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Solutions详解

Probability and Statistics Foundations · Solutions概率与统计基础 · 详解

Companion to the Practice Set · Mark-by-mark walkthroughs · SAT / AP-Feeder / ON / AB styles练习题配套答案 · 逐分讲解 · SAT / AP 衔接 / 安 / 阿省考风格

EASY易 MEDIUM中 HARD难 🇺🇸 US美 🇨🇦 ON安 🇨🇦 AB阿 SAT-style MCQSAT 风格选择题 AP-feeder FRQAP 衔接简答题 ON MDM4U-style安省 MDM4U 风格 AB 30-2-style阿省 30-2 风格 Honors for US A2 / AB 30-2美国 A2 / 阿省 30-2 荣誉级
PART I  ·  SHORT RESPONSE · SOLUTIONS第一部分  ·  短答题 · 详解SAT MCQ + ON/AB short answer · 18 marksSAT 选择题 + 安 / 阿省考短答 · 共 18 分

Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Sample Spaces样本空间 · HSS-CP.A.1 [3 marks][3 分]

Two fair dice; $P($ sum $= 8)$.两枚均匀骰子；求 $P($ 点数和 $= 8)$。

Answer:答案：  (B)  $5/36$

(a) Sample space and event size样本空间与事件大小 M1·A1·A1

The sample space of two ordered dice has $|S| = 6 \cdot 6 = 36$ equally-likely outcomes. The event $A = \{(d_{1}, d_{2}) : d_{1} + d_{2} = 8\}$ is两枚有序骰子的样本空间有 $|S| = 6 \cdot 6 = 36$ 个等可能结果。事件 $A = \{(d_{1}, d_{2}) : d_{1} + d_{2} = 8\}$ 为 $$ A = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\}, \qquad |A| = 5. $$ Hence $P(A) = |A|/|S| = 5/36$, which matches option (B).故 $P(A) = |A|/|S| = 5/36$，对应选项 (B)。
Why the wrong choices fail.干扰项分析。
  • (A) $4/36$, off-by-one: the student lists $(2,6), (3,5), (5,3), (6,2)$ but forgets $(4,4)$, where the two dice show the same face.：差一错误，学生列出 $(2,6), (3,5), (5,3), (6,2)$ 却漏掉两枚骰子同点的 $(4,4)$。
  • (C) $6/36 = 1/6$, the count for sum $= 7$ (the modal sum), not $8$. A classic "anchored on the middle" trap.：这是和 $= 7$（众数和）的计数，而非 $8$。典型的"锚定在中间"陷阱。
  • (D) $8/36$, the student conflates the target sum $8$ with the count of outcomes, $|A| = 8$. The target value is not the count.：学生把目标和 $8$ 误当成结果个数 $|A| = 8$。目标数值不是计数。
Two ordered dice means $36$ outcomes, not $21$, and $(4, 4)$ is its own outcome.两枚有序骰子有 $36$ 个结果，而非 $21$ 个，且 $(4, 4)$ 自成一个结果。 A common SAT pitfall: students who think of dice as unordered pairs $\{d_{1}, d_{2}\}$ get $21$ pairs (the $15$ off-diagonal pairs counted once each, plus the $6$ doubles), but those $21$ are not equally likely; off-diagonal pairs are twice as likely as doubles. The classical formula $P(A) = |A|/|S|$ requires equally-likely outcomes, so always model dice as ordered pairs. The bell-shaped sum distribution ($2$ and $12$ each rare, $7$ most common, $5/36$ for $8$ and for $6$) drops out for free.常见 SAT 陷阱：把骰子当成无序对 $\{d_{1}, d_{2}\}$ 的学生得到 $21$ 对（$15$ 个非对角对各计一次，加 $6$ 个重复点），但这 $21$ 对并非等可能；非对角对的可能性是重复点的两倍。经典公式 $P(A) = |A|/|S|$ 要求结果等可能，故骰子永远要建模为有序对。钟形的和分布（$2$ 与 $12$ 各自稀有，$7$ 最常见，$8$ 与 $6$ 均为 $5/36$）随之自然得出。
Q2EASY易 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §1 Complement Rule补集法则 · HSS-CP.A.1 [3 marks][3 分]

$P($ at least one tail in $5$ flips $)$.$5$ 次投掷中 $P($ 至少一次反面 $)$。

Answer:答案：  (D)  $31/32$

(a) Apply the complement rule套用补集法则 M1·A1

"At least one tail" is the complement of "no tails", i.e. all five flips are heads. For independent fair flips,"至少一次反面"是"没有反面"（即五次全为正面）的补集。对于独立的均匀投掷， $$ P(\text{no tails}) = P(\text{all heads}) = \bigl(\tfrac{1}{2}\bigr)^{5} = \tfrac{1}{32}. $$

(b) Subtract from $1$用 $1$ 去减 A1

$$ P(\text{at least one tail}) = 1 - P(\text{no tails}) = 1 - \tfrac{1}{32} = \tfrac{31}{32}. $$ This matches option (D).对应选项 (D)。
Why the wrong choices fail.干扰项分析。
  • (A) $1/32$, the probability of "no tails" (i.e. all heads), which is the complement of the event asked for.：这是"没有反面"（即全为正面）的概率，恰是所求事件的补集。
  • (B) $5/32$, $P(X = 1)$ for $X \sim B(5, 1/2)$, the probability of exactly one tail. "Exactly" $\ne$ "at least".：$X \sim B(5, 1/2)$ 的 $P(X = 1)$，即恰好一次反面的概率。"恰好" $\ne$ "至少"。
  • (C) $1/2$, the probability one specific flip lands tails. The asker is about any of the $5$ flips landing tails, not a single one.：某一特定投掷出现反面的概率。问的是 $5$ 次中任一次出现反面，而非单独一次。
"At least one" is the complement-rule fingerprint."至少一个"是补集法则的标志。 Whenever you see "at least one $X$ in $n$ independent trials" for an event with probability $p$ per trial, write $$ P(\text{at least one}) = 1 - (1 - p)^{n}, $$ because the only thing easier than "at least one $X$" to compute is its complement, "no $X$ at all", which factorises by independence. The direct sum $P(X = 1) + P(X = 2) + \cdots + P(X = n)$ is always more work. The complement trick reappears verbatim in Q12(c), and on the AP Stats exam every May, so lock it in.每当看到"在 $n$ 次独立试验中至少一次 $X$"、且每次概率为 $p$ 时，写下 $$ P(\text{at least one}) = 1 - (1 - p)^{n}, $$ 因为比"至少一次 $X$"更易算的唯有其补集"完全没有 $X$"，后者凭独立性可直接相乘分解。直接求和 $P(X = 1) + P(X = 2) + \cdots + P(X = n)$ 总是更费力。这个补集技巧在第 12 题 (c) 与每年五月的 AP 统计考试中原样重现，务必牢记。
Q3MEDIUM中 🇺🇸 US美 SAT-style MCQSAT 风格选择题 §3 Independence Test独立性检验 · HSS-CP.A.2 [4 marks][4 分]

$P(A) = 0.6$, $P(B) = 0.5$, $P(A \cup B) = 0.8$. Decide independence.$P(A) = 0.6$、$P(B) = 0.5$、$P(A \cup B) = 0.8$。判断是否独立。

Answer:答案：  (B)  Independent, $P(A \cap B) = P(A)P(B) = 0.3$独立，$P(A \cap B) = P(A)P(B) = 0.3$

(a) Recover $P(A \cap B)$ from the addition rule由加法法则反求 $P(A \cap B)$ M1·A1

The addition rule gives $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Rearrange:加法法则给出 $P(A \cup B) = P(A) + P(B) - P(A \cap B)$。移项： $$ P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.8 = 0.3. $$

