{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-descriptive-statistics-01",
  "mko_id": "mko-descriptive-statistics",
  "title": {
    "zh-Hant": "描述統計：保留重數計算平均數、中位數、全距與眾數集合",
    "en": "Descriptive statistics: mean, median, range and mode set with repetitions kept"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-descriptive-statistics-01",
    "source": {
      "mko_id": "mko-descriptive-statistics",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "bb188b4a4f67bc4e07da170bd9ea1b92aa0a51719cac24e1570cb156d8f76204",
      "english_sha256": "54dad0cd96ba57e6ace6bc691c6895d71b99feca5b90a348806c619d275baffe"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "numeric_evaluation",
      "finite_sum_rules",
      "frac_rules",
      "order_basics",
      "finite_counting_basics"
    ],
    "goal": {
      "zh-Hant": "對觀測列 D=(4,10,1,3,8,1,6,3)，保留重複值，求觀測數 n、算術平均、中位數、全距與本輪約定的眾數集合；證明一般性質「最小值 ≤ 平均 ≤ 最大值」並用它核對；最後說明若先去重會得到不同（錯誤）的平均與中位數。",
      "en": "For the observation list D=(4,10,1,3,8,1,6,3), keeping repeated values, find the number of observations n, the arithmetic mean, the median, the range and the mode set under this entry's convention; prove the general fact 'minimum ≤ mean ≤ maximum' and use it as a check; finally show that deduplicating first would give a different (wrong) mean and median."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "D=(4,10,1,3,8,1,6,3) 是原創合成的整數觀測列；它是列表，保留順序與重複值，不是去重集合（MKO 假設 a1、a2）。",
          "en": "D=(4,10,1,3,8,1,6,3) is an original synthetic list of integer observations; it is a list that keeps order and repeated values, not a deduplicated set (MKO assumptions a1, a2)."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "眾數集合約定：所有出現次數達最大值的數，按遞增列出、每值一次（MKO 陳述與假設 a3）。",
          "en": "Mode-set convention: all values whose frequency is maximal, listed in increasing order, each once (MKO statement and assumption a3)."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "本題只作描述摘要，不假定 D 是某母體的獨立樣本（MKO 假設 a3）。",
          "en": "This task is a descriptive summary only; D is not assumed to be an independent sample from a population (MKO assumption a3)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "G2"
        ],
        "before_tex": "D=(d_i)_{i=1}^{n},\\ n>0",
        "after_tex": "\\bar d=\\tfrac{1}{n}\\sum_{i=1}^{n}d_i,\\ \\tilde d,\\ R=\\max D-\\min D,\\ \\operatorname{Mo}(D)",
        "relation": "definition",
        "rule": "mko_descstat_def",
        "conditions": [
          "n>0"
        ],
        "explanation": {
          "zh-Hant": "寫出 MKO 的四個定義：平均是總和除以 n；中位數在排序後取（奇數 n 取中央值，偶數 n 取中間兩值的平均）；全距是最大值減最小值；眾數集合依 G2 的約定。所有計算都在保留重數的列表上做。",
          "en": "State the MKO's four definitions: the mean is the sum divided by n; the median is taken after sorting (the middle value for odd n, the mean of the two middle values for even n); the range is the maximum minus the minimum; the mode set follows the convention of G2. Every computation is done on the list with repetitions kept."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "G2",
            "mko_descstat_def"
          ],
          "explanation": {
            "zh-Hant": "此步只寫出定義，不推出新事實；n>0 是平均有定義的前提。",
            "en": "This step only states the definitions and derives no new fact; n>0 is what makes the mean defined."
          }
        },
        "leaf_basis": "mko_descstat_def + def_unfold"
      },
      {
        "id": "S02",
        "uses": [
          "G1"
        ],
        "before_tex": "D=(4,10,1,3,8,1,6,3)",
        "after_tex": "n=8>0",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐項計數，D 有 8 個位置，重複的 1 與 3 各占兩個位置，所以 n=8，不是不同值的個數 6。n>0，平均與中位數有定義。",
          "en": "Counting positions, D has 8 entries; the repeated values 1 and 3 each occupy two positions, so n=8, not the number 6 of distinct values. Since n>0, the mean and the median are defined."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "G1",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "計的是列表位置，不是不同值；這正是 MKO 假設 a1 的要求。",
            "en": "What is counted is list positions, not distinct values; this is exactly what MKO assumption a1 requires."
