Practice Questions · SAT · AP-Feeder · ON / AB Provincial Styles练习题集 · SAT 风格 · AP 衔接 · 安大略 / 阿尔伯塔省考风格
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安大略 MDM4U 风格AB 30-2-style阿尔伯塔 30-2 风格Honors for US A2 / AB 30-2美 A2 / 阿 30-2 荣誉级
Name:姓名：Date:日期：
PART I · SHORT RESPONSE第一部分 · 短答题SAT-style MCQ + ON/AB short answer · 18 marksSAT 风格选择题 + 安/阿省考短答 · 共 18 分
Section A · Short ResponseA 部分 · 短答题
Mix of multiple-choice and short-answer items. For MCQs, circle the letter; show enough work in the margin that a marker could verify. For short-answer items, write probabilities as exact fractions or to four decimal places when irrational. No calculator on Q1–Q3; calculator permitted on Q4–Q5.本节包含选择题与短答题。选择题请圈出字母答案，并在空白处写下足以让阅卷人核对的过程。短答题中概率请写成精确分数，或在无理时保留四位小数。Q1–Q3 不可使用计算器；Q4–Q5 可用计算器。
Two events $A$ and $B$ have $P(A) = 0.6$, $P(B) = 0.5$, and $P(A \cup B) = 0.8$. Which of the following statements is true?两个事件 $A$ 与 $B$ 满足 $P(A) = 0.6$、$P(B) = 0.5$、$P(A \cup B) = 0.8$。下列哪一项陈述正确？
(A)$A$ and $B$ are mutually exclusive.$A$ 与 $B$ 互斥。
(B)$A$ and $B$ are independent because $P(A \cap B) = P(A)P(B)$.$A$ 与 $B$ 独立，因为 $P(A \cap B) = P(A)P(B)$。
(C)$A$ and $B$ are dependent because $P(A \cap B) \ne P(A)P(B)$.$A$ 与 $B$ 不独立，因为 $P(A \cap B) \ne P(A)P(B)$。
(D)The given information is insufficient to decide.所给信息不足以判断。
A bag contains $7$ red and $5$ blue marbles. Two marbles are drawn at random without replacement.一个袋中装有 $7$ 颗红球和 $5$ 颗蓝球。不放回地随机抽取两颗球。
(a)Find $P($ first red and second red $)$. State the conditional probability used.求 $P($ 第一次红 且 第二次红 $)$，并写出所用到的条件概率。[2]
(b)Find $P($ both same colour $)$.求 $P($ 两球同色 $)$。[2]
Q5MEDIUM中🇨🇦 AB阿AB 30-2-style阿尔伯塔 30-2 风格§6 z-Scores$z$ 分数 · Math 30-2 Stats GO 1.6[4 marks][4 分]
A standardised test has scores modelled by a normal distribution with mean $\mu = 500$ and standard deviation $\sigma = 80$, i.e. $X \sim N(500, 80^{2})$.某标准化考试成绩服从正态分布，均值 $\mu = 500$，标准差 $\sigma = 80$，即 $X \sim N(500, 80^{2})$。
(a)Compute the $z$-score of a raw score of $620$.求原始分 $620$ 的 $z$ 分数。[2]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + ON MDM4U-style · 35 marksAP 衔接简答题 + 安大略 MDM4U 风格 · 共 35 分
Section B · Extended ResponseB 部分 · 简答题
Show every algebraic step. State assumptions (in particular, independence) explicitly before applying $P(A \cap B) = P(A)P(B)$. For Bayes-type problems, label prior, likelihood, and posterior. Calculator permitted throughout Part II.每一步代数运算都要写出。在使用 $P(A \cap B) = P(A)P(B)$ 之前，先明确写出所做假设（尤其是独立性）。对于贝叶斯类问题，请标注先验、似然与后验。第二部分全程可用计算器。
Q6HARD难Honors for US A2 / AB 30-2美 A2 / 阿 30-2 荣誉级🇺🇸 US美AP-feeder FRQAP 衔接简答题§2 Bayes' Rule贝叶斯法则 · HSS-CP.A.3 / .B.6[8 marks][8 分]
A rare allergy affects $2\%$ of a population, so $P(D) = 0.02$. A screening test for the allergy has sensitivity $P(+ \mid D) = 0.95$ and false-positive rate $P(+ \mid D^{c}) = 0.04$.某种罕见过敏症在某人群中的患病率为 $2\%$，即 $P(D) = 0.02$。针对该过敏的筛查测试灵敏度为 $P(+ \mid D) = 0.95$，假阳性率为 $P(+ \mid D^{c}) = 0.04$。