(b) Run the independence test进行独立性检验 M1·A1

Compare $P(A \cap B)$ to $P(A) P(B)$:将 $P(A \cap B)$ 与 $P(A) P(B)$ 比较： $$ P(A) P(B) = 0.6 \cdot 0.5 = 0.30 = P(A \cap B). $$ The two are equal, so $A$ and $B$ are independent by CCSSM HSS-CP.A.2. This matches option (B).两者相等，故由 CCSSM HSS-CP.A.2，$A$ 与 $B$ 独立。对应选项 (B)。
Why the wrong choices fail.干扰项分析。
  • (A) Mutually exclusive互斥, would require $P(A \cap B) = 0$, but $P(A \cap B) = 0.3 \ne 0$. Mutually exclusive and independent are different, and two events with positive probability cannot be both.：须满足 $P(A \cap B) = 0$，但 $P(A \cap B) = 0.3 \ne 0$。互斥与独立不同，两个正概率事件不可能同时兼具。
  • (C) Dependent不独立, would require $P(A \cap B) \ne P(A) P(B)$; here they are equal, so the test fails to detect dependence.：须满足 $P(A \cap B) \ne P(A) P(B)$；此处两者相等，检验未检出相关。
  • (D) Insufficient information信息不足, the addition rule pins $P(A \cap B)$ down uniquely from the three given numbers, so the question is fully determined.：加法法则已由三个已知值唯一确定 $P(A \cap B)$，故题目完全可解。
Independent and mutually exclusive are opposites, not synonyms.独立与互斥是对立而非同义。 If $A$ and $B$ are mutually exclusive (disjoint), then knowing $B$ has occurred tells you $A$ has not occurred, $P(A \mid B) = 0$, which is maximal dependence. Conversely, if $A$ and $B$ are independent, then $P(A \mid B) = P(A)$, $B$ tells you nothing about $A$. Two events with $P(A), P(B) > 0$ can be one or the other or neither, but never both. The SAT/AP loves seeding "mutually exclusive" as a distractor on independence questions because students conflate the two; the moment you see "exclusive" on an independence problem, check the intersection.若 $A$ 与 $B$ 互斥（不相交），则已知 $B$ 发生就意味着 $A$ 没有发生，$P(A \mid B) = 0$，这是最大程度的相关。反之，若 $A$ 与 $B$ 独立，则 $P(A \mid B) = P(A)$，$B$ 对 $A$ 毫无信息。两个 $P(A), P(B) > 0$ 的事件可以是其中之一，也可以都不是，但绝不可能两者兼具。SAT/AP 喜欢在独立性题里放"互斥"当干扰项，因为学生常混淆二者；一看到独立性题里出现"互斥"，就去检查交集。
Q4MEDIUM中 🇨🇦 ON安 ON MDM4U-style安省 MDM4U 风格 §2 Conditional Probability条件概率 · HSS-CP.B.6 [4 marks][4 分]

$7$R, $5$B in a bag; draw $2$ without replacement. (a) $P($ RR $)$. (b) $P($ same colour $)$.袋中 $7$ 红、$5$ 蓝；不放回地抽 $2$ 个。(a) $P($ 红红 $)$。(b) $P($ 同色 $)$。

Answer:答案：  (a) $P(\text{RR}) = \tfrac{7}{22}$  ·  (b) $P(\text{same}) = \tfrac{31}{66}$

(a) Multiplication rule with conditional含条件概率的乘法法则 M1·A1

Without replacement, the second draw depends on the first. The multiplication rule reads不放回时，第二次抽取依赖第一次。乘法法则为 $$ P(R_{1} \cap R_{2}) = P(R_{1}) \cdot P(R_{2} \mid R_{1}). $$ Initially $|S| = 12$ marbles ($7$R + $5$B), so $P(R_{1}) = 7/12$. Given $R_{1}$ occurred, $6$ reds remain in $11$ marbles, so $P(R_{2} \mid R_{1}) = 6/11$. Hence初始 $|S| = 12$ 个球（$7$ 红 + $5$ 蓝），故 $P(R_{1}) = 7/12$。在 $R_{1}$ 发生后，$11$ 个球中余 $6$ 红，故 $P(R_{2} \mid R_{1}) = 6/11$。于是 $$ P(\text{RR}) = \frac{7}{12} \cdot \frac{6}{11} = \frac{42}{132} = \frac{7}{22}. $$

(b) Add the two same-colour branches把两条同色分支相加 M1·A1

"Both same colour" is the union of disjoint events RR and BB. For BB: $P(B_{1}) = 5/12$, $P(B_{2} \mid B_{1}) = 4/11$, so"两个同色"是不相交事件 RR 与 BB 的并。对 BB：$P(B_{1}) = 5/12$，$P(B_{2} \mid B_{1}) = 4/11$，故 $$ P(\text{BB}) = \frac{5}{12} \cdot \frac{4}{11} = \frac{20}{132} = \frac{5}{33}. $$ Add the two disjoint branches:把两条不相交分支相加： $$ P(\text{same}) = P(\text{RR}) + P(\text{BB}) = \frac{42}{132} + \frac{20}{132} = \frac{62}{132} = \frac{31}{66}. $$
Multiply along a branch; add across branches.沿分支相乘，跨分支相加。 A tree diagram is the universal organiser for sequential-event problems. Each branch is an intersection (multiply the conditional probabilities along it), and disjoint branches at the same level get added. The without-replacement giveaway: the denominator in the second factor shrinks by $1$, and the numerator of the matching colour shrinks by $1$. If you instead wrote $\bigl(\tfrac{7}{12}\bigr)^{2}$ for $P(\text{RR})$, you assumed independence (with-replacement), and your answer would be wrong by exactly the amount of dependence introduced by the shrinking sample space. ON MDM4U markers explicitly want the conditional probability $P(R_{2} \mid R_{1}) = 6/11$ named, not just numerically computed.树状图是顺序事件问题的通用整理工具。每条分支是一个交（沿分支把条件概率相乘），同层的不相交分支相加。不放回的标志是：第二个因子的分母减 $1$，对应颜色的分子也减 $1$。若你把 $P(\text{RR})$ 写成 $\bigl(\tfrac{7}{12}\bigr)^{2}$，就是假设了独立（放回），误差恰为样本空间缩小所引入的相关程度。安省 MDM4U 评分明确要求写出条件概率 $P(R_{2} \mid R_{1}) = 6/11$，而非仅算出数值。
Q5MEDIUM中 🇨🇦 AB阿 AB 30-2-style阿省 30-2 风格 §6 z-Scoresz 分数 · Math 30-2 Stats GO 1.6 [4 marks][4 分]

$X \sim N(500, 80^{2})$. (a) $z$ at $620$. (b) $P(X > 660)$ via $68$-$95$-$99.7$.$X \sim N(500, 80^{2})$。(a) $620$ 处的 $z$。(b) 用 $68$-$95$-$99.7$ 求 $P(X > 660)$。

Answer:答案：  (a) $z = 1.5$  ·  (b) $P(X > 660) \approx 0.025$ (i.e. $2.5\%$)（即 $2.5\%$）

(a) Apply the $z$-score formula套用 $z$ 分数公式 M1·A1

The $z$-score measures the number of standard deviations a raw value sits above (or below) the mean:$z$ 分数衡量原始值在均值之上（或之下）多少个标准差： $$ z = \frac{x - \mu}{\sigma} = \frac{620 - 500}{80} = \frac{120}{80} = 1.5. $$ So a raw score of $620$ is $1.5$ standard deviations above the mean.故原始分 $620$ 在均值之上 $1.5$ 个标准差。