          }
        },
        "leaf_basis": "finite_counting_basics"
      },
      {
        "id": "S03",
        "uses": [
          "G1",
          "finite_sum_rules"
        ],
        "before_tex": "\\sum_{i=1}^{8}d_i",
        "after_tex": "4+10+1+3+8+1+6+3=36",
        "relation": "equivalent",
        "rule": "finite_sum_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "依有限和的遞迴規則逐項加入，部分和依次為 4、14、15、18、26、27、33、36。值 1 與值 3 各出現兩次，每筆觀測都加入一次。",
          "en": "Add the terms one at a time by the recursive rule for finite sums; the partial sums are 4, 14, 15, 18, 26, 27, 33, 36. The value 1 is added twice, and so is 3; each observation is added once."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "finite_sum_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確整數計算（N1），可獨立重算，雙向成立。",
            "en": "An exact integer computation (N1) that can be recomputed, two-way."
          }
        },
        "leaf_basis": "finite_sum_rules + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S01",
          "S02",
          "S03"
        ],
        "before_tex": "\\bar d=\\tfrac{36}{8}",
        "after_tex": "\\bar d=\\tfrac{9}{2}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "8\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "平均 = 36/8。分子分母同除以 gcd(36,8)=4（frac_rules，a/b=(ak)/(bk)，k≠0）得 9/2；gcd(9,2)=1，已是最簡分數（MKO 假設 a2）。核對：8·(9/2)=36，即各觀測與平均之差的總和為 36−8·(9/2)=0。",
          "en": "The mean is 36/8. Dividing numerator and denominator by gcd(36,8)=4 (frac_rules, a/b=(ak)/(bk) with k≠0) gives 9/2; gcd(9,2)=1, so it is reduced (MKO assumption a2). Check: 8·(9/2)=36, so the deviations from the mean sum to 36−8·(9/2)=0."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反向：分子分母同乘 4 回到 36/8；需要 8≠0 與 4≠0。",
            "en": "Backward: multiply numerator and denominator by 4 to return to 36/8; this needs 8≠0 and 4≠0."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S05",
        "uses": [
          "G1",
          "perm_invariance"
        ],
        "before_tex": "(4,10,1,3,8,1,6,3)",
        "after_tex": "(s_{(k)})_{k=1}^{8}=(1,1,3,3,4,6,8,10)",
        "relation": "implies",
        "rule": "perm_invariance",
        "conditions": [],
        "explanation": {
          "zh-Hant": "把 D 由小到大排列。核對它只改順序：在兩個列表中，1 與 3 各出現 2 次，4、6、8、10 各 1 次，總數 8；總和仍是 36。依 MKO 假設 a1，排序只改順序，不刪去重複值。",
          "en": "Arrange D in increasing order. Check that only the order changed: in both lists 1 and 3 occur twice each and 4, 6, 8, 10 once each, 8 entries in all, and the sum is still 36. By MKO assumption a1, sorting changes only the order and does not remove repeated values."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "perm_invariance",
            "finite_counting_basics"
          ],
          "explanation": {
            "zh-Hant": "需要的是每個值的出現次數相同（N5、N8）；排序後的列表一般不等於原列表，所以只宣稱「同一組觀測」，不宣稱兩列表相等。",
            "en": "What is needed is that each value occurs equally often (N5, N8); the sorted list is in general not equal to the original list, so only 'the same observations' is claimed, not equality of the lists."
