{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-0079",
  "pilot_number": "R0053-27204",
  "number": 27204,
  "family_id": "qf-mod49-clt-application-boundary",
  "title": {
    "zh-Hant": "第 27204 題：有限變異數IID中央極限定理",
    "en": "Question 27204: Declared Theorem Application, Failed Hypotheses and Finite Evidence Have Different Roles"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-0079",
    "batch_id": "batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-central-limit-theorem-finite-variance-iid"
    ],
    "title_zh": "有限變異數IID中央極限定理",
    "stem_zh": "明示無限IID、共同mu、0<sigma^2<infinity與sigma*sqrt(n)標準化責任。操作=clt-application-boundary；資料={\"mode\":\"wrong_scaling\",\"model\":{\"values\":[1,2],\"probabilities\":[[1,2],[1,2]],\"process_certificate\":\"infinite_iid\",\"n\":5}}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(wrong_scaling, 3/2, 1/4, 5, denominator=n*sigma, required=sigma*sqrt(n), 否)",
      "explanation_zh": "Wave6第十二層核對=(wrong_scaling, 3/2, 1/4, 5, denominator=n*sigma, required=sigma*sqrt(n), 否)；依完整明示資料獨立重算；finite moments與binomial lattice完整重算但不作normal approximation或limit evidence；non-IID為不適用，zero variance、wrong scaling與finite histogram不得冒充CLT。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 1,
      "skills": [
        "有限變異數IID中央極限定理"
      ]
    },
    "variant": {
      "family_id": "qf-mod49-clt-application-boundary",
      "seed": 202611181079,
      "parameters": {
        "mode": "wrong_scaling",
        "model": {
          "values": [
            1,
            2
          ],
          "probabilities": [
            [
              1,
              2
            ],
            [
              1,
              2
            ]
          ],
          "process_certificate": "infinite_iid",
          "n": 5
        },
        "domain": "Exact_finite_IID_limit_theorem_boundary",
        "operation": "clt-application-boundary",
        "semantic_template": "clt-application-boundary|{mode:\"wrong_scaling\",model:{law:[[1,[1,2]],[2,[1,2]]],n:5,process_certificate:\"infinite_iid\"}}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須主端重算；不構成一般limit theorem、coverage、number-theory或language theorem證明、人工審定、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-15",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-0079",
    "batch_id": "batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "461606279384502fa06d8dbcf33ceca9c81d9d457f61207be7051f8ed267a9e8",
      "family_id": "qf-mod49-clt-application-boundary",
      "seed": 202611181079,
      "semantic_template": "clt-application-boundary|{mode:\"wrong_scaling\",model:{law:[[1,[1,2]],[2,[1,2]]],n:5,process_certificate:\"infinite_iid\"}}"
    },
    "fields": {
      "title": "Declared Theorem Application, Failed Hypotheses and Finite Evidence Have Different Roles",
      "stem": "Use mode=wrong_scaling and complete model={\"values\":[1,2],\"probabilities\":[[1,2],[1,2]],\"process_certificate\":\"infinite_iid\",\"n\":5}. Give (mode, mean, variance, n, proposed_denominator, required_denominator, proposed_scaling_claim).",
      "answer": {
        "kind": "expression",
        "value": "(wrong_scaling, 3/2, 1/4, 5, denominator=n*sigma, required=sigma*sqrt(n), No)",
        "explanation": "symbolic_application invokes the named CLT conditionally on the declared infinite IID law and positive variance; it is not a proof performed here. Different marginals exclude this IID theorem only. Zero variance makes this normalization undefined. n*sigma is the wrong scaling. One finite histogram does not establish a distributional limit. Keep the five branch-specific field layouts. The exact result, retaining the requested order, is (wrong_scaling, 3/2, 1/4, 5, denominator=n*sigma, required=sigma*sqrt(n), No)."