(b) Read off the tail probability via the empirical rule用经验法则读出尾部概率 M1·A1

The raw score $660$ has $z = (660 - 500)/80 = 2$. Under $X \sim N(\mu, \sigma^{2})$, the $68$-$95$-$99.7$ rule says原始分 $660$ 的 $z = (660 - 500)/80 = 2$。在 $X \sim N(\mu, \sigma^{2})$ 下，$68$-$95$-$99.7$ 法则给出 $$ P(|X - \mu| \le 2 \sigma) \approx 0.95, \qquad \text{so} \qquad P(X > \mu + 2\sigma) \approx \frac{1 - 0.95}{2} = 0.025. $$ Hence $P(X > 660) \approx 0.025$, i.e. about $2.5\%$.故 $P(X > 660) \approx 0.025$，即约 $2.5\%$。
The empirical rule is the AB Math 30-2 / FPM go-to estimator.经验法则是阿省 Math 30-2 / FPM 的首选估算器。 The $68$-$95$-$99.7$ rule gives quick tail probabilities at exactly $\pm 1\sigma$, $\pm 2\sigma$, $\pm 3\sigma$, and by symmetry, each tail outside $\pm 2\sigma$ holds half of the $5\%$ that's not in the middle $95\%$. The standard normal table $\Phi(2.00) = 0.9772$ gives the more precise $P(X > 660) = 1 - 0.9772 = 0.0228$, but $0.025$ is the curriculum-expected estimate when only the empirical rule is invoked. AB indicator 1.6 explicitly requires this kind of "use the rule to estimate, then if asked, refine with technology" two-step.$68$-$95$-$99.7$ 法则在恰好 $\pm 1\sigma$、$\pm 2\sigma$、$\pm 3\sigma$ 处给出快速尾部概率，并由对称性，$\pm 2\sigma$ 之外每条尾部各占不在中间 $95\%$ 之内那 $5\%$ 的一半。标准正态表 $\Phi(2.00) = 0.9772$ 给出更精确的 $P(X > 660) = 1 - 0.9772 = 0.0228$，但仅用经验法则时课程期望的估值是 $0.025$。阿省指标 1.6 明确要求这种"先用法则估算，必要时再用技术精算"的两步法。
PART II  ·  EXTENDED RESPONSE · SOLUTIONS第二部分  ·  拓展题 · 详解AP-feeder FRQ + ON MDM4U-style · 35 marksAP 衔接简答题 + 安省 MDM4U 风格 · 共 35 分

Section B · Worked SolutionsB 部分 · 详细解答

Q6HARD难Honors for US A2 / AB 30-2美国 A2 / 阿省 30-2 荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Bayes' Rule贝叶斯公式 · HSS-CP.A.3 / .B.6 [8 marks][8 分]

$P(D) = 0.02$; $P(+ \mid D) = 0.95$; $P(+ \mid D^{c}) = 0.04$. (a) Identify prior + likelihoods. (b) $P(+)$. (c) Posterior $P(D \mid +)$. (d) Interpret.$P(D) = 0.02$；$P(+ \mid D) = 0.95$；$P(+ \mid D^{c}) = 0.04$。(a) 确定先验与似然。(b) $P(+)$。(c) 后验 $P(D \mid +)$。(d) 解释。

Answer:答案：  (a) prior $P(D) = 0.02$; likelihoods $0.95$, $0.04$先验 $P(D) = 0.02$；似然 $0.95$、$0.04$  ·  (b) $P(+) = 0.0582$  ·  (c) $P(D \mid +) \approx 0.3265$  ·  (d) base-rate fallacy — rarity of $D$ swamps high sensitivity基率谬误 — $D$ 的稀有性淹没了高灵敏度

(a) Label the Bayesian quantities标注各贝叶斯量 A1

Translate the prose into the three Bayesian roles:把文字转译为三个贝叶斯角色：
  • Prior:先验： $P(D) = 0.02$ (baseline allergy prevalence; what we believed before seeing the test result).$P(D) = 0.02$（过敏基线患病率；看到检测结果之前我们的信念）。
  • Likelihood under $D$:$D$ 下的似然： $P(+ \mid D) = 0.95$ (the test's sensitivity, true-positive rate).$P(+ \mid D) = 0.95$（检测灵敏度，即真阳性率）。
  • Likelihood under $D^{c}$:$D^{c}$ 下的似然： $P(+ \mid D^{c}) = 0.04$ (the false-positive rate; complement of specificity).$P(+ \mid D^{c}) = 0.04$（假阳性率；特异度的补）。

(b) Total probability of a positive test阳性检测的全概率 M1·A1

The events $D$ and $D^{c}$ partition the sample space, so the law of total probability gives事件 $D$ 与 $D^{c}$ 划分样本空间，故由全概率公式 $$ P(+) = P(+ \mid D) P(D) + P(+ \mid D^{c}) P(D^{c}) = 0.95 \cdot 0.02 + 0.04 \cdot 0.98. $$ Compute: $0.95 \cdot 0.02 = 0.019$ and $0.04 \cdot 0.98 = 0.0392$. Hence计算：$0.95 \cdot 0.02 = 0.019$，$0.04 \cdot 0.98 = 0.0392$。故 $$ P(+) = 0.019 + 0.0392 = 0.0582. $$

(c) Bayes' rule for the posterior用贝叶斯公式求后验 M1·A1·A1

Bayes' rule reads贝叶斯公式为 $$ P(D \mid +) = \frac{P(+ \mid D) \, P(D)}{P(+)} = \frac{0.019}{0.0582} \approx 0.3265. $$ So even after a positive screen, the posterior probability of having the allergy is about $32.65\%$, well under half.故即便筛查呈阳性，患过敏的后验概率也仅约 $32.65\%$，远不到一半。

(d) Interpret the base-rate fallacy解释基率谬误 R1·A1

Decompose the numerator and denominator of (c) into "true positives" vs. "false positives":将 (c) 的分子与分母分解为"真阳性"与"假阳性"：
  • True positives: $0.95 \cdot 0.02 = 0.019$ (sensitivity times prior).真阳性：$0.95 \cdot 0.02 = 0.019$（灵敏度乘先验）。
  • False positives: $0.04 \cdot 0.98 = 0.0392$ (false-positive rate times prior of not having the allergy).假阳性：$0.04 \cdot 0.98 = 0.0392$（假阳性率乘未患过敏的先验）。
Because the disease is rare ($P(D) = 0.02$), the pool of healthy people ($P(D^{c}) = 0.98$) is $49$ times larger than the pool of sick people. Even a low $4\%$ false-positive rate applied to that huge healthy pool produces $0.0392$ false positives, more than double the $0.019$ true positives. The posterior $0.019/0.0582 \approx 32.65\%$ reflects this imbalance. Base-rate fallacy: reasoning from the conditional $P(+ \mid D) = 0.95$ to the (mis)conclusion "positive test means $95\%$ chance of allergy" ignores the prior $P(D)$ entirely, that's what Bayes corrects.因为该病罕见（$P(D) = 0.02$），健康人群（$P(D^{c}) = 0.98$）的规模是患病人群的 $49$ 倍。即便仅 $4\%$ 的假阳性率，作用于这庞大的健康人群也产生 $0.0392$ 个假阳性，是 $0.019$ 个真阳性的两倍多。后验 $0.019/0.0582 \approx 32.65\%$ 正反映了这种失衡。基率谬误：从条件概率 $P(+ \mid D) = 0.95$ 推出"阳性即有 $95\%$ 概率过敏"这一（错误）结论，完全忽略了先验 $P(D)$，而这正是贝叶斯所要纠正的。
Posterior $\propto$ likelihood $\times$ prior; rarity wins when sensitivity is finite.后验 $\propto$ 似然 $\times$ 先验；当灵敏度有限时，稀有性占上风。 The qualitative law: as $P(D)$ shrinks, $P(D \mid +)$ shrinks too even when sensitivity is high. To dominate a $4\%$ false-positive rate at $P(D) = 0.02$ you'd need sensitivity well above the practical ceiling, the math is structural, not a quirk of these numbers. AP Stats and HSS-CP.A.3 explicitly call out that conditional probabilities are not symmetric: $P(D \mid +) \ne P(+ \mid D)$, and the Bayes inversion is the lone formula that exchanges them correctly. Whenever you see a "rare disease + sensitive test" prompt, set up the $2 \times 2$ table or the Bayes ratio before reading the multiple-choice answers.定性规律：随着 $P(D)$ 减小，即便灵敏度很高，$P(D \mid +)$ 也随之减小。要在 $P(D) = 0.02$ 时压过 $4\%$ 的假阳性率，所需灵敏度远超实际上限——这是结构性的数学，而非这组数字的偶然。AP 统计与 HSS-CP.A.3 明确指出条件概率不对称：$P(D \mid +) \ne P(+ \mid D)$，而贝叶斯反演是唯一能正确互换二者的公式。每当看到"罕见病 + 高灵敏检测"的题目，先列出 $2 \times 2$ 表或贝叶斯比值，再去看选择项。
Q7HARD难Honors for US A2 / AB 30-2美国 A2 / 阿省 30-2 荣誉级 🇨🇦 ON安 ON MDM4U-style安省 MDM4U 风格 §4 Discrete RV: $E(X)$, $\mathrm{Var}(X)$离散随机变量：$E(X)$、$\mathrm{Var}(X)$ · HSS-MD.A.2–3 [9 marks][9 分]