          }
        },
        "leaf_basis": "perm_invariance + finite_counting_basics"
      },
      {
        "id": "S06",
        "uses": [
          "S01",
          "S02",
          "S05"
        ],
        "before_tex": "\\tilde d,\\ n=8\\ \\text{even}",
        "after_tex": "\\tilde d=\\tfrac{s_{(4)}+s_{(5)}}{2}=\\tfrac{3+4}{2}=\\tfrac{7}{2}",
        "relation": "equivalent",
        "rule": "mko_descstat_def",
        "conditions": [
          "n\\ \\text{even}"
        ],
        "explanation": {
          "zh-Hant": "n=8 是偶數，中間兩個位置是 n/2=4 與 n/2+1=5。排序後 s_(4)=3、s_(5)=4，中位數是兩者的平均 (3+4)/2=7/2。它介於 3 與 4 之間，不是任何一筆觀測。",
          "en": "n=8 is even, so the two middle positions are n/2=4 and n/2+1=5. After sorting, s_(4)=3 and s_(5)=4, and the median is their mean (3+4)/2=7/2. It lies between 3 and 4 and is not one of the observations."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_descstat_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "依定義展開後是精確計算（N6），雙向成立；位置必須取自排序後的列表。",
            "en": "After unfolding the definition it is an exact computation (N6), two-way; the positions must be taken in the sorted list."
          }
        },
        "leaf_basis": "mko_descstat_def + frac_rules + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S01",
          "S05"
        ],
        "before_tex": "R=\\max D-\\min D",
        "after_tex": "R=10-1=9",
        "relation": "equivalent",
        "rule": "mko_descstat_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "排序後最小值是第一項 1，最大值是最後一項 10；排序不改變觀測的集合，所以它們也是 D 的最小值與最大值。全距 10−1=9。",
          "en": "After sorting, the minimum is the first entry 1 and the maximum is the last entry 10; sorting does not change which observations are present, so these are also the minimum and maximum of D. The range is 10−1=9."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "mko_descstat_def",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "精確計算（N7），雙向成立。",
            "en": "An exact computation (N7), two-way."
          }
        },
        "leaf_basis": "mko_descstat_def + order_basics + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "G1",
          "S05",
          "finite_counting_basics"
        ],
        "before_tex": "f(v)\\ \\text{for } v\\in\\{1,3,4,6,8,10\\}",
        "after_tex": "f(1)=2,\\ f(3)=2,\\ f(4)=f(6)=f(8)=f(10)=1",
        "relation": "implies",
        "rule": "finite_counting_basics",
        "conditions": [],
        "explanation": {
          "zh-Hant": "逐值計數出現次數。核對：2+2+1+1+1+1=8=n，每筆觀測恰好被算一次。計頻率時必須保留觀測重數（MKO 第二段）。",
          "en": "Count how often each value occurs. Check: 2+2+1+1+1+1=8=n, so every observation is counted exactly once. Frequencies must be counted with repetitions kept (MKO second paragraph)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "finite_counting_basics",
            "S02"
          ],
          "explanation": {
            "zh-Hant": "頻率總和等於 n 是必要核對（N8）；不宣稱頻率表能還原原本的順序。",
            "en": "The frequencies summing to n is a necessary check (N8); it is not claimed that the frequency table recovers the original order."
          }
        },
        "leaf_basis": "finite_counting_basics + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S08",
          "G2"
        ],
        "before_tex": "\\max_v f(v)=2",
        "after_tex": "\\operatorname{Mo}(D)=\\{1,3\\}",
        "relation": "implies",
        "rule": "mko_descstat_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "最大頻率是 2，達到它的值有 1 與 3。依 G2 的約定，兩者都屬於眾數集合，按遞增列出、每值一次：{1,3}。漏掉並列者是 MKO 列出的誤解。",
          "en": "The maximum frequency is 2, attained by 1 and 3. By the convention of G2 both belong to the mode set, listed in increasing order, each once: {1,3}. Omitting a tied value is a misconception the MKO lists."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S08",
            "G2",
            "mko_descstat_def"
          ],
          "explanation": {
            "zh-Hant": "需要完整的頻率表；只看第一個達到最大頻率的值會漏掉 3。",
            "en": "It needs the complete frequency table; looking only at the first value with maximum frequency would miss 3."