      },
      "skills": [
        "Declared Theorem Application, Failed Hypotheses and Finite Evidence Have Different Roles"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "明示無限IID、共同mu、0<sigma^2<infinity與sigma*sqrt(n)標準化責任。操作=clt-application-boundary；資料={\"mode\":\"wrong_scaling\",\"model\":{\"values\":[1,2],\"probabilities\":[[1,2],[1,2]],\"process_certificate\":\"infinite_iid\",\"n\":5}}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use mode=wrong_scaling and complete model={\"values\":[1,2],\"probabilities\":[[1,2],[1,2]],\"process_certificate\":\"infinite_iid\",\"n\":5}. Give (mode, mean, variance, n, proposed_denominator, required_denominator, proposed_scaling_claim)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "source roles = [clt-application-boundary]",
      "after": "derived = [[mode, mean, variance, n, proposed_denominator, required_denominator, proposed_scaling_claim]]",
      "relation": "implies",
      "rule_id": "source_ordered_fields_without_JS",
      "conditions": [],
      "explanation": {
        "zh-Hant": "此頁直接列完整來源欄序與mode特定角色，不靠JavaScript或由答案外形猜介面；No、undefined、infinity各保原意。",
        "en": "List the complete source field order and mode-specific roles without JavaScript or answer-shape inference;No,undefined and infinity retain their distinct meanings."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "source roles = [infinite_iid]",
      "after": "derived = [ASSUME infinite sequence is IID with this common marginal]",
      "relation": "implies",
      "rule_id": "CLT_infinite_iid_is_EXACT_GIVEN_premise_not_constructed_process",
      "conditions": [],
      "explanation": {
        "zh-Hant": "infinite_iid是明示無限過程前提token，並非這個有限計算已構造或驗證無限independence。後續使用該前提作條件式數學推論。",
        "en": "infinite_iid is an explicit infinite-process premise token,not a construction or finite verification of infinite independence.Subsequent reasoning is conditional on that premise."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL law mass sum0 := sum(0, 1/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL law mass sum0 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL law mass sum0，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL law mass sum0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "0",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL law mass sum0"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL law mass sum1 := sum(1/2, 1/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL law mass sum1 = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL law mass sum1，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL law mass sum1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1/2"
        ],
        "output": "1",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL law mass sum1"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "source roles = [CLT ORIGINAL conditional marginal, [1, 2], [[1, 2], [1, 2]]]",
      "after": "derived = [[[1, 1/2], [2, 1/2]], 1]",
      "relation": "implies",
      "rule_id": "CLT_ORIGINAL_value_probability_pairs_NUMERIC_order_normalized",
      "conditions": [],
      "explanation": {
        "zh-Hant": "原始values與probabilities先配對再numeric排序，保留零機率label並核全部law質量1。這是有限marginal qualification，不是由samples證無限IID。",
        "en": "Pair ORIGINAL values/probabilities before numeric sorting,retaining zero-mass labels and checking total mass1.This qualifies a finite marginal,not an infinite IID process inferred from samples."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL value times probability0 := mul(1, 1/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL value times probability0 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL value times probability0，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL value times probability0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL value times probability0"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum0 := sum(0, 1/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum0 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum0，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "0",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum0"
      }
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "CLT ORIGINAL conditional marginal exact value square0 := mul(1, 1)",
      "after": "CLT ORIGINAL conditional marginal exact value square0 = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal exact value square0，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal exact value square0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1"
        ],
        "output": "1",
        "label": "CLT ORIGINAL conditional marginal exact value square0"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL square times probability0 := mul(1, 1/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL square times probability0 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL square times probability0，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL square times probability0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL square times probability0"
      }
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum0 := sum(0, 1/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum0 = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum0，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum0 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "0",
          "1/2"
        ],
        "output": "1/2",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum0"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL value times probability1 := mul(2, 1/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL value times probability1 = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL value times probability1，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL value times probability1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "2",
          "1/2"
        ],
        "output": "1",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL value times probability1"
      }
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum1 := sum(1/2, 1)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum1 = 3/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum1，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "1"
        ],
        "output": "3/2",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL ascending mean sum1"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "CLT ORIGINAL conditional marginal exact value square1 := mul(2, 2)",
      "after": "CLT ORIGINAL conditional marginal exact value square1 = 4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal exact value square1，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal exact value square1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "2",
          "2"
        ],
        "output": "4",
        "label": "CLT ORIGINAL conditional marginal exact value square1"
      }