$X$ takes values $0, 1, 2, 3, 4$ with $P = 0.10, 0.25, 0.30, 0.25, 0.10$. (a) Validity. (b) $E(X)$. (c) $E(X^{2})$. (d) $\mathrm{Var}(X)$ + $\sigma_{X}$. (e) Interpret.$X$ 取值 $0, 1, 2, 3, 4$，对应 $P = 0.10, 0.25, 0.30, 0.25, 0.10$。(a) 有效性。(b) $E(X)$。(c) $E(X^{2})$。(d) $\mathrm{Var}(X)$ 与 $\sigma_{X}$。(e) 解释。

Answer:答案：  (a) valid有效  ·  (b) $E(X) = 2.00$  ·  (c) $E(X^{2}) = 5.30$  ·  (d) $\mathrm{Var}(X) = 1.30$, $\sigma_{X} \approx 1.1402$  ·  (e) long-run average $2$; typical deviation $\approx 1.14$长期平均 $2$；典型偏差 $\approx 1.14$

(a) Validity check有效性检验 A1

All five probabilities are non-negative, and五个概率均非负，且 $$ \sum_{x=0}^{4} P(X = x) = 0.10 + 0.25 + 0.30 + 0.25 + 0.10 = 1.00. \;\checkmark $$ So the distribution is valid (CCSSM HSS-MD.A.1).故该分布有效（CCSSM HSS-MD.A.1）。

(b) Expected value期望值 M1·A1

Apply $E(X) = \sum x \, P(X = x)$:套用 $E(X) = \sum x \, P(X = x)$： $$ E(X) = 0(0.10) + 1(0.25) + 2(0.30) + 3(0.25) + 4(0.10) = 0 + 0.25 + 0.60 + 0.75 + 0.40 = 2.00. $$

(c) Expected value of $X^{2}$$X^{2}$ 的期望值 M1·A1

$$ E(X^{2}) = 0(0.10) + 1(0.25) + 4(0.30) + 9(0.25) + 16(0.10) = 0 + 0.25 + 1.20 + 2.25 + 1.60 = 5.30. $$

(d) Variance and standard deviation方差与标准差 M1·A1·A1

Using the computational formula:使用计算公式： $$ \mathrm{Var}(X) = E(X^{2}) - [E(X)]^{2} = 5.30 - 2.00^{2} = 5.30 - 4.00 = 1.30. $$ $$ \sigma_{X} = \sqrt{\mathrm{Var}(X)} = \sqrt{1.30} \approx 1.1402. $$

(e) Interpret解释 A1

$E(X) = 2$ is the long-run mean: if you observed many independent copies of $X$ and averaged them, the sample mean would tend to $2$. $\sigma_{X} \approx 1.14$ is the typical deviation from that mean: most observed values sit within roughly $\pm 1.14$ of $2$, i.e. in the rough range $[0.86, 3.14]$ which captures the bulk of the distribution.$E(X) = 2$ 是长期均值：若观测 $X$ 的许多独立副本并取平均，样本均值会趋于 $2$。$\sigma_{X} \approx 1.14$ 是相对该均值的典型偏差：多数观测值落在 $2$ 的约 $\pm 1.14$ 范围内，即大致区间 $[0.86, 3.14]$，囊括分布的主体。
$\mathrm{Var}(X) = E(X^{2}) - [E(X)]^{2}$ is the computational shortcut for the defining $E[(X - \mu)^{2}]$.$\mathrm{Var}(X) = E(X^{2}) - [E(X)]^{2}$ 是定义式 $E[(X - \mu)^{2}]$ 的计算捷径。 The expanded form $E[(X - \mu)^{2}]$ is conceptually cleaner but algebraically heavier: you'd compute five deviations $(x - 2)^{2}$, multiply by the probabilities, and sum. The shortcut needs only two sums, $\sum x P$ and $\sum x^{2} P$. MDM4U markers happily award the same A1 for either path, but the computational formula is the one used in the binomial proof $\mathrm{Var}(X) = np(1-p)$ (which factors $E(X^{2})$ via $E[X(X-1)] + E(X)$, see Q8). Lock the shortcut in early; you'll use it in §5 binomial and in the standard-normal moment derivation in IB Math AA HL.展开式 $E[(X - \mu)^{2}]$ 概念上更清晰但代数上更繁重：要算五个偏差 $(x - 2)^{2}$、乘概率、再求和。捷径只需两个求和 $\sum x P$ 与 $\sum x^{2} P$。MDM4U 评分对两条路径同样给 A1，但计算公式正是二项式证明 $\mathrm{Var}(X) = np(1-p)$ 所用的（它通过 $E[X(X-1)] + E(X)$ 分解 $E(X^{2})$，见第 8 题）。尽早把这一捷径记牢；你会在 §5 二项分布以及 IB Math AA HL 的标准正态矩推导中用到它。
Q8MEDIUM中 🇺🇸 US美 🇨🇦 AB阿 AP-feeder FRQAP 衔接简答题 §5 Binomial Distribution二项分布 · HSS-MD.A.3 / Math 30-1 Binomial [9 marks][9 分]

$X \sim B(10, 0.8)$. (a) PMF. (b) $P(X = 8)$. (c) $P(X \ge 9)$. (d) $E(X)$, $\mathrm{Var}(X)$, $\sigma_{X}$. (e) BINS conditions.$X \sim B(10, 0.8)$。(a) 概率质量函数。(b) $P(X = 8)$。(c) $P(X \ge 9)$。(d) $E(X)$、$\mathrm{Var}(X)$、$\sigma_{X}$。(e) BINS 条件。

Answer:答案：  (a) $P(X = k) = \dbinom{10}{k} (0.8)^{k} (0.2)^{10-k}$  ·  (b) $\approx 0.3020$  ·  (c) $\approx 0.3758$  ·  (d) $E(X) = 8$, $\mathrm{Var}(X) = 1.6$, $\sigma_{X} \approx 1.2649$  ·  (e) all four BINS conditions metBINS 四项条件全部满足