          }
        },
        "leaf_basis": "mko_descstat_def + order_basics"
      },
      {
        "id": "S10",
        "uses": [
          "S04",
          "S07",
          "L_mean_bounds"
        ],
        "before_tex": "1\\le d_i\\le 10\\ \\text{for every } i",
        "after_tex": "1\\le \\tfrac{9}{2}\\le 10",
        "relation": "theorem_application",
        "rule": "L_mean_bounds",
        "conditions": [
          "n=8>0"
        ],
        "explanation": {
          "zh-Hant": "每筆觀測都介於最小值 1 與最大值 10 之間，依引理 L_mean_bounds（m=1、M=10），平均 9/2 必須介於 1 與 10 之間；確實 1≤9/2≤10。這是對 S04 的合理性核對，不能取代 S03、S04 的計算。引理在 S10.1 到 S10.3 證明。",
          "en": "Every observation lies between the minimum 1 and the maximum 10, so by Lemma L_mean_bounds (m=1, M=10) the mean 9/2 must lie between 1 and 10; indeed 1≤9/2≤10. This is a plausibility check of S04 and cannot replace the computation in S03 and S04. The lemma is proved in S10.1 to S10.3."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_mean_bounds",
            "S07"
          ],
          "explanation": {
            "zh-Hant": "引理需要 n>0 與逐項的上下界；不宣稱反向，即落在 [1,10] 的數不一定是平均。",
            "en": "The lemma needs n>0 and termwise bounds; no converse is claimed: a number in [1,10] need not be the mean."
          }
        },
        "leaf_basis": "L_mean_bounds + numeric_evaluation"
      },
      {
        "id": "S10.1",
        "uses": [
          "finite_sum_rules",
          "order_basics"
        ],
        "before_tex": "m\\le d_i\\ \\text{for every } i",
        "after_tex": "nm=\\sum_{i=1}^{n}m\\le\\sum_{i=1}^{n}d_i",
        "relation": "implies",
        "rule": "finite_sum_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "對項數 k 由 0 到 n 逐步加入：k=0 時兩邊都是 0。若 Σ_{i≤k} m ≤ Σ_{i≤k} d_i，又 m≤d_{k+1}，依 order_basics（兩邊各加一個數，再用傳遞：u≤v、w≤z ⇒ u+w≤v+w≤v+z）得 Σ_{i≤k+1} m ≤ Σ_{i≤k+1} d_i。依數學歸納到 k=n。左邊 n 個 m 相加等於 nm（finite_sum_rules，Σ c·1=c·n）。",
          "en": "Add the terms one at a time for k from 0 to n: at k=0 both sides are 0. If Σ_{i≤k} m ≤ Σ_{i≤k} d_i and m≤d_{k+1}, then by order_basics (adding a number to both sides and using transitivity: u≤v and w≤z give u+w≤v+w≤v+z) Σ_{i≤k+1} m ≤ Σ_{i≤k+1} d_i. By induction this reaches k=n. The left side, n copies of m, equals nm (finite_sum_rules, Σ c·1=c·n)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "finite_sum_rules",
            "order_basics",
            "nat_induction"
          ],
          "explanation": {
            "zh-Hant": "需要每一項都滿足 m≤d_i；只要有一項小於 m，結論可能失敗。不宣稱反向。",
            "en": "Every term must satisfy m≤d_i; if one term were below m the conclusion could fail. No converse is claimed."
          }
        },
        "leaf_basis": "finite_sum_rules + order_basics + nat_induction"
      },
      {
        "id": "S10.2",
        "uses": [
          "S10.1",
          "order_basics",
          "mul_inverse_nonzero"
        ],
        "before_tex": "nm\\le\\sum_{i=1}^{n}d_i,\\ n>0",
        "after_tex": "m\\le\\tfrac{1}{n}\\sum_{i=1}^{n}d_i=\\bar d",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "n>0"
        ],
        "explanation": {
          "zh-Hant": "n>0，所以 1/n 存在且 1/n>0。依 order_basics，不等式兩邊同乘正數 1/n 方向不變：m=(nm)/n≤(Σd_i)/n=d̄。",
          "en": "Since n>0, 1/n exists and 1/n>0. By order_basics, multiplying both sides of an inequality by the positive number 1/n keeps its direction: m=(nm)/n≤(Σd_i)/n=d̄."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10.1",
            "order_basics",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "必須 n>0：乘負數會反向，n=0 時平均無定義。",
            "en": "n>0 is essential: multiplying by a negative number would reverse the inequality, and for n=0 the mean is undefined."