(a)Identify the prior probability $P(D)$ and the two likelihoods $P(+ \mid D)$ and $P(+ \mid D^{c})$ explicitly.分别明确写出先验概率 $P(D)$ 和两个似然 $P(+ \mid D)$、$P(+ \mid D^{c})$。[1]
(b)Use the law of total probability to compute $P(+)$.利用全概率公式计算 $P(+)$。[2]
(c)Apply Bayes' rule to compute the posterior probability $P(D \mid +)$. Round to four decimal places.用贝叶斯法则计算后验概率 $P(D \mid +)$，结果保留四位小数。[3]
(d)Interpret the result in context. Comment on the base-rate fallacy: why does a positive test on a $95\%$-sensitive screen still yield a posterior under $50\%$?结合情境解释结果，并评论基率谬误：为何在灵敏度 $95\%$ 的筛查上测出阳性，后验仍低于 $50\%$？[2]
Q7HARD难Honors for US A2 / AB 30-2美 A2 / 阿 30-2 荣誉级🇨🇦 ON安ON MDM4U-style安大略 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]
A discrete random variable $X$ has the probability distribution given below.离散随机变量 $X$ 的概率分布如下表。
$x$
$0$
$1$
$2$
$3$
$4$
$P(X = x)$
$0.10$
$0.25$
$0.30$
$0.25$
$0.10$
(a)Verify that the distribution is valid: each $P(X = x) \ge 0$ and $\sum P(X = x) = 1$.验证该分布有效：每个 $P(X = x) \ge 0$，且 $\sum P(X = x) = 1$。[1]
(b)Compute the expected value $E(X) = \sum x P(X = x)$.求期望 $E(X) = \sum x P(X = x)$。[2]
(d)Use the computational formula $\mathrm{Var}(X) = E(X^{2}) - [E(X)]^{2}$ to compute the variance and the standard deviation $\sigma_{X} = \sqrt{\mathrm{Var}(X)}$.利用计算公式 $\mathrm{Var}(X) = E(X^{2}) - [E(X)]^{2}$ 求方差，并求标准差 $\sigma_{X} = \sqrt{\mathrm{Var}(X)}$。[3]
(e)Interpret $E(X)$ and $\sigma_{X}$ in one sentence each (long-run-average vs. typical-deviation).分别用一句话解释 $E(X)$ 与 $\sigma_{X}$ 的含义（长期均值 vs. 典型偏离）。[1]
A free-throw shooter makes each shot independently with probability $p = 0.8$. Let $X$ count the number of made shots in $n = 10$ attempts, so $X \sim B(10, 0.8)$.一名罚球手每次罚球的命中相互独立，且命中概率 $p = 0.8$。设 $X$ 为 $n = 10$ 次罚球中命中的次数，则 $X \sim B(10, 0.8)$。
(a)State the binomial PMF $P(X = k) = \dbinom{n}{k} p^{k} (1-p)^{n-k}$ with $n$ and $p$ substituted.将 $n$ 与 $p$ 代入，写出二项分布的概率质量函数 $P(X = k) = \dbinom{n}{k} p^{k} (1-p)^{n-k}$。[1]
(b)Compute $P(X = 8)$ exactly, then to four decimals.先精确计算 $P(X = 8)$，再保留四位小数。[2]
(d)State $E(X) = np$ and $\mathrm{Var}(X) = np(1-p)$ for this $X$, and compute the standard deviation $\sigma_{X}$.写出此 $X$ 的 $E(X) = np$ 与 $\mathrm{Var}(X) = np(1-p)$，并求标准差 $\sigma_{X}$。[2]
(e)State the four binomial conditions (BINS: Binary outcomes, Independent trials, fixed Number of trials, constant Success probability) and verify each holds for this shooter.写出二项模型的四个条件（BINS：B—二元结局、I—独立试验、N—试验次数固定、S—成功概率恒定），并对该罚球手逐项验证。[2]
A school surveys $200$ students about whether they own a bicycle and whether they walk to school. The results are tabulated below.某学校对 $200$ 名学生进行调查，记录其是否拥有自行车以及是否步行上学。结果如下表。
Walks to school步行上学
Does not walk不步行
Total合计
Owns a bike有自行车
$45$
$75$
$120$
No bike无自行车
$15$
$65$
$80$
Total合计
$60$
$140$
$200$
(a)Find $P($ owns a bike $)$ and $P($ walks $)$.求 $P($ 有自行车 $)$ 与 $P($ 步行 $)$。[2]
(b)Find $P($ owns a bike $\cap$ walks $)$.求 $P($ 有自行车 $\cap$ 步行 $)$。[1]