(a) Binomial PMF二项分布概率质量函数 A1

$$ P(X = k) = \binom{n}{k} p^{k} (1-p)^{n-k} = \binom{10}{k} (0.8)^{k} (0.2)^{10-k}, \quad k = 0, 1, \ldots, 10. $$

(b) $P(X = 8)$ M1·A1

$$ P(X = 8) = \binom{10}{8} (0.8)^{8} (0.2)^{2} = 45 \cdot 0.16777216 \cdot 0.04. $$ Compute step by step: $0.16777216 \cdot 0.04 = 0.0067108864$, and $45 \cdot 0.0067108864 \approx 0.30199$. Round to four decimals: $P(X = 8) \approx 0.3020$.逐步计算：$0.16777216 \cdot 0.04 = 0.0067108864$，且 $45 \cdot 0.0067108864 \approx 0.30199$。保留四位小数：$P(X = 8) \approx 0.3020$。

(c) $P(X \ge 9)$ M1·A1

$$ P(X = 9) = \binom{10}{9} (0.8)^{9} (0.2)^{1} = 10 \cdot 0.134217728 \cdot 0.2 \approx 0.2684. $$ $$ P(X = 10) = (0.8)^{10} \approx 0.1074. $$ $$ P(X \ge 9) = 0.2684 + 0.1074 = 0.3758. $$

(d) Binomial moments二项分布的矩 M1·A1

$E(X) = np = 10(0.8) = 8$. $\mathrm{Var}(X) = np(1-p) = 10(0.8)(0.2) = 1.6$. $\sigma_{X} = \sqrt{1.6} \approx 1.2649$.

(e) BINS conditionsBINS 条件 A1·A1

The four binomial-model assumptions:二项模型的四项假设：
  • Binary outcomes: each free throw is a make or a miss. $\checkmark$（二值结果）：每次罚球非中即不中。$\checkmark$
  • Independent trials: the prompt asserts "each shot independently". $\checkmark$（独立试验）：题目声明"每次投篮相互独立"。$\checkmark$
  • Number of trials fixed in advance: $n = 10$. $\checkmark$（试验次数预先固定）：$n = 10$。$\checkmark$
  • Success probability constant: $p = 0.8$ on every shot. $\checkmark$（成功概率恒定）：每次投篮 $p = 0.8$。$\checkmark$
All four hold, so the binomial model is appropriate for $X$.四项均成立，故二项模型适用于 $X$。
$E(X) = np$ is the indicator-variable proof you should keep close.$E(X) = np$ 是应当牢记的示性变量证明。 Write $X = \sum_{i=1}^{n} I_{i}$ where $I_{i} = 1$ if trial $i$ is a success and $0$ otherwise. Then $E(I_{i}) = p$ for each trial, and linearity of expectation gives $E(X) = \sum E(I_{i}) = np$, no independence needed. For the variance, $\mathrm{Var}(I_{i}) = p(1-p)$, and by independence of the trials the variances add: $\mathrm{Var}(X) = np(1-p)$. The independence hypothesis appears in the variance but not in the mean, and AP Stats explicitly distinguishes the two. The indicator trick is also how you'd prove these for sums of arbitrary independent Bernoullis with different $p_{i}$, the workhorse of next-step inferential statistics.令 $X = \sum_{i=1}^{n} I_{i}$，其中第 $i$ 次试验成功则 $I_{i} = 1$，否则为 $0$。则每次 $E(I_{i}) = p$，由期望的线性性得 $E(X) = \sum E(I_{i}) = np$，无需独立性。对方差，$\mathrm{Var}(I_{i}) = p(1-p)$，凭试验独立性方差可加：$\mathrm{Var}(X) = np(1-p)$。独立性假设出现在方差中而非均值中，AP 统计明确区分二者。示性变量技巧同样可证明各 $p_{i}$ 不同的任意独立伯努利之和的情形，这是后续推断统计的主力工具。
Q9MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §2 Two-Way Tables双向表 · HSS-CP.B.6 [9 marks][9 分]

$200$ students; bike/no-bike $\times$ walk/no-walk. (a) Marginals. (b) Intersection. (c) Both conditional probabilities. (d) Independence test.$200$ 名学生；有车/无车 $\times$ 走读/非走读。(a) 边缘概率。(b) 交。(c) 两个条件概率。(d) 独立性检验。

Answer:答案：  (a) $P(B) = 0.6$, $P(W) = 0.3$  ·  (b) $P(B \cap W) = 0.225$  ·  (c) $P(W \mid B) = 0.375$, $P(B \mid W) = 0.75$  ·  (d) not independent: $0.225 \ne 0.18$不独立：$0.225 \ne 0.18$

(a) Marginal probabilities from row / column totals由行/列合计求边缘概率 A1·A1

Let $B = $ "owns bike", $W = $ "walks to school". Row totals give令 $B = $ "有自行车"，$W = $ "走读"。行合计给出 $$ P(B) = \frac{120}{200} = 0.6, \qquad P(W) = \frac{60}{200} = 0.3. $$

(b) Joint probability from the intersection cell由交叉格求联合概率 A1

The cell at the intersection of "owns bike" and "walks" is $45$, out of $200$:"有自行车"与"走读"交叉格为 $45$，总数 $200$： $$ P(B \cap W) = \frac{45}{200} = 0.225. $$

(c) Conditional probabilities via row / column restriction用行/列限制求条件概率 M1·A1·A1

Row restriction:行限制： condition on "owns bike" (row total $120$), then within that row ask what fraction walks (cell $45$):以"有自行车"为条件（行合计 $120$），再在该行内问走读所占比例（格 $45$）： $$ P(W \mid B) = \frac{45}{120} = 0.375 = \frac{P(B \cap W)}{P(B)} = \frac{0.225}{0.6}. \;\checkmark $$ Column restriction:列限制： condition on "walks" (column total $60$), then within that column ask what fraction owns a bike (cell $45$):以"走读"为条件（列合计 $60$），再在该列内问有自行车所占比例（格 $45$）： $$ P(B \mid W) = \frac{45}{60} = 0.75 = \frac{P(B \cap W)}{P(W)} = \frac{0.225}{0.3}. \;\checkmark $$ Note $P(W \mid B) \ne P(B \mid W)$, conditional probabilities are not symmetric in their arguments.注意 $P(W \mid B) \ne P(B \mid W)$，条件概率对其两个参数并不对称。