          }
        },
        "leaf_basis": "order_basics + mul_inverse_nonzero + frac_rules"
      },
      {
        "id": "S10.3",
        "uses": [
          "S10.1",
          "S10.2"
        ],
        "before_tex": "d_i\\le M\\ \\text{for every } i",
        "after_tex": "\\bar d\\le M",
        "relation": "implies",
        "rule": "order_basics",
        "conditions": [
          "n>0"
        ],
        "explanation": {
          "zh-Hant": "與 S10.1、S10.2 相同的論證，把不等式方向換成 d_i≤M：Σd_i≤nM，再同乘 1/n>0 得 d̄≤M。S10.1 到 S10.3 合起來證得引理 L_mean_bounds。",
          "en": "The same argument as S10.1 and S10.2 with the inequalities d_i≤M: Σd_i≤nM, then multiply by 1/n>0 to get d̄≤M. S10.1 to S10.3 together prove Lemma L_mean_bounds."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S10.1",
            "S10.2",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "同樣需要逐項上界與 n>0。",
            "en": "It likewise needs termwise upper bounds and n>0."
          }
        },
        "leaf_basis": "order_basics + finite_sum_rules + mul_inverse_nonzero"
      },
      {
        "id": "S11",
        "uses": [
          "S04",
          "S06",
          "G1"
        ],
        "before_tex": "\\bar d=\\tfrac{9}{2},\\ \\tilde d=\\tfrac{7}{2}",
        "after_tex": "\\bar d\\notin D,\\ \\tilde d\\notin D",
        "relation": "implies",
        "rule": "numeric_evaluation",
        "conditions": [],
        "explanation": {
          "zh-Hant": "9/2 是最簡分數且分母為 2（9 除以 2 餘 1），不是整數；D 的每筆觀測都是整數，所以平均不是任何一筆觀測。同理 7/2 不是整數，中位數也不是任何一筆觀測。MKO 第一段：平均不一定是某筆觀測，偶數個資料的中位數可能在中間兩值之間。",
          "en": "9/2 is in lowest terms with denominator 2 (9 leaves remainder 1 on division by 2), so it is not an integer; every observation in D is an integer, so the mean is none of the observations. Likewise 7/2 is not an integer, so the median is none of the observations either. MKO first paragraph: the mean need not be an observation, and for even n the median may lie between the two middle values."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S06",
            "divides_def"
          ],
          "explanation": {
            "zh-Hant": "只宣稱這兩個摘要不在 D 中；不宣稱平均或中位數一般不能是觀測值。",
            "en": "It claims only that these two summaries are not in D; it does not claim that a mean or median can never be an observation."
          }
        },
        "leaf_basis": "numeric_evaluation + divides_def"
      },
      {
        "id": "S12",
        "uses": [
          "S04",
          "S06",
          "G1"
        ],
        "before_tex": "\\{1,3,4,6,8,10\\}\\ \\text{(deduplicated)}",
        "after_tex": "\\tfrac{32}{6}=\\tfrac{16}{3}\\neq\\tfrac{9}{2},\\quad \\tfrac{4+6}{2}=5\\neq\\tfrac{7}{2}",
        "relation": "implies",
        "rule": "mko_descstat_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若先把 D 去重成 {1,3,4,6,8,10}（6 個值），平均會是 (1+3+4+6+8+10)/6=32/6=16/3，中位數會是 (4+6)/2=5，兩者都和 S04、S06 不同。所以去重確實改變結果，違反 MKO 假設 a1；正確答案以保留重數的 D 為準。去重不改變非空有限觀測列的最小值與最大值，所以全距必定不變（本例為10−1=9）；但這不表示去重對平均、中位數與頻率等摘要無害。",
          "en": "If D were first deduplicated to {1,3,4,6,8,10} (6 values), the mean would be (1+3+4+6+8+10)/6=32/6=16/3 and the median (4+6)/2=5, both different from S04 and S06. So deduplication really changes the results and violates MKO assumption a1; the correct answers are those for D with repetitions kept. Deduplication preserves the minimum and maximum of any nonempty finite observation list, so it always preserves the range (10−1=9 here); this does not make it harmless for summaries such as the mean, median, or frequencies."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "S04",
            "S06",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "反例只需要一組具體數字（N12 到 N14）；它說明去重一般不保持平均與中位數。",
            "en": "A counterexample needs only one concrete instance (N12 to N14); it shows that deduplication does not in general preserve the mean and the median."
