, is (25/26, (2*pi)^(-1/2)*exp(-625/1352), phi(x)=(2*pi)^(-1/2)*exp(-x^2/2), normalization_is_a_named_Gaussian_integral_fact_not_a_finite_evaluation)."},"skills":["The Standard-Normal Density Value Remains an Exact Symbolic Expression"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0026","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30acd45ef82336aedbcf1367467358e0cd8fbe142b45492dab76a191bacbd7af","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144026,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[1,2]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=1/2 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(1/2, -1/2, (2*pi)^(-1/2)*exp(-1/8), (2*pi)^(-1/2)*exp(-1/8), Yes, phi(x)/phi(0)=exp(-1/8), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (1/2, -1/2, (2*pi)^(-1/2)*exp(-1/8), (2*pi)^(-1/2)*exp(-1/8), Yes, phi(x)/phi(0)=exp(-1/8), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0027","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cc0a67c29060524d77879054ac7ef6add868186cc67c4f09b02deef0988c9d0","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144027,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-2,3]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-2/3 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-2/3, 2/3, (2*pi)^(-1/2)*exp(-2/9), (2*pi)^(-1/2)*exp(-2/9), Yes, phi(x)/phi(0)=exp(-2/9), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-2/3, 2/3, (2*pi)^(-1/2)*exp(-2/9), (2*pi)^(-1/2)*exp(-2/9), Yes, phi(x)/phi(0)=exp(-2/9), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0028","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2914f0fcc6e7da4f4d881f1461d3d34bcd7b4c4a2fbda04672cd63c438418ce5","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144028,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[3,4]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=3/4 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(3/4, -3/4, (2*pi)^(-1/2)*exp(-9/32), (2*pi)^(-1/2)*exp(-9/32), Yes, phi(x)/phi(0)=exp(-9/32), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (3/4, -3/4, (2*pi)^(-1/2)*exp(-9/32), (2*pi)^(-1/2)*exp(-9/32), Yes, phi(x)/phi(0)=exp(-9/32), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0029","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"013ec79c9cf93784967b50bcf970b3fdcf8802d532dcdbbab09291e87ca0c696","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144029,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-4,5]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-4/5 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-4/5, 4/5, (2*pi)^(-1/2)*exp(-8/25), (2*pi)^(-1/2)*exp(-8/25), Yes, phi(x)/phi(0)=exp(-8/25), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-4/5, 4/5, (2*pi)^(-1/2)*exp(-8/25), (2*pi)^(-1/2)*exp(-8/25), Yes, phi(x)/phi(0)=exp(-8/25), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0030","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51ef6459a7af9358d4c48b94fe53a02252a3e1278c86ed6211c23aeed9eb0b15","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144030,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[5,6]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=5/6 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(5/6, -5/6, (2*pi)^(-1/2)*exp(-25/72), (2*pi)^(-1/2)*exp(-25/72), Yes, phi(x)/phi(0)=exp(-25/72), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (5/6, -5/6, (2*pi)^(-1/2)*exp(-25/72), (2*pi)^(-1/2)*exp(-25/72), Yes, phi(x)/phi(0)=exp(-25/72), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0031","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ee0babf0aae667e884a303da977d9609fd9ea443d9fbd97133b984164f32aef","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144031,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-6,7]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-6/7 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-6/7, 6/7, (2*pi)^(-1/2)*exp(-18/49), (2*pi)^(-1/2)*exp(-18/49), Yes, phi(x)/phi(0)=exp(-18/49), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-6/7, 6/7, (2*pi)^(-1/2)*exp(-18/49), (2*pi)^(-1/2)*exp(-18/49), Yes, phi(x)/phi(0)=exp(-18/49), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0032","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"171b0fae572f67fa3a51c20cc689e89788bbfd728747cd69e47066b28c2abef2","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144032,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[7,8]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=7/8 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(7/8, -7/8, (2*pi)^(-1/2)*exp(-49/128), (2*pi)^(-1/2)*exp(-49/128), Yes, phi(x)/phi(0)=exp(-49/128), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (7/8, -7/8, (2*pi)^(-1/2)*exp(-49/128), (2*pi)^(-1/2)*exp(-49/128), Yes, phi(x)/phi(0)=exp(-49/128), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0033","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8b2171d2e92eca07fc5a4a8ff887247efea3307df83ed48f2d36367ceb1a757","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144033,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-8,9]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-8/9 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-8/9, 8/9, (2*pi)^(-1/2)*exp(-32/81), (2*pi)^(-1/2)*exp(-32/81), Yes, phi(x)/phi(0)=exp(-32/81), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-8/9, 8/9, (2*pi)^(-1/2)*exp(-32/81), (2*pi)^(-1/2)*exp(-32/81), Yes, phi(x)/phi(0)=exp(-32/81), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0034","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc823f699bf683d549f7e86433a9622ea672a0c45c06ddfd6d7250f58345d3f7","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144034,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[9,10]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=9/10 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(9/10, -9/10, (2*pi)^(-1/2)*exp(-81/200), (2*pi)^(-1/2)*exp(-81/200), Yes, phi(x)/phi(0)=exp(-81/200), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (9/10, -9/10, (2*pi)^(-1/2)*exp(-81/200), (2*pi)^(-1/2)*exp(-81/200), Yes, phi(x)/phi(0)=exp(-81/200), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0035","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e3c74e9c4ff14e4e939b028a8b007b51794b6e1da5ed2416e0bb6b97055cd426","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144035,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-10,11]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-10/11 