(d) Independence test独立性检验 M1·A1·R1

Compare $P(B \cap W) = 0.225$ to $P(B) P(W) = 0.6 \cdot 0.3 = 0.180$:将 $P(B \cap W) = 0.225$ 与 $P(B) P(W) = 0.6 \cdot 0.3 = 0.180$ 比较： $$ P(B \cap W) = 0.225 \;\;\ne\;\; 0.180 = P(B) P(W). $$ Therefore $B$ and $W$ are not independent. The discrepancy direction $0.225 > 0.180$ means bike-owners are more likely to walk than the marginal would predict , a positive association.故 $B$ 与 $W$ 不独立。差异方向 $0.225 > 0.180$ 表明有自行车者走读的可能性比边缘概率预测的更高，即正相关。
Conditional probability $=$ row-restriction $=$ joint $/$ marginal, all three say the same thing.条件概率 $=$ 行限制 $=$ 联合 $/$ 边缘，三者说的是同一件事。 CCSSM HSS-CP.B.6 explicitly names the "row / column restriction" interpretation because it grounds the abstract formula $P(A \mid B) = P(A \cap B)/P(B)$ in something visual: shrink the sample space from the whole table to one slice, then renormalise. The two-way-table workflow is the AP Stats default for any "given that" prompt: (i) build the table from counts; (ii) restrict to the conditioning row or column; (iii) report the relative frequency. Same answer, no formula required. The independence test in (d) collapses to "joint cell vs. (row $\times$ column) outer product", and any $\chi^{2}$ test of independence in AP Stats is just this comparison scaled up.CCSSM HSS-CP.B.6 明确提出"行/列限制"解释，因为它把抽象公式 $P(A \mid B) = P(A \cap B)/P(B)$ 落到可视化的操作：把样本空间从整张表缩到一个切片，再重新归一化。双向表流程是 AP 统计处理任何"在……条件下"题目的默认方法：(i) 由计数建表；(ii) 限制到条件所在的行或列；(iii) 报告相对频率。答案相同，且不需公式。(d) 中的独立性检验归结为"联合格 vs.（行 $\times$ 列）外积"，AP 统计中任何 $\chi^{2}$ 独立性检验不过是这一比较的放大版。
PART III  ·  MODELING / APPLIED · SOLUTIONS第三部分  ·  建模 / 应用 · 详解Universal · 28 marks通用 · 共 28 分

Section C · Modeling and ApplicationsC 部分 · 建模与应用

Q10MEDIUM中 🇨🇦 AB阿 AB 30-2-style阿省 30-2 风格 §6 Normal Distribution正态分布 · Math 30-2 Stats GO 1 [9 marks][9 分]

$X \sim N(165, 7^{2})$ cm. (a) 1-2-3 SD intervals. (b) $P(158 \le X \le 172)$, $P(151 \le X \le 179)$. (c) $z(176)$, $z(150)$. (d) Tail probabilities. (e) $P(150 \le X \le 176)$ + interpret.$X \sim N(165, 7^{2})$ cm。(a) 1-2-3 个标准差区间。(b) $P(158 \le X \le 172)$、$P(151 \le X \le 179)$。(c) $z(176)$、$z(150)$。(d) 尾部概率。(e) $P(150 \le X \le 176)$ 并解释。

Answer:答案：  (a) $[158, 172]$, $[151, 179]$, $[144, 186]$  ·  (b) $\approx 0.68$, $\approx 0.95$  ·  (c) $z(176) \approx 1.57$, $z(150) \approx -2.14$  ·  (d) $P(X > 176) \approx 0.0580$, $P(X < 150) \approx 0.0161$  ·  (e) $\approx 0.9259$

(a) Empirical-rule intervals经验法则区间 A1

With $\mu = 165$, $\sigma = 7$:取 $\mu = 165$、$\sigma = 7$：
  • $\mu \pm 1\sigma = [165 - 7,\; 165 + 7] = [158, 172]$.
  • $\mu \pm 2\sigma = [165 - 14,\; 165 + 14] = [151, 179]$.
  • $\mu \pm 3\sigma = [165 - 21,\; 165 + 21] = [144, 186]$.

(b) Empirical-rule probabilities经验法则概率 A1·A1

The interval $[158, 172]$ is exactly $\mu \pm 1\sigma$, so $P(158 \le X \le 172) \approx 0.68$.
The interval $[151, 179]$ is exactly $\mu \pm 2\sigma$, so $P(151 \le X \le 179) \approx 0.95$.区间 $[158, 172]$ 恰为 $\mu \pm 1\sigma$，故 $P(158 \le X \le 172) \approx 0.68$。
区间 $[151, 179]$ 恰为 $\mu \pm 2\sigma$，故 $P(151 \le X \le 179) \approx 0.95$。

(c) $z$-scores 分数 M1·A1

$$ z(176) = \frac{176 - 165}{7} = \frac{11}{7} \approx 1.5714 \approx 1.57. $$ $$ z(150) = \frac{150 - 165}{7} = \frac{-15}{7} \approx -2.1429 \approx -2.14. $$

(d) Tail probabilities via $z$-table用 $z$ 表求尾部概率 M1·A1

Standard-normal CDF lookups:查标准正态 CDF：
  • $\Phi(1.57) \approx 0.9418$, so $P(X > 176) = 1 - \Phi(1.57) \approx 0.0582$. (Using the more precise $z = 1.5714$ gives $\approx 0.0580$.)，故 $P(X > 176) = 1 - \Phi(1.57) \approx 0.0582$。（用更精确的 $z = 1.5714$ 得 $\approx 0.0580$。）
  • $\Phi(-2.14) \approx 0.0162$, so $P(X < 150) \approx 0.0161$. (Equivalent to $1 - \Phi(2.14)$ by symmetry.)，故 $P(X < 150) \approx 0.0161$。（由对称性等于 $1 - \Phi(2.14)$。）

(e) Middle probability + interpretation中间概率与解释 M1·A1

The interval $[150, 176]$ is the complement of the two tails:区间 $[150, 176]$ 是两条尾部的补： $$ P(150 \le X \le 176) = 1 - P(X < 150) - P(X > 176) \approx 1 - 0.0161 - 0.0580 = 0.9259. $$ Context:情境： roughly $92.6\%$ of women in this population have heights between $150$ cm and $176$ cm. The upper tail (above $176$ cm) is much thinner than the lower tail because $176$ is $1.57\sigma$ above the mean while $150$ is $2.14\sigma$ below; the interval is not centred on $\mu = 165$.该总体中约 $92.6\%$ 的女性身高在 $150$ cm 到 $176$ cm 之间。上尾（高于 $176$ cm）比下尾薄得多，因为 $176$ 在均值之上 $1.57\sigma$，而 $150$ 在其下 $2.14\sigma$；该区间并非以 $\mu = 165$ 为中心。
The $z$-score reduces every normal problem to one table.$z$ 分数把每个正态问题归约到一张表。 The standard-normal CDF $\Phi(z) = P(Z \le z)$ is the only function you ever need to look up; everything else is arithmetic. Symmetry shortcut: $\Phi(-z) = 1 - \Phi(z)$ for any $z$, so you only need the positive half of the table. AB Math 30-2 indicators 1.7–1.9 explicitly want students to "solve contextual problems using $z$-scores and a standard normal table or technology"; the two-step "standardise $\to$ look up" workflow is the curriculum-named method. Whenever the interval is not symmetric about $\mu$, like $[150, 176]$ here, compute each tail separately and subtract from $1$; do not try to apply the empirical rule to an off-centre interval.标准正态 CDF $\Phi(z) = P(Z \le z)$ 是你唯一需要查的函数；其余皆是算术。对称捷径：对任意 $z$ 有 $\Phi(-z) = 1 - \Phi(z)$，故只需表的正半部分。阿省 Math 30-2 指标 1.7–1.9 明确要求学生"用 $z$ 分数与标准正态表或技术解决情境问题"；"标准化 $\to$ 查表"的两步法是课程指定的方法。每当区间相对 $\mu$ 不对称，如此处的 $[150, 176]$，应分别计算两条尾部再用 $1$ 去减；切勿对偏心区间套用经验法则。
Q11MEDIUM中 🇨🇦 ON安 ON MDM4U-style安省 MDM4U 风格 §7 Descriptive Statistics描述统计 · HSS-ID.A.2–3 [10 marks][10 分]

Data: $4, 6, 7, 7, 8, 8, 8, 9, 9, 9, 10$ ($n = 11$). (a) Mean / median / mode. (b) Range / $Q_{1}$ / $Q_{3}$ / IQR. (c) Sample SD. (d) $1.5 \cdot$ IQR outlier check. (e) Skew.数据：$4, 6, 7, 7, 8, 8, 8, 9, 9, 9, 10$（$n = 11$）。(a) 均值 / 中位数 / 众数。(b) 极差 / $Q_{1}$ / $Q_{3}$ / 四分位距。(c) 样本标准差。(d) $1.5 \cdot$ 四分位距异常值检查。(e) 偏度。