          }
        },
        "leaf_basis": "mko_descstat_def + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S02",
          "S04",
          "S06",
          "S07",
          "S09",
          "S10",
          "G3"
        ],
        "before_tex": "\\text{tasks of the goal}",
        "after_tex": "n=8,\\ \\bar d=\\tfrac{9}{2},\\ \\tilde d=\\tfrac{7}{2},\\ R=9,\\ \\operatorname{Mo}(D)=\\{1,3\\}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "合併：n=8，平均 9/2，中位數 7/2，全距 9，眾數集合 {1,3}；平均通過 1≤9/2≤10 的核對。依 G3，這些數只描述 D 本身，不推出母體平均、機率模型或因果結論。",
          "en": "Combine: n=8, mean 9/2, median 7/2, range 9, mode set {1,3}; the mean passes the check 1≤9/2≤10. By G3 these numbers describe D itself only and imply no population mean, probability model or causal conclusion."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S02",
            "S04",
            "S06",
            "S07",
            "S09",
            "S10",
            "S12"
          ],
          "explanation": {
            "zh-Hant": "目標 G：題目要求的各摘要與核對。足夠條件 H：S02 到 S12 的結論；H⇒G 只用合取。H 由 G1 到 G3 正向推出，沒有把 G 當作前提。",
            "en": "Goal G: the summaries and checks asked for. Sufficient condition H: the conclusions of S02 to S12; H⇒G uses only conjunction. H is derived forwards from G1 to G3; G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "對 D=(4,10,1,3,8,1,6,3)：n=8，平均 36/8=9/2，排序後中位數 (3+4)/2=7/2，全距 10−1=9，眾數集合 {1,3}；且 1≤9/2≤10。若先去重會誤得平均 16/3、中位數 5。這些摘要只描述 D，不構成任何母體推論。",
      "en": "For D=(4,10,1,3,8,1,6,3): n=8, mean 36/8=9/2, median after sorting (3+4)/2=7/2, range 10−1=9, mode set {1,3}; and 1≤9/2≤10. Deduplicating first would wrongly give mean 16/3 and median 5. These summaries describe D only and support no population inference."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "中位數是 D 的第 4、5 項的平均 (3+8)/2=11/2",
          "en": "The median is the mean of the 4th and 5th entries of D, (3+8)/2=11/2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "中位數要在排序後取；排序後第 4、5 項是 3 與 4，中位數 7/2。",
          "en": "The median is taken after sorting; the sorted 4th and 5th entries are 3 and 4, so the median is 7/2."
        }
      },
      {
        "claim": {
          "zh-Hant": "平均是 (1+3+4+6+8+10)/6=16/3",
          "en": "The mean is (1+3+4+6+8+10)/6=16/3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "這是去重後的平均；保留重數得 36/8=9/2。",
          "en": "That is the mean after deduplication; with repetitions kept it is 36/8=9/2."
        }
      },
      {
        "claim": {
          "zh-Hant": "眾數集合是 {1}",
          "en": "The mode set is {1}"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "3 也出現 2 次，達到最大頻率；眾數集合是 {1,3}。",
          "en": "3 also occurs twice, the maximum frequency; the mode set is {1,3}."
        }
      },
      {
        "claim": {
          "zh-Hant": "D 的平均是 9/2，所以產生 D 的母體平均也是 9/2",
          "en": "The mean of D is 9/2, so the population that produced D has mean 9/2"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "描述摘要只描述 D，本題沒有母體或抽樣模型（MKO 假設 a3）。",
          "en": "A descriptive summary describes D only; there is no population or sampling model here (MKO assumption a3)."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "finite_counting_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。",
          "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements."
        }
      },
      {
        "id": "def_unfold",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。",
          "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true."