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-10/11, 10/11, (2*pi)^(-1/2)*exp(-50/121), (2*pi)^(-1/2)*exp(-50/121), Yes, phi(x)/phi(0)=exp(-50/121), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-10/11, 10/11, (2*pi)^(-1/2)*exp(-50/121), (2*pi)^(-1/2)*exp(-50/121), Yes, phi(x)/phi(0)=exp(-50/121), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0036","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14e061efba629aa65c42a9b109e901d47314c7b7b57685ebe47342a17f64fd2d","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144036,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[11,12]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=11/12 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(11/12, -11/12, (2*pi)^(-1/2)*exp(-121/288), (2*pi)^(-1/2)*exp(-121/288), Yes, phi(x)/phi(0)=exp(-121/288), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (11/12, -11/12, (2*pi)^(-1/2)*exp(-121/288), (2*pi)^(-1/2)*exp(-121/288), Yes, phi(x)/phi(0)=exp(-121/288), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0037","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e3a4489b6b16923858716e4fe1f9861434de79a795388d071f7890a43f0808f","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144037,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-12,13]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-12/13 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-12/13, 12/13, (2*pi)^(-1/2)*exp(-72/169), (2*pi)^(-1/2)*exp(-72/169), Yes, phi(x)/phi(0)=exp(-72/169), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-12/13, 12/13, (2*pi)^(-1/2)*exp(-72/169), (2*pi)^(-1/2)*exp(-72/169), Yes, phi(x)/phi(0)=exp(-72/169), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0038","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9b0db8ec51b25581dc59ba6ddce9c9b6190e495be9aa1261070b742bd04b2df","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144038,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[13,14]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=13/14 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(13/14, -13/14, (2*pi)^(-1/2)*exp(-169/392), (2*pi)^(-1/2)*exp(-169/392), Yes, phi(x)/phi(0)=exp(-169/392), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (13/14, -13/14, (2*pi)^(-1/2)*exp(-169/392), (2*pi)^(-1/2)*exp(-169/392), Yes, phi(x)/phi(0)=exp(-169/392), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0039","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f00eddd7a325efebf2c5a1fddb9cb3c59028cc68612f6e929586ccb6b9abb761","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144039,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-14,15]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-14/15 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-14/15, 14/15, (2*pi)^(-1/2)*exp(-98/225), (2*pi)^(-1/2)*exp(-98/225), Yes, phi(x)/phi(0)=exp(-98/225), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-14/15, 14/15, (2*pi)^(-1/2)*exp(-98/225), (2*pi)^(-1/2)*exp(-98/225), Yes, phi(x)/phi(0)=exp(-98/225), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0040","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f796f2e20095a3828819632d6be4d049887e9e3f673d9ac89af0d140c3073b83","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144040,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[15,16]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=15/16 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(15/16, -15/16, (2*pi)^(-1/2)*exp(-225/512), (2*pi)^(-1/2)*exp(-225/512), Yes, phi(x)/phi(0)=exp(-225/512), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (15/16, -15/16, (2*pi)^(-1/2)*exp(-225/512), (2*pi)^(-1/2)*exp(-225/512), Yes, phi(x)/phi(0)=exp(-225/512), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0041","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca6c843eccf4b86d0f6cc0592d2e3b82e13cfe43ec60eb411c9ff751972c43da","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144041,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-16,17]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-16/17 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-16/17, 16/17, (2*pi)^(-1/2)*exp(-128/289), (2*pi)^(-1/2)*exp(-128/289), Yes, phi(x)/phi(0)=exp(-128/289), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-16/17, 16/17, (2*pi)^(-1/2)*exp(-128/289), (2*pi)^(-1/2)*exp(-128/289), Yes, phi(x)/phi(0)=exp(-128/289), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0042","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2af8f2f9854fb9a09024b301120d3dc2d9dd678abb2f0e8c3c16674b444a62a9","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144042,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[17,18]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=17/18 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(17/18, -17/18, (2*pi)^(-1/2)*exp(-289/648), (2*pi)^(-1/2)*exp(-289/648), Yes, phi(x)/phi(0)=exp(-289/648), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (17/18, -17/18, (2*pi)^(-1/2)*exp(-289/648), (2*pi)^(-1/2)*exp(-289/648), Yes, phi(x)/phi(0)=exp(-289/648), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0043","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"781719143d0e9530f8215ca7e083e5bfdbe0b4376cc3f1d781a68fecacc99d17","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144043,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-18,19]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-18/19 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-18/19, 18/19, (2*pi)^(-1/2)*exp(-162/361), (2*pi)^(-1/2)*exp(-162/361), Yes, phi(x)/phi(0)=exp(-162/361), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-18/19, 18/19, (2*pi)^(-1/2)*exp(-162/361), (2*pi)^(-1/2)*exp(-162/361), Yes, phi(x)/phi(0)=exp(-162/361), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0044","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6995e3c18e3bb55ec061a7175c228b79c741beea4f5216f66fbec20ff3f8cf54","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144044,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[19,20]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=19/20 