Answer:答案：  (a) $\bar{x} = 85/11 \approx 7.7273$, median $= 8$, modes $8$ and $9$ (bimodal)$\bar{x} = 85/11 \approx 7.7273$，中位数 $= 8$，众数 $8$ 与 $9$（双峰）  ·  (b) range $= 6$, $Q_{1} = 7$, $Q_{3} = 9$, IQR $= 2$极差 $= 6$，$Q_{1} = 7$，$Q_{3} = 9$，四分位距 $= 2$  ·  (c) $s \approx 1.6788$  ·  (d) fences $[4, 12]$; no outliers栅栏 $[4, 12]$；无异常值  ·  (e) slightly left-skewed轻微左偏

(a) Centre measures集中趋势量 M1·A1

Sum: $4 + 6 + 7 + 7 + 8 + 8 + 8 + 9 + 9 + 9 + 10 = 85$.求和：$4 + 6 + 7 + 7 + 8 + 8 + 8 + 9 + 9 + 9 + 10 = 85$。 $$ \bar{x} = \frac{85}{11} \approx 7.7273. $$ With $n = 11$ (odd), the median is the $6$th value of the sorted list: median $= 8$.
Frequencies: $4(1), 6(1), 7(2), 8(3), 9(3), 10(1)$. Both $8$ and $9$ appear $3$ times: bimodal, modes $= 8$ and $9$.$n = 11$（奇数），中位数是排序后第 $6$ 个值：中位数 $= 8$。
频数：$4(1), 6(1), 7(2), 8(3), 9(3), 10(1)$。$8$ 与 $9$ 各出现 $3$ 次：双峰，众数 $= 8$ 与 $9$。

(b) Spread measures离散程度量 M1·A1·A1

Range $= \max - \min = 10 - 4 = 6$.
Lower half (first $5$ values, excluding the median): $4, 6, 7, 7, 8$. Median of lower half $= Q_{1} = 7$.
Upper half (last $5$ values): $8, 9, 9, 9, 10$. Median of upper half $= Q_{3} = 9$.
$\mathrm{IQR} = Q_{3} - Q_{1} = 9 - 7 = 2$.极差 $= \max - \min = 10 - 4 = 6$。
下半部分（前 $5$ 个值，不含中位数）：$4, 6, 7, 7, 8$。下半部分中位数 $= Q_{1} = 7$。
上半部分（后 $5$ 个值）：$8, 9, 9, 9, 10$。上半部分中位数 $= Q_{3} = 9$。
$\mathrm{IQR} = Q_{3} - Q_{1} = 9 - 7 = 2$。

(c) Sample standard deviation样本标准差 M1·A1

Deviations from $\bar{x} = 85/11$, squared (use common denominator $11$ for arithmetic precision):相对 $\bar{x} = 85/11$ 的偏差的平方（用公分母 $11$ 以保证运算精度）： $$ \sum_{i} (x_{i} - \bar{x})^{2} = \frac{1681 + 361 + 64 + 64 + 9 + 9 + 9 + 196 + 196 + 196 + 625}{121} = \frac{3410}{121}. $$ (Numerators: $|4 - 85/11| = 41/11$ so $(41)^{2} = 1681$; $|6 - 85/11| = 19/11$ so $361$; $|7 - 85/11| = 8/11$ so $64$ (twice); $|8 - 85/11| = 3/11$ so $9$ (three times); $|9 - 85/11| = 14/11$ so $196$ (three times); $|10 - 85/11| = 25/11$ so $625$.)（分子：$|4 - 85/11| = 41/11$ 故 $(41)^{2} = 1681$；$|6 - 85/11| = 19/11$ 故 $361$；$|7 - 85/11| = 8/11$ 故 $64$（两次）；$|8 - 85/11| = 3/11$ 故 $9$（三次）；$|9 - 85/11| = 14/11$ 故 $196$（三次）；$|10 - 85/11| = 25/11$ 故 $625$。） $$ s^{2} = \frac{1}{n - 1} \sum (x_{i} - \bar{x})^{2} = \frac{3410}{121 \cdot 10} = \frac{341}{121} \approx 2.8182. $$ $$ s = \sqrt{2.8182} \approx 1.6788. $$

(d) Outlier check异常值检查 M1·A1

Fences:栅栏： $$ Q_{1} - 1.5 \cdot \mathrm{IQR} = 7 - 3 = 4, \qquad Q_{3} + 1.5 \cdot \mathrm{IQR} = 9 + 3 = 12. $$ A data point is an outlier if $x < 4$ or $x > 12$. The minimum is exactly $4$ (on the fence, not below) and the maximum $10$ is well inside the upper fence. No outliers.若 $x < 4$ 或 $x > 12$ 则为异常值。最小值恰为 $4$（在栅栏上，未越过）且最大值 $10$ 远在上栅栏之内。无异常值。

(e) Skew偏度 A1

$\bar{x} \approx 7.7273$ sits below the median $= 8$. The mean is pulled toward the long lower tail (one value at $4$, one at $6$, while the upper tail is short and dense). So the distribution is mildly left-skewed (negatively skewed): mean $<$ median.$\bar{x} \approx 7.7273$ 位于中位数 $= 8$ 之下。均值被拉向较长的下尾（一个值在 $4$、一个在 $6$，而上尾短而密集）。故分布轻微左偏（负偏）：均值 $<$ 中位数。
Centre measures answer different questions; pick the one matching the prompt.不同集中趋势量回答不同问题；选与题意相符的那个。 Mean is sensitive to extremes (any value of $4$ pulls $\bar{x}$ down by $4/11$ relative to a value of $8$); median is robust (a single change to the minimum from $4$ to $0$ leaves the median at $8$); mode finds the most common value (here ambiguous, two modes, the data is bimodal). HSS-ID.A.3 explicitly asks students to "interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers)." The mean/median gap is the standard diagnostic: $|\bar{x} - \text{median}| / s$ is a quick skew detector, here about $(8 - 7.73)/1.68 \approx 0.16$, small, so the skew is mild. Provincial markers reward the explicit comparison sentence in (e).均值对极端值敏感（任一 $4$ 相对 $8$ 把 $\bar{x}$ 拉低 $4/11$）；中位数稳健（把最小值从 $4$ 改为 $0$ 也不改变中位数 $8$）；众数找最常见值（此处含糊，两个众数，数据双峰）。HSS-ID.A.3 明确要求学生"结合数据集情境解读形状、中心与离散的差异，并考虑极端数据点（异常值）的可能影响。"均值/中位数之差是标准诊断：$|\bar{x} - \text{median}| / s$ 是快速偏度判据，此处约 $(8 - 7.73)/1.68 \approx 0.16$，较小，故偏度轻微。省考评分认可 (e) 中明确的比较句。
Q12HARD难Honors for US A2 / AB 30-2美国 A2 / 阿省 30-2 荣誉级 🇺🇸 US美 🇨🇦 AB阿 AP-feeder FRQAP 衔接简答题 §5–§6 Binomial + Normal二项 + 正态 · HSS-MD.A.3 / HSS-IC.A [9 marks][9 分]