(c)Find $P($ walks $\mid$ owns a bike $)$ and $P($ owns a bike $\mid$ walks $)$. Show the row-restriction (or column-restriction) interpretation explicitly.求 $P($ 步行 $\mid$ 有自行车 $)$ 与 $P($ 有自行车 $\mid$ 步行 $)$，并明确写出"行限制（或列限制）"的解释方式。[3]
(d)Are the events "owns a bike" and "walks to school" independent? Use the test $P(A \cap B) \stackrel{?}{=} P(A)P(B)$ and justify your conclusion numerically.事件"有自行车"与"步行上学"是否独立？用检验 $P(A \cap B) \stackrel{?}{=} P(A)P(B)$ 并以数值论证结论。[3]
PART III · MODELING / APPLIED第三部分 · 建模与应用Universal · 28 marks通用题型 · 共 28 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Name your variables (with units) before writing equations. For normal-distribution work, sketch the bell curve and shade the region of interest before reaching for the $z$-table or calculator. Calculator permitted throughout Part III. For Q12 you may use either binomial-exact or normal-approximation reasoning where indicated.在写方程前，先定义变量名（含单位）。处理正态分布问题时，请先画出钟形曲线并标出所求区域，再使用 $z$ 表或计算器。第三部分全程可用计算器。Q12 在标明处可使用二项精确计算或正态近似两种思路。
Q10MEDIUM中🇨🇦 AB阿AB 30-2-style阿尔伯塔 30-2 风格§6 Normal Distribution正态分布 · Math 30-2 Stats GO 1[9 marks][9 分]
The heights of adult women in a population are modelled by a normal distribution $X \sim N(165, 7^{2})$ cm, so $\mu = 165$ cm and $\sigma = 7$ cm.某人群成年女性身高服从正态分布 $X \sim N(165, 7^{2})$ cm，即 $\mu = 165$ cm，$\sigma = 7$ cm。
(a)State the three intervals on either side of $\mu$ corresponding to the $68$-$95$-$99.7$ rule.写出对应 $68$-$95$-$99.7$ 法则、以 $\mu$ 为中心的三个区间。[1]
(b)Use the $68$-$95$-$99.7$ rule to estimate $P(158 \le X \le 172)$ and $P(151 \le X \le 179)$.利用 $68$-$95$-$99.7$ 法则估算 $P(158 \le X \le 172)$ 与 $P(151 \le X \le 179)$。[2]
(c)Compute the $z$-score of a height of $176$ cm and the $z$-score of a height of $150$ cm.分别计算身高 $176$ cm 与 $150$ cm 的 $z$ 分数。[2]
(d)Using the $z$-table, find $P(X > 176)$ and $P(X < 150)$ to four decimals.查 $z$ 表求 $P(X > 176)$ 与 $P(X < 150)$，结果保留四位小数。[2]
(e)Find $P(150 \le X \le 176)$ to four decimals, and interpret the result in context.求 $P(150 \le X \le 176)$，保留四位小数，并结合情境解释。[2]
A teacher records the quiz scores (out of $10$) of $11$ students on a recent quiz:某老师记录了 $11$ 名学生在最近一次小测中的成绩（满分 $10$ 分）：
$$ 4, \;6, \;7, \;7, \;8, \;8, \;8, \;9, \;9, \;9, \;10. $$
(a)Compute the mean $\bar{x}$, the median, and the mode.求均值 $\bar{x}$、中位数与众数。[2]
(b)Compute the range, and the quartiles $Q_{1}$ and $Q_{3}$. State the IQR $= Q_{3} - Q_{1}$.求极差，以及四分位数 $Q_{1}$ 与 $Q_{3}$，并写出四分位距（IQR） $= Q_{3} - Q_{1}$。[3]
(c)Compute the sample standard deviation $s = \sqrt{\dfrac{1}{n-1} \sum (x_{i} - \bar{x})^{2}}$ to four decimal places.求样本标准差 $s = \sqrt{\dfrac{1}{n-1} \sum (x_{i} - \bar{x})^{2}}$，保留四位小数。[2]
(d)Use the $1.5 \cdot \mathrm{IQR}$ rule to identify any outliers: a data point $x$ is an outlier if $x < Q_{1} - 1.5 \cdot \mathrm{IQR}$ or $x > Q_{3} + 1.5 \cdot \mathrm{IQR}$. State the fences and list any outliers.用 $1.5 \cdot \mathrm{IQR}$ 法则判定异常值：当 $x < Q_{1} - 1.5 \cdot \mathrm{IQR}$ 或 $x > Q_{3} + 1.5 \cdot \mathrm{IQR}$ 时为异常值。写出上下围栏并列出所有异常值。[2]