        }
      },
      {
        "id": "nat_induction",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "數學歸納原理：對從 n₀ 起的命題族 P(n)，若 P(n₀) 成立，且對每個 n≥n₀ 都有 P(n)⇒P(n+1)，則所有 n≥n₀ 都有 P(n)。",
          "en": "Principle of induction: for a family P(n) starting at n₀, if P(n₀) holds and P(n)⇒P(n+1) for every n≥n₀, then P(n) holds for every n≥n₀."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "divides_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "整除：對整數 a、b，a∣b 表示存在整數 k 使 b=ak。",
          "en": "Divisibility: for integers a, b, a∣b means there is an integer k with b=ak."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_descstat_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-descriptive-statistics",
          "sha256": "bb188b4a4f67bc4e07da170bd9ea1b92aa0a51719cac24e1570cb156d8f76204",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "給非空有限實數觀測列 D，算術平均為 Σdi/n；排序後奇數 n 的中位數是中央值，偶數 n 是中間兩值的平均；全距為最大值減最小值。本輪眾數集合約定為所有出現頻率達最大值的數，按遞增列出；各數均只出現一次時，全部不同觀測值都是此約定的頻率最大者。",
          "en": "For a nonempty finite real observation list D of length n, preserve repeated values. The mean is the sum divided by n. After sorting, the median is the middle value for odd n and the mean of the two middle values for even n. The range is max(D)−min(D). In this entry, the mode set contains all values with maximum frequency, listed in increasing order, including all distinct observed values when every frequency is one."
        }
      },
      {
        "id": "perm_invariance",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "重新排列有限列表的項不改變其和，也不改變每個值的出現次數（有限和的一般交換律，可由 add_assoc、add_comm 歸納得到）。",
          "en": "Rearranging the entries of a finite list changes neither its sum nor the number of times each value occurs (generalized commutativity of finite sums, obtainable from add_assoc and add_comm by induction)."
        }
      },
      {
        "id": "L_mean_bounds",
        "kind": "lemma",
        "proved_in": [
          "S10.1",
          "S10.2",
          "S10.3"
        ],
        "statement": {
          "zh-Hant": "對非空有限實數列 (d_i)_{i=1}^{n}，若 m≤d_i≤M 對每個 i 成立，則 m≤(Σd_i)/n≤M。",
          "en": "For a nonempty finite real list (d_i)_{i=1}^{n}, if m≤d_i≤M for every i, then m≤(Σd_i)/n≤M."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S03",
        "lhs": "4+10+1+3+8+1+6+3",
        "rhs": "36"
      },
      {
        "id": "N2",
        "step": "S04",
        "lhs": "36/8",
        "rhs": "9/2"
      },
      {
        "id": "N3",
        "step": "S04",
        "lhs": "gcd(36,8)",
        "rhs": "4"
      },
      {
        "id": "N4",
        "step": "S04",
        "lhs": "gcd(9,2)",
        "rhs": "1"
      },
      {
        "id": "N5",
        "step": "S05",
        "lhs": "1+1+3+3+4+6+8+10",
        "rhs": "36"
      },
      {
        "id": "N6",
        "step": "S06",
        "lhs": "(3+4)/2",
        "rhs": "7/2"
      },
      {
        "id": "N7",
        "step": "S07",
        "lhs": "10-1",
        "rhs": "9"
      },
      {
        "id": "N8",
        "step": "S08",
        "lhs": "2+2+1+1+1+1",
        "rhs": "8"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "1",
        "relation": "<=",
        "rhs": "9/2"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "9/2",
        "relation": "<=",
        "rhs": "10"
      },
      {
        "id": "N11",
        "step": "S04",
        "lhs": "36-8*(9/2)",
        "rhs": "0"
      },
      {
        "id": "N12",
        "step": "S12",
        "lhs": "(1+3+4+6+8+10)/6",
        "rhs": "16/3"
      },
      {
        "id": "N13",
        "step": "S12",
        "lhs": "16/3",
        "relation": "!=",
        "rhs": "9/2"
      },
      {
        "id": "N14",
        "step": "S12",
        "lhs": "(4+6)/2",
        "rhs": "5"
      },
      {
        "id": "N15",
        "lhs": "(3+8)/2",
        "relation": "!=",
        "rhs": "7/2"
      },
      {
        "id": "N16",