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL square times probability1 := mul(4, 1/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL square times probability1 = 2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL square times probability1，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL square times probability1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "4",
          "1/2"
        ],
        "output": "2",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL square times probability1"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum1 := sum(1/2, 2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum1 = 5/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum1，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum1 with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/2",
          "2"
        ],
        "output": "5/2",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL ascending second moment sum1"
      }
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL mean square := mul(3/2, 3/2)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL mean square = 9/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL mean square，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL mean square with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "3/2",
          "3/2"
        ],
        "output": "9/4",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL mean square"
      }
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "CLT ORIGINAL conditional marginal ORIGINAL second minus mean square := sub(5/2, 9/4)",
      "after": "CLT ORIGINAL conditional marginal ORIGINAL second minus mean square = 1/4",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐原始角色與已證中間量精確代值；不以近似值、索引位址或自洽但換過的操作數取代來源。",
        "en": "Substitute original roles and established intermediates exactly;no approximation,slot substitution or internally consistent changed operand."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 CLT ORIGINAL conditional marginal ORIGINAL second minus mean square，不以小數近似替代。",
          "en": "Recompute CLT ORIGINAL conditional marginal ORIGINAL second minus mean square with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sub",
        "inputs": [
          "5/2",
          "9/4"
        ],
        "output": "1/4",
        "label": "CLT ORIGINAL conditional marginal ORIGINAL second minus mean square"
      }
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "source roles = [CLT ORIGINAL conditional marginal, [[1, 1/2], [2, 1/2]]]",
      "after": "derived = [3/2, 5/2, 1/4]",
      "relation": "implies",
      "rule_id": "CLT_ALL_finite_marginal_moments_not_a_process_or_limit_proof",
      "conditions": [],
      "explanation": {
        "zh-Hant": "E[X]、E[X²]與Var(X)=E[X²]-E[X]²都用同一完整law。value²先保精確教學量，再乘prob才照來源檢profile，不把null outcome的大value²額外拒絕。有限矩不證independence、無限過程或分布極限。",
        "en": "Use the SAME complete law for E[X],E[X²] and Var(X)=E[X²]-E[X]².Keep value² as an exact teaching intermediate and apply the source profile to its product with probability,not an extra rejection for a large zero-mass value square.Finite moments prove neither independence,an infinite process nor a distributional limit."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "source roles = [wrong_scaling, 3/2, 1/4, 5]",
      "after": "derived = [wrong_scaling, 3/2, 1/4, 5, denominator=n*sigma, required=sigma*sqrt(n), 否]",
      "relation": "implies",
      "rule_id": "CLT_original_wrong_scale_or_finite_evidence_inference_boundary",
      "conditions": [],
      "explanation": {
        "zh-Hant": "標準化分母須sigma*sqrt(n)，因Var(S_n)=n*sigma²；n*sigma作為一般n的公式不同，但n=1可巧合相同，No不宣稱本次兩個數必不等。單一有限histogram即使看起來normal，也不證所有n的分布極限。",
        "en": "The denominator must be sigma*sqrt(n),since Var(S_n)=n*sigma².The general formula n*sigma differs,but coincides at n=1;No does not assert numerical inequality in every instance.One finite histogram,even if normal-looking,does not prove a distributional limit."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "回查原始完整資料、步驟操作數與本步前提；有限核對、條件式定理引用與一般形式證明分開。",
          "en": "Reverse-check the complete original data,step operands and premises;finite checks,conditional named-theorem use and general formal proof remain distinct."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "result": [
      "wrong_scaling",
      "3/2",
      "1/4",
      5,
      "denominator=n*sigma",
      "required=sigma*sqrt(n)",
      false
    ]
  },
  "conclusion": {
    "canonical_value": "(wrong_scaling, 3/2, 1/4, 5, denominator=n*sigma, required=sigma*sqrt(n), 否)",
    "explanation": {
      "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。",
      "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged."
    }
  },
  "definition_note": {
    "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。",
    "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations",
      "batch_path": "public/data/questions/batches/batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations/questions.jsonl",
      "batch_sha256": "6eaac14f97913d20b3a00dc8becf4943e2b1791210c8c5bbf3a89ecced69c6d1",
      "original_line": 79,
      "original_line_sha256": "87fc33fb1ecc47f0e847c4d9f9e569f180c6d780292205ba58a51fd4e1c41c46",
      "question_compact_sha256": "461606279384502fa06d8dbcf33ceca9c81d9d457f61207be7051f8ed267a9e8",
      "english_row_compact_sha256": "f64cd7ce5870a9cd049e985a43d747b7c6f800da581788b9c0606793d118b8f9",
      "english_manifest_path": "public/data/i18n/editions/batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261015-advance-20261118-r01-central-limit-theorem-finite-variance-iid-foundations.en.jsonl"
    },
    "mko_bindings": {
      "mko-central-limit-theorem-finite-variance-iid": {
        "sha256": "c37700292553db11f9702a86ba165e9287d7830d81ee8351eff28de1c07ea9fe",
        "english_sha256": "951e71ab789a556b03eeb44c3472b5e89e3ac252d23992bc8abf538f39a494cb",
        "original_path": "public/data/mko/central-limit-theorem-finite-variance-iid.json",
        "english_path": "public/data/i18n/editions/mko-central-limit-theorem-finite-variance-iid.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "a54409c0bc3f68730e1072c6d3d772970462c1bc7ffc203b4030a6136a610283",
    "original_candidate_sha256": "b0010c9764d4d4189fb7448f4c95aee36d0f1dc65c69767fcde488610760772c",
    "original_candidate_line_sha256": "b0010c9764d4d4189fb7448f4c95aee36d0f1dc65c69767fcde488610760772c"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。",
      "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261026-010-p54",
    "review_date": "2026-10-26",
    "owner_authorized_reader_publication": true,
    "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "1b7a62514347b82518305d5feb4369a52d7d44c442a0f2b131e9eb3cd7b45f62",
      "question_first_witness_sha256": "79dc2a1077081f3ea2a03f75e32778163bf05e814cdf4c29ed86e567295f9ee1"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 19,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "e3b2e93db0db17397ec30f6704bf3e0bfae3da12c24d933d6bc39955f66e46d6",
      "arithmetic_steps": 14
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}