$X \sim B(20, 0.05)$. (a) State the model. (b) $P(X = 0)$, $P(X = 1)$. (c) $P(X \ge 2)$ via complement. (d) $E(X)$, $\mathrm{SD}(X)$. (e) $Y \sim B(400, 0.05)$, estimate $P(11 \le Y \le 29)$ via normal approx + empirical rule. (f) Normal-approximation rule of thumb.$X \sim B(20, 0.05)$。(a) 写出模型。(b) $P(X = 0)$、$P(X = 1)$。(c) 用补集求 $P(X \ge 2)$。(d) $E(X)$、$\mathrm{SD}(X)$。(e) $Y \sim B(400, 0.05)$，用正态近似 + 经验法则估计 $P(11 \le Y \le 29)$。(f) 正态近似经验准则。

Answer:答案：  (a) $X \sim B(20, 0.05)$  ·  (b) $P(X = 0) \approx 0.3585$, $P(X = 1) \approx 0.3774$  ·  (c) $P(X \ge 2) \approx 0.2641$  ·  (d) $E(X) = 1$, $\mathrm{SD}(X) \approx 0.9747$  ·  (e) $E(Y) = 20$, $\mathrm{SD}(Y) \approx 4.359$; $P(11 \le Y \le 29) \approx 0.95$  ·  (f) use when $np \ge 10$ and $n(1-p) \ge 10$当 $np \ge 10$ 且 $n(1-p) \ge 10$ 时使用

(a) Binomial model二项模型 A1

Each bulb is defective independently with probability $p = 0.05$, and the inspector samples a fixed $n = 20$. So $X \sim B(20, 0.05)$ (BINS: binary, independent, $n$ fixed, $p$ constant).每只灯泡以概率 $p = 0.05$ 独立地为次品，检验员抽取固定的 $n = 20$ 只。故 $X \sim B(20, 0.05)$（BINS：二值、独立、$n$ 固定、$p$ 恒定）。

(b) PMF at $k = 0$ and $k = 1$$k = 0$ 与 $k = 1$ 处的概率质量 A1

$$ P(X = 0) = \binom{20}{0} (0.05)^{0} (0.95)^{20} = (0.95)^{20} \approx 0.3585. $$ $$ P(X = 1) = \binom{20}{1} (0.05)^{1} (0.95)^{19} = 20 \cdot 0.05 \cdot (0.95)^{19} = 1 \cdot (0.95)^{19} \approx 0.3774. $$ (Numerical: $(0.95)^{19} = (0.95)^{20} / 0.95 \approx 0.3585 / 0.95 \approx 0.3774$.)（数值：$(0.95)^{19} = (0.95)^{20} / 0.95 \approx 0.3585 / 0.95 \approx 0.3774$。）

(c) Complement trick for "at least two"用补集技巧求"至少两个" M1·A1

$$ P(X \ge 2) = 1 - P(X = 0) - P(X = 1) \approx 1 - 0.3585 - 0.3774 = 0.2641. $$ So about $26.41\%$ of samples have $2$ or more defects, more than one in four batches will trip a "two defects" inspection threshold.故约 $26.41\%$ 的样本含 $2$ 个或更多次品，即超过四分之一的批次会触发"两个次品"的检验阈值。

(d) Mean and SD of $X$$X$ 的均值与标准差 M1·A1

$$ E(X) = np = 20 \cdot 0.05 = 1.00, \qquad \mathrm{Var}(X) = np(1-p) = 20 \cdot 0.05 \cdot 0.95 = 0.95. $$ $$ \mathrm{SD}(X) = \sqrt{0.95} \approx 0.9747. $$

(e) Normal approximation for $Y$$Y$ 的正态近似 M1·A1

$E(Y) = 400 \cdot 0.05 = 20$ and $\mathrm{Var}(Y) = 400 \cdot 0.05 \cdot 0.95 = 19$, so $\mathrm{SD}(Y) = \sqrt{19} \approx 4.3589$. Use the normal approximation $Y \approx N(20, 19)$ and recognise the asked interval relative to $\mu_{Y} \pm 2 \sigma_{Y}$:$E(Y) = 400 \cdot 0.05 = 20$ 且 $\mathrm{Var}(Y) = 400 \cdot 0.05 \cdot 0.95 = 19$，故 $\mathrm{SD}(Y) = \sqrt{19} \approx 4.3589$。采用正态近似 $Y \approx N(20, 19)$，并将所问区间对照 $\mu_{Y} \pm 2 \sigma_{Y}$： $$ \mu_{Y} - 2 \sigma_{Y} \approx 20 - 8.72 = 11.28, \qquad \mu_{Y} + 2 \sigma_{Y} \approx 20 + 8.72 = 28.72. $$ The interval $[11, 29]$ is approximately $\mu_{Y} \pm 2\sigma_{Y}$ (slightly wider). By the $68$-$95$-$99.7$ rule,区间 $[11, 29]$ 约等于 $\mu_{Y} \pm 2\sigma_{Y}$（略宽）。由 $68$-$95$-$99.7$ 法则， $$ P(11 \le Y \le 29) \;\approx\; P(\mu_{Y} - 2\sigma_{Y} \le Y \le \mu_{Y} + 2\sigma_{Y}) \;\approx\; 0.95. $$

(f) Normal approximation rule of thumb正态近似经验准则 A1

The normal approximation $Y \approx N(np, np(1-p))$ is appropriate when both $np \ge 10$ and $n(1-p) \ge 10$ (the "large-sample" rule of thumb taught in AP Stats and AB Math 30-2). Check for our $Y$: $np = 20 \ge 10$ and $n(1-p) = 380 \ge 10$, both satisfied, so the approximation is justified. For $X \sim B(20, 0.05)$ in parts (a)–(d), $np = 1$ fails the rule and the normal approximation would be poor, exactly why parts (b)–(c) use the exact binomial PMF.当 $np \ge 10$ 且 $n(1-p) \ge 10$ 时，正态近似 $Y \approx N(np, np(1-p))$ 适用（AP 统计与阿省 Math 30-2 教授的"大样本"经验准则）。对我们的 $Y$ 验证：$np = 20 \ge 10$ 且 $n(1-p) = 380 \ge 10$，两者皆满足，故近似成立。对 (a)–(d) 中的 $X \sim B(20, 0.05)$，$np = 1$ 不满足准则，正态近似会很差，这正是 (b)–(c) 使用精确二项概率质量的原因。
The binomial transitions to the normal as $np$, $n(1-p)$ both grow.随着 $np$、$n(1-p)$ 同时增大，二项分布过渡到正态分布。 This is the de Moivre–Laplace special case of the central limit theorem: a sum of $n$ independent Bernoullis is approximately normal once neither tail is too thin. The $np \ge 10$, $n(1-p) \ge 10$ rule of thumb is calibrated so the binomial PMF is well-approximated by the normal density across the bulk of the distribution; for our $Y \sim B(400, 0.05)$ it holds, and the empirical-rule estimate $0.95$ for $\mu \pm 2\sigma$ matches an exact binomial computation to about $0.5\%$. For our $X \sim B(20, 0.05)$, $np = 1 \ll 10$, and the normal approximation would be terrible (the distribution is sharply right-skewed), so always check the rule before invoking the normal. AP Stats grades this as a separate point: state the rule, verify it, then apply the approximation.这是中心极限定理的棣莫弗–拉普拉斯特例：当两条尾部都不过薄时，$n$ 个独立伯努利之和近似正态。$np \ge 10$、$n(1-p) \ge 10$ 的经验准则经过校准，使二项概率质量在分布主体上能被正态密度良好逼近；对我们的 $Y \sim B(400, 0.05)$ 它成立，且经验法则给出的 $\mu \pm 2\sigma$ 估值 $0.95$ 与精确二项计算相差约 $0.5\%$。对我们的 $X \sim B(20, 0.05)$，$np = 1 \ll 10$，正态近似会很糟（分布严重右偏），故在调用正态前务必检验准则。AP 统计将此单列一分：陈述准则、验证、然后再应用近似。