(e)Compare mean and median: explain in one sentence whether the distribution is roughly symmetric, left-skewed, or right-skewed, citing the relative positions of $\bar{x}$ and the median.比较均值与中位数：用一句话说明分布是大致对称、左偏还是右偏，依据是 $\bar{x}$ 与中位数的相对位置。[1]
Q12HARD难Honors for US A2 / AB 30-2美 A2 / 阿 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]
A factory produces light bulbs. Historically, $5\%$ of bulbs are defective; assume defects across bulbs are independent. A quality-control inspector samples $n = 20$ bulbs from a production run. Let $X$ count defective bulbs in the sample.某工厂生产灯泡。历史数据显示 $5\%$ 的灯泡有瑕疵；设各灯泡是否有瑕疵相互独立。某质检员从一批产品中抽取 $n = 20$ 只灯泡。设 $X$ 为该样本中有瑕疵的灯泡数。
(a)State the distribution of $X$ in the form $X \sim B(n, p)$.以 $X \sim B(n, p)$ 的形式写出 $X$ 的分布。[1]
(b)Compute $P(X = 0)$ (no defects) and $P(X = 1)$ to four decimals.计算 $P(X = 0)$（无瑕疵）与 $P(X = 1)$，保留四位小数。[1]
(c)Compute $P(X \ge 2)$ using the complement: $P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$. Give your answer to four decimals.用补集求 $P(X \ge 2)$：$P(X \ge 2) = 1 - P(X = 0) - P(X = 1)$，结果保留四位小数。[2]
(d)Compute $E(X) = np$ and $\mathrm{SD}(X) = \sqrt{np(1-p)}$ for this sample.求本样本的 $E(X) = np$ 与 $\mathrm{SD}(X) = \sqrt{np(1-p)}$。[2]
(e)The inspector now examines a larger run of $n = 400$ bulbs. Let $Y$ be the number of defective bulbs in the larger run, $Y \sim B(400, 0.05)$. State $E(Y)$ and $\mathrm{SD}(Y)$, and use a normal approximation $Y \approx N(\mu_{Y}, \sigma_{Y}^{2})$ together with the $68$-$95$-$99.7$ rule to estimate $P(11 \le Y \le 29)$ (use the rule of thumb that this interval is approximately $\mu_{Y} \pm 2 \sigma_{Y}$).质检员转而检查更大批次的 $n = 400$ 只灯泡。设 $Y$ 为这批中有瑕疵的数量，$Y \sim B(400, 0.05)$。写出 $E(Y)$ 与 $\mathrm{SD}(Y)$，并用正态近似 $Y \approx N(\mu_{Y}, \sigma_{Y}^{2})$ 结合 $68$-$95$-$99.7$ 法则估算 $P(11 \le Y \le 29)$（提示：该区间近似为 $\mu_{Y} \pm 2 \sigma_{Y}$）。[2]
(f)Comment in one sentence on when the normal approximation $Y \approx N(np, np(1-p))$ is appropriate (cite the standard rule of thumb $np \ge 10$ and $n(1-p) \ge 10$).用一句话说明正态近似 $Y \approx N(np, np(1-p))$ 何时适用（引用经验法则 $np \ge 10$ 且 $n(1-p) \ge 10$）。[1]
🇺🇸 US Common Core美国共同核心HSS-CP.A.1–5 · HSS-CP.B.6–7 · HSS-MD.A.1–4 (+) · HSS-IC.A.1–2 · HSS-ID.A.2–3
🇨🇦 Alberta阿尔伯塔Math 30-1 Perm / Comb / Binomial Theorem (combinatorial backbone) · Math 30-2 Probability GO 1–3 · Math 30-2 Statistics GO 1 (normal, $\sigma$, $z$-scores)Math 30-1 排列 / 组合 / 二项式定理（组合学骨架）· Math 30-2 概率 GO 1–3 · Math 30-2 统计 GO 1（正态分布、$\sigma$、$z$ 分数）
Honors flag荣誉级标记Bayes (§2) and full discrete variance (§4) are honors for US Algebra II / Stats and AB Math 30-2. Core for ON MDM4U and AP Stats. MBF3C is college prep and does not cover this material.贝叶斯法则（§2）与完整的离散方差（§4）对美国 Algebra II / Stats 与阿尔伯塔 Math 30-2 而言属于荣誉级，对安大略 MDM4U 与 AP Stats 则是核心内容。MBF3C 是大专衔接课，不涉及此部分。
Full 4-column Syllabus Map lives in ../Study Guides/Unit_13_Probability_and_Statistics_Foundations.html.完整的四列大纲对照表见 ../Study Guides/Unit_13_Probability_and_Statistics_Foundations.html。