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(19/20, -19/20, (2*pi)^(-1/2)*exp(-361/800), (2*pi)^(-1/2)*exp(-361/800), Yes, phi(x)/phi(0)=exp(-361/800), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (19/20, -19/20, (2*pi)^(-1/2)*exp(-361/800), (2*pi)^(-1/2)*exp(-361/800), Yes, phi(x)/phi(0)=exp(-361/800), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0045","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a60796da7a77327990453ba5a7cc57c6b8a96e305d22fcf6e5464aa84045731","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144045,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-20,21]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-20/21 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-20/21, 20/21, (2*pi)^(-1/2)*exp(-200/441), (2*pi)^(-1/2)*exp(-200/441), Yes, phi(x)/phi(0)=exp(-200/441), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-20/21, 20/21, (2*pi)^(-1/2)*exp(-200/441), (2*pi)^(-1/2)*exp(-200/441), Yes, phi(x)/phi(0)=exp(-200/441), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0046","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a0d05bce440607e05883e1c95c6c4e36e6dce7bee8a02b98ecbdb7878d54adb","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144046,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[21,22]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=21/22 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(21/22, -21/22, (2*pi)^(-1/2)*exp(-441/968), (2*pi)^(-1/2)*exp(-441/968), Yes, phi(x)/phi(0)=exp(-441/968), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (21/22, -21/22, (2*pi)^(-1/2)*exp(-441/968), (2*pi)^(-1/2)*exp(-441/968), Yes, phi(x)/phi(0)=exp(-441/968), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0047","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49fda9238db1a435dd742ece41fc6562a7fb289d89854d96e96294cc325eb37c","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144047,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-22,23]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-22/23 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-22/23, 22/23, (2*pi)^(-1/2)*exp(-242/529), (2*pi)^(-1/2)*exp(-242/529), Yes, phi(x)/phi(0)=exp(-242/529), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-22/23, 22/23, (2*pi)^(-1/2)*exp(-242/529), (2*pi)^(-1/2)*exp(-242/529), Yes, phi(x)/phi(0)=exp(-242/529), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0048","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06f19037214d174c1bb35643b4609438cd908a0a703351c3cd7cd5c02f9ab473","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144048,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[23,24]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=23/24 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(23/24, -23/24, (2*pi)^(-1/2)*exp(-529/1152), (2*pi)^(-1/2)*exp(-529/1152), Yes, phi(x)/phi(0)=exp(-529/1152), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (23/24, -23/24, (2*pi)^(-1/2)*exp(-529/1152), (2*pi)^(-1/2)*exp(-529/1152), Yes, phi(x)/phi(0)=exp(-529/1152), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0049","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"769b95df67a6c6e3fa0f224b53d498db38fb2d2053c43c4a873ccdeb27ba78a2","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144049,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[-24,25]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-24/25 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(-24/25, 24/25, (2*pi)^(-1/2)*exp(-288/625), (2*pi)^(-1/2)*exp(-288/625), Yes, phi(x)/phi(0)=exp(-288/625), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (-24/25, 24/25, (2*pi)^(-1/2)*exp(-288/625), (2*pi)^(-1/2)*exp(-288/625), Yes, phi(x)/phi(0)=exp(-288/625), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0050","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5aea603fbdea3e232fd933e479f2f5919413ed803f0c7b3b69e3a27e5c60b6a0","family_id":"qf-mod45-normaldist-evenness-ratio","seed":202611144050,"semantic_template":"qf-mod45-normaldist-evenness-ratio/{\"x\":[25,26]}"},"fields":{"title":"Evenness and the Density Ratio Follow from the Squared Argument","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=25/26 and −x. Give (x, negative_x, density_at_x, density_at_negative_x, even, density_ratio, certificate).","answer":{"kind":"expression","value":"(25/26, -25/26, (2*pi)^(-1/2)*exp(-625/1352), (2*pi)^(-1/2)*exp(-625/1352), Yes, phi(x)/phi(0)=exp(-625/1352), evenness_follows_from_x_squared)","explanation":"The density expressions agree because the exponent uses x^2. The normalization factor cancels in phi(x)/phi(0)=exp(−x^2/2). Keep these symbolic density identities separate from normalization and CDF values. The exact result, retaining the requested order, is (25/26, -25/26, (2*pi)^(-1/2)*exp(-625/1352), (2*pi)^(-1/2)*exp(-625/1352), Yes, phi(x)/phi(0)=exp(-625/1352), evenness_follows_from_x_squared)."},"skills":["Evenness and the Density Ratio Follow from the Squared Argument"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0051","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea5976c6a95577740887761d90ba52b931d922aa7530fedeed0c5e8810418ee3","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144051,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-12,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-12. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-12, Phi(12)=1-Phi(-12), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-12, Phi(12)=1-Phi(-12), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0052","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3545b708e2dfd86998de168295cf03ae1f90d9c05f3f76a76437010c0d092eb4","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144052,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-11,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-11. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-11, Phi(11)=1-Phi(-11), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-11, Phi(11)=1-Phi(-11), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0053","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80c521645d8d3e192ca5a204fdb20484315eccee214935742c4cbfd593babd9c","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144053,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-10,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-10. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-10, Phi(10)=1-Phi(-10), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-10, Phi(10)=1-Phi(-10), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0054","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4767f9d9e05652914f95ff4f233f76f8433b0c06d493bb4c7c38cfe65ff7eda","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144054,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-9,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-9. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-9, Phi(9)=1-Phi(-9), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-9, Phi(9)=1-Phi(-9), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0055","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f794700d737906affed44594bc576551faa343dc0996e9699ece9b5c2a5f5b17","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144055,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-8,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-8. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-8, Phi(8)=1-Phi(-8), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-8, Phi(8)=1-Phi(-8), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0056","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd644f4619b548bffd2591df1f47737078a4940ef5671d99327bde72215c86dc","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144056,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-7,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-7. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-7, Phi(7)=1-Phi(-7), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-7, Phi(7)=1-Phi(-7), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0057","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b25cdea6613a968ed698c8496ec59acdbeeb190c2c75d4cc46c28b77336b459","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144057,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-6,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-6. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-6, Phi(6)=1-Phi(-6), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-6, Phi(6)=1-Phi(-6), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0058","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f65c052c2938e62ca67ce04729f84ad286784d8299711f82d744db24601ca902","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144058,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-5,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-5. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-5, Phi(5)=1-Phi(-5), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-5, Phi(5)=1-Phi(-5), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0059","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"385d68205996a4b41d429d01b7e303e3cc2a4134eb6817ebf4d85a1dd45b00ee","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144059,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-4,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-4. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-4, Phi(4)=1-Phi(-4), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-4, Phi(4)=1-Phi(-4), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0060","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f883747ef989df33fb7c480e9452c141f35adc2063dccd9e3848ed1249826596","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144060,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-3,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-3. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-3, Phi(3)=1-Phi(-3), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-3, Phi(3)=1-Phi(-3), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0061","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"466d5028673672c6aa4475f6a88298a161d465d4601596982eea89eaeacca6ae","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144061,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-2,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-2. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-2, Phi(2)=1-Phi(-2), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-2, Phi(2)=1-Phi(-2), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0062","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2889ae29860de2576cb1ea0e06cd98b3ae858444ab828eabee0a904c604ae422","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144062,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[-1,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=-1. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(-1, Phi(1)=1-Phi(-1), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (-1, Phi(1)=1-Phi(-1), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0063","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad82f728864f609c3824d0de2c7e53ef5951e10adf6a99f33a8ab19ce9dae4c1","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144063,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[0,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=0. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(0, Phi(0)=1-Phi(0), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (0, Phi(0)=1-Phi(0), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0064","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"06f7a419f73707af54c42b3d01abe5c8879c07aef7a2d0e819dc1ff723016f38","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144064,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[1,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=1. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(1, Phi(-1)=1-Phi(1), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (1, Phi(-1)=1-Phi(1), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0065","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21db80cdfd206720d4672841174e3c1475e0039f8b92622abba6bef69c9de023","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144065,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[2,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=2. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(2, Phi(-2)=1-Phi(2), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (2, Phi(-2)=1-Phi(2), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0066","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a59f72dec3ddf886922ee98e6cd745d12d0967c25f9d67ad9a9af846fb59496d","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144066,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[3,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=3. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(3, Phi(-3)=1-Phi(3), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (3, Phi(-3)=1-Phi(3), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0067","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16336499c719158b85e852ad6f935d3d79a558eefa784a397718d24bccf36270","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144067,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[4,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=4. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(4, Phi(-4)=1-Phi(4), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (4, Phi(-4)=1-Phi(4), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0068","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"929afdaff3421fe37492a3e5d08cd5f95969ee1eb5d9a7532cb2ccf909cef208","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144068,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[5,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=5. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(5, Phi(-5)=1-Phi(5), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (5, Phi(-5)=1-Phi(5), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0069","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ae25d041ccbf0f9cfab11512973580c3d3dbd24cb8e30074e183f28d612c404","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144069,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[6,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=6. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(6, Phi(-6)=1-Phi(6), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (6, Phi(-6)=1-Phi(6), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0070","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e4ca7ecc3113482caf1d65327bdce1d4a122eac47e02d401a10fc0b5e78189f","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144070,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[7,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=7. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(7, Phi(-7)=1-Phi(7), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (7, Phi(-7)=1-Phi(7), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0071","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68d8ad51369154f2e5cf454b021e6dbfc7e651dc366f9ca92f66a2554f65b879","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144071,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[8,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=8. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(8, Phi(-8)=1-Phi(8), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (8, Phi(-8)=1-Phi(8), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0072","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e256d2d9f2bccd3fec68bd09728192c9115974bc8fd6a571c2b573a9eb81773","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144072,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[9,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=9. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(9, Phi(-9)=1-Phi(9), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (9, Phi(-9)=1-Phi(9), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0073","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b08297d2adf72d5b1885d8d63083699237c7ff77a99f2fb491afba8aade909a","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144073,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[10,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=10. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(10, Phi(-10)=1-Phi(10), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (10, Phi(-10)=1-Phi(10), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0074","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7394c6dcb812aa1368d84f8047fd7f6e0be581586fd7feb5e4d1a1abdb5b0aa","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144074,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[11,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=11. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(11, Phi(-11)=1-Phi(11), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (11, Phi(-11)=1-Phi(11), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0075","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bee7d3bd08a97d233aa5a60e6746d1d36b5299943da61d3573e9ed4ca8d5cda0","family_id":"qf-mod45-normaldist-cdf-symmetry","seed":202611144075,"semantic_template":"qf-mod45-normaldist-cdf-symmetry/{\"x\":[12,1]}"},"fields":{"title":"Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Use x=12. Give (x, CDF_symmetry_identity, CDF_at_zero, declared_symmetry_certificate, boundary).","answer":{"kind":"expression","value":"(12, Phi(-12)=1-Phi(12), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)","explanation":"With the declared normalized even law, Phi(−x)=1−Phi(x) and Phi(0)=1/2. Do not interpret this as computing a general nonzero elementary CDF value or as proving normalization from finitely many evaluations. The exact result, retaining the requested order, is (12, Phi(-12)=1-Phi(12), Phi(0)=1/2, Yes, declared_normalization_plus_even_density_implies_CDF_symmetry_but_no_nonzero_elementary_CDF_value)."},"skills":["Declared Normalization and Even Density Give CDF Symmetry, Not Arbitrary Exact Values"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0076","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f76fd227df122a3f9f3e1069840f8ce4bec8a16ccc093049d1295b595289f59b","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144076,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":10,\"window\":[1,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=1>0, panel count=10. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(1, 10, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (1, 10, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0077","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b5e98a1e5f3100b4adebf3cc1d9570d4aa5f00c2ea467a46b76cb8485eacf5a","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144077,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":11,\"window\":[2,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=2>0, panel count=11. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(2, 11, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (2, 11, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0078","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbf230d7d4351cbb6f0c2e97c6208fb2bdfdc44619825b87d663327822aeac7f","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144078,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":12,\"window\":[3,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=3>0, panel count=12. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(3, 12, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (3, 12, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0079","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d8bf75af969ec6833f86652ce13706442fef461ba7ebc9428d3644ce446d582","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144079,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":13,\"window\":[4,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=4>0, panel count=13. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(4, 13, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (4, 13, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0080","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27e9b854a9cb3f49bcae4890995b797fb1bfdb331411bda01c7bbdf42c5ed165","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144080,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":14,\"window\":[5,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=5>0, panel count=14. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(5, 14, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (5, 14, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0081","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db56890ae449030732638995595b9e5959e8f4f14b76d8a00ec7dc3f7ebc8364","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144081,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":15,\"window\":[6,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=6>0, panel count=15. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(6, 15, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (6, 15, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0082","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cb31a53333c3b61a09e8d485e67c8f1bae49c2ff50d939538bacf1584b6d64e9","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144082,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":16,\"window\":[7,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=7>0, panel count=16. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(7, 16, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (7, 16, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0083","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c562db832eff11aeca45adb61b688652c104333508dbbc1a0323037b4bb0d80","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144083,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":17,\"window\":[8,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=8>0, panel count=17. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(8, 17, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (8, 17, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0084","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"08e288ed888153dcdaa7dde001fdf5900aa25d596d1eed88424d5703357c6690","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144084,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":18,\"window\":[9,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=9>0, panel count=18. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(9, 18, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (9, 18, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0085","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e62bfb610fb57bb167cb5f57f30ac5f119baf310098ba5ace2dcf9694d859be","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144085,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":19,\"window\":[10,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=10>0, panel count=19. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(10, 19, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (10, 19, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0086","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6edcae74beea9303d78b9fa3300d230fe958164f0608d411cd810923ccceeb02","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144086,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":20,\"window\":[11,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=11>0, panel count=20. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(11, 20, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (11, 20, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0087","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93e09687cd7bec9974460e1bfb166a1b5e364596e0f4ed62ad0f6f1887a328d4","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144087,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":21,\"window\":[12,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=12>0, panel count=21. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(12, 21, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (12, 21, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0088","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b06715ebeb2c0500c967b36486cf86b85e959e9709b222340a61514e1d34f01a","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144088,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":22,\"window\":[13,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=13>0, panel count=22. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(13, 22, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (13, 22, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0089","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a153f3997f3de59717f5c8950946f06e554b9b1f13935b5c9a34a58ebd0efb86","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144089,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":23,\"window\":[14,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=14>0, panel count=23. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(14, 23, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (14, 23, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0090","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3eac9d5afecd1ef4ef2067d3e9d5be09204f32dfceac1b4b81558726a2353916","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144090,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":24,\"window\":[15,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=15>0, panel count=24. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(15, 24, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (15, 24, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0091","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4de0eb9f86427c7a737bcda092f5aae635a73249887c783183c32af12b7fcab","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144091,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":25,\"window\":[16,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=16>0, panel count=25. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(16, 25, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (16, 25, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0092","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0da7ded6f862ad86fabc960b43fb336c8a9803ee85fd7fe11caa7b9bfaa6a413","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144092,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":26,\"window\":[17,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=17>0, panel count=26. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(17, 26, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (17, 26, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0093","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11ec24927a222a964e9c2fe6d50789d936378f3f68a3e50812772f2ebddc1826","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144093,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":27,\"window\":[18,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=18>0, panel count=27. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(18, 27, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (18, 27, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0094","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e880339730473ac1e50a62db233106d70d873956ad6e02819a49350e5db49aa","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144094,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":28,\"window\":[19,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=19>0, panel count=28. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(19, 28, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (19, 28, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0095","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"97003c6b367e6a3199fbf45696b1805491f60254654b2e760655028e0059c28f","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144095,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":29,\"window\":[20,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=20>0, panel count=29. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(20, 29, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (20, 29, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0096","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a3ef2ec777933fa56684824ee6dde621829fca11159d84c37fa4a9298f43323","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144096,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":30,\"window\":[21,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=21>0, panel count=30. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(21, 30, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (21, 30, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0097","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d14c0433cd8fb1fc68cc99c4539fc9d0967a3e9953b74e1e10e687c4d6e2fb3","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144097,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":31,\"window\":[22,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=22>0, panel count=31. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(22, 31, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (22, 31, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0098","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"012a27036785cb9e715224b99afb6aa5afc83088317994f1d6f692985662117c","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144098,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":32,\"window\":[23,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=23>0, panel count=32. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(23, 32, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (23, 32, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0099","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c3313c7214a92db8e0808ea484b1a07c4730941f98b600a3d6745e0a51825c7","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144099,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":33,\"window\":[24,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=24>0, panel count=33. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(24, 33, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (24, 33, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261114-r04-standard-normal-distribution-0100","batch_id":"batch-20261015-advance-20261114-r04-standard-normal-distribution-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75e50d15781592843e8a354f7ef97bc03f4e4e88eeaa364003fce0907929af19","family_id":"qf-mod45-normaldist-normalization-moment-boundary","seed":202611144100,"semantic_template":"qf-mod45-normaldist-normalization-moment-boundary/{\"panels\":34,\"window\":[25,1]}"},"fields":{"title":"Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments","stem":"Use the declared standard-normal law with Gaussian density phi(x)=(2*pi)^(-1/2)*exp(-x^2/2) and its continuous CDF Phi. Gaussian normalization, mean0 and variance1 remain explicitly named theorem facts, not results of finite quadrature in this slice. Declared finite window=25>0, panel count=34. Give (window, panels, finite_method_scope, finite_method_proof_claim, normalization_named_fact, mean_named_fact, variance_named_fact, boundary).","answer":{"kind":"expression","value":"(25, 34, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)","explanation":"This boundary records finite-method scope only; it does not even supply a quadrature result. Retain the No proof claim separately from named normalization/mean/variance facts. Those facts need their own theorem-level derivations, not a larger finite window or panel count. The exact result, retaining the requested order, is (25, 34, finite_quadrature_only, No, named_fact:integral_R_phi=1, named_fact:E[Z]=0, named_fact:Var(Z)=1, finite_window_or_panel_count_does_not_prove_normalization_or_moments)."},"skills":["Finite Windows and Panel Counts Do Not Prove Gaussian Normalization or Moments"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}