        "step": "S11",
        "lhs": "9%2",
        "rhs": "1"
      },
      {
        "id": "N17",
        "step": "S10.2",
        "vars": [
          "a",
          "b",
          "c",
          "m"
        ],
        "degree": 1,
        "lhs": "(a+b+c)/3-m",
        "rhs": "((a-m)+(b-m)+(c-m))/3"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "只處理非空有限的整數觀測列與 MKO 定義的四個摘要；完整證明平均數介於最小值與最大值之間。不處理樣本變異數、其他分位數或圖表，不從摘要推論母體、機率律或因果，也不否定其他教材「無眾數」的用語；不改動 canonical MKO 或英文 edition。",
      "en": "Treats only a nonempty finite list of integer observations and the four summaries defined by the MKO; fully proves that the mean lies between the minimum and the maximum. It does not treat sample variance, other quantiles or charts, draws no population, probability or causal conclusion from the summaries, and does not reject other textbooks' phrase 'no mode'; it does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "D=(d_i)_{i=1}^{n}",
        "保留順序與重數的觀測列",
        "the observation list, with order and repetitions"
      ],
      [
        "n",
        "觀測數",
        "the number of observations"
      ],
      [
        "\\bar d",
        "算術平均",
        "the arithmetic mean"
      ],
      [
        "s_{(k)}",
        "排序後第 k 個值",
        "the k-th value after sorting"
      ],
      [
        "f(v)",
        "值 v 在 D 中出現的次數",
        "the number of times v occurs in D"
      ],
      [
        "m,\\ M",
        "最小值與最大值",
        "the minimum and the maximum"
      ]
    ],
    "md_error_controls": [
      {
        "title": "不要先去重 / Do not deduplicate first",
        "zh": "MKO 的誤解之一是先去重再算。去重後平均變成 16/3、中位數變成 5（S12），都不是 D 的摘要。重複值是資料的一部分：兩個 1 表示有兩筆觀測等於 1。",
        "en": "One misconception the MKO names is deduplicating before computing. After deduplication the mean becomes 16/3 and the median 5 (S12), neither of which summarizes D. Repeated values are part of the data: two 1s mean two observations equal to 1."
      },
      {
        "title": "並列的最大頻率都要列入 / Every tied value of maximal frequency belongs to the mode set",
        "zh": "1 與 3 都出現 2 次，眾數集合是 {1,3}，不是 {1}。若所有值都只出現一次，依本輪約定全部不同值都在眾數集合中；其他教材可能稱為「無眾數」，這是用語慣例不同，不是矛盾（MKO 假設 a3）。",
        "en": "1 and 3 both occur twice, so the mode set is {1,3}, not {1}. If every value occurred once, under this entry's convention all distinct values would be in the mode set; other textbooks may call that 'no mode', which is a different convention, not a contradiction (MKO assumption a3)."
      },
      {
        "title": "中位數要在排序後取 / Take the median after sorting",
        "zh": "未排序的 D 第 4、5 項是 3 與 8，平均 11/2，這不是中位數。必須先排序得 (1,1,3,3,4,6,8,10)，再取第 4、5 項 3 與 4，得 7/2（S06）。",
        "en": "In the unsorted D the 4th and 5th entries are 3 and 8, whose mean 11/2 is not the median. One must first sort to (1,1,3,3,4,6,8,10) and then take the 4th and 5th entries 3 and 4, giving 7/2 (S06)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "perm_invariance（有限和的一般交換律與排序保留頻率）作為前提使用，未在本稿證明。",
        "en": "perm_invariance (generalized commutativity of finite sums and preservation of frequencies under sorting) is used as a prerequisite and is not proved here."
      },
      {
        "zh-Hant": "樣本變異數、其他分位數、圖表與機率矩不在本稿範圍；描述摘要與機率期望的型別區別保留給後續切片。",
        "en": "Sample variance, other quantiles, charts and probabilistic moments are outside this guide; the type distinction between descriptive summaries and probabilistic expectations is left to later slices."
      },
      {
        "zh-Hant": "MKO 的 formal next（有限列表求和、排序中位與頻數最大集合的橋接）未做。",
        "en": "The MKO's formal next step (bridging finite list sums, sorted medians and maximal-frequency sets) is not done."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "de10c74e217f968eb9b391752651fd9f7f2626b798bfb8e87c9af6eadff7e2e3",
    "original_source_sha256": "de10c74e217f968eb9b391752651fd9f7f2626b798bfb8e87c9af6eadff7e2e3",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}