v1.1 — 12 questions · 91 marks across IB Paper 1A / 1B / 2 / 3 HL. HL-only material (variance, continuous PDFs, linear transformations of RVs) marked with `HL` pill. Companion Solutions file in Solutions/.v1.1 —— 12 道题 · 91 分，覆盖 IB Paper 1A / 1B / 2 / 3 HL。仅 HL 的内容（方差、连续型概率密度函数（pdf）、随机变量的线性变换）以 `HL` 标签标出。配套答案文件位于 Solutions/ 目录。
PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 22 marks不可使用计算器 · 简答题 · 22 分
Section A — Short ResponseA 节 —— 简答题
Leave answers as exact fractions where natural. No calculator permitted; where the normal distribution appears, any required tabulated value is provided in the question.能写成精确分数时，请保留精确分数形式。不可使用计算器；涉及正态分布（normal distribution）时，题目中会给出所需的查表值。
Q1EASYPaper 1A4.7 Discrete RV / $E(X)$[6 marks]
A discrete random variable $X$ has the probability distribution离散型随机变量（discrete random variable）$X$ 的概率分布如下：
$x$
$1$
$2$
$3$
$4$
$P(X = x)$
$0.2$
$0.3$
$k$
$0.1$
(a)Find $k$.求 $k$。[2]
(b)Find $P(X \le 2)$.求 $P(X \le 2)$。[1]
(c)Find $E(X)$.求期望（E(X)，expected value）。[3]
Q2EASYPaper 1A4.7 Fair Game[5 marks]
A fair coin is tossed. If it lands heads, the player wins $\$5$; if tails, the player loses $\$3$.抛掷一枚均匀硬币（fair coin）。若出现正面，玩家赢 $\$5$；若出现反面，玩家输 $\$3$。
(a)Find the expected profit per toss.求每次抛掷的期望收益（expected profit）。[2]
(b)State whether the game is fair, and explain.判断该游戏是否为公平博弈（fair game），并说明理由。[1]
(c)Find the win amount $\$w$ for heads (the loss for tails staying at $\$3$) that would make the game fair.求正面时的赢款 $\$w$（反面输款保持为 $\$3$），使博弈成为公平博弈。[2]
Q3MEDIUMPaper 1A4.8 Binomial Exact[6 marks]
A biased coin lands heads with probability $\dfrac{1}{3}$. It is tossed $3$ times. Let $X$ be the number of heads.一枚有偏硬币（biased coin）出现正面的概率为 $\dfrac{1}{3}$。抛掷 $3$ 次，设 $X$ 为正面出现的次数。
(a)Write out the probability distribution of $X$ in a table, as exact fractions over $27$.以分母为 $27$ 的精确分数形式，列表写出 $X$ 的概率分布。[3]
(b)Find $E(X)$ either from the table or by quoting $np$.通过列表计算或直接引用 $np$，求 $E(X)$。[1]
(c)Find $P(X \ge 1)$.求 $P(X \ge 1)$。[2]
Q4MEDIUMPaper 1A4.9 Normal (with tabulated $z$)[5 marks]
$X \sim N(50, 100)$. You are given $P(Z < 1) = 0.8413$, where $Z = \dfrac{X - \mu}{\sigma}$.$X \sim N(50, 100)$。已知 $P(Z < 1) = 0.8413$，其中 $Z = \dfrac{X - \mu}{\sigma}$ 为标准化（z-score）后的变量。
(a)Find the $z$-score corresponding to $x = 60$.求 $x = 60$ 对应的 z 值（z-score）。[1]
(b)Hence find $P(X < 60)$.由此求 $P(X < 60)$。[1]
(c)Use symmetry of the standard normal to find $P(40 < X < 60)$.利用标准正态分布（standard normal）的对称性，求 $P(40 < X < 60)$。[3]
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 23 marks不可使用计算器 · 长答题 · 23 分
Section B — Extended ResponseB 节 —— 长答题
Q5MEDIUMPaper 1B4.8 Binomial Cumulative[7 marks]
A field-hockey player has scoring rate $\dfrac{1}{2}$ per penalty stroke. She takes $5$ independent strokes. Let $X$ be the number of scores.某曲棍球队员每次罚球的得分率为 $\dfrac{1}{2}$。她独立地罚 $5$ 次球，设 $X$ 为得分次数，则 $X$ 服从二项分布（binomial distribution，B(n, p)）。
(a)Find $P(X = 2)$ as an exact fraction over $32$.以分母为 $32$ 的精确分数形式，求 $P(X = 2)$。[2]
(b)Find $P(X \le 1)$.求 $P(X \le 1)$。[2]
(c)Find $P(X \ge 3)$.求 $P(X \ge 3)$。[2]
(d)Use the symmetry of $\text{B}(5, \tfrac{1}{2})$ to explain why $P(X \ge 3) = P(X \le 2)$ — and check this against your answers.利用 $\text{B}(5, \tfrac{1}{2})$ 的对称性，说明为何 $P(X \ge 3) = P(X \le 2)$，并用上面的答案进行验证。[1]
Q6HARDPaper 1BAHL 4.11 Variance[8 marks]HL
A discrete random variable $X$ has distribution离散型随机变量 $X$ 的分布为
$x$
$-1$
$0$
$1$
$2$
$P(X = x)$
$\dfrac{1}{8}$
$\dfrac{3}{8}$
$\dfrac{3}{8}$
$\dfrac{1}{8}$
(a)Verify that the probabilities sum to $1$.验证各概率之和为 $1$。[1]
(b)Find $E(X)$.求 $E(X)$。[2]
(c)Find $E(X^2)$ and hence $\text{Var}(X)$ using $\text{Var}(X) = E(X^2) - [E(X)]^2$.求 $E(X^2)$，并由此利用 $\text{Var}(X) = E(X^2) - [E(X)]^2$ 求方差（Var(X)）。[3]
A continuous random variable $X$ has probability density function
$$ f(x) \;=\; \begin{cases} kx & \text{if } 0 \le x \le 2, \\ 0 & \text{otherwise}. \end{cases} $$连续型随机变量（continuous random variable）$X$ 的概率密度函数（probability density function，pdf）为
$$ f(x) \;=\; \begin{cases} kx & \text{if } 0 \le x \le 2, \\ 0 & \text{otherwise}. \end{cases} $$
(a)Show that $k = \dfrac{1}{2}$.证明 $k = \dfrac{1}{2}$。[3]
(b)Find $P(X > 1)$.求 $P(X > 1)$。[2]
(c)Find the median $m$ of $X$.求 $X$ 的中位数（median）$m$。[3]
PART III · PAPER 2第三部分 · 第二卷Calculator (GDC) · 15 marks可使用计算器（GDC） · 15 分
Section C — Paper 2 (Calculator)C 节 —— 第二卷（可用计算器）
Use GDC normal-distribution routines (normalcdf and invNorm) where appropriate. State $\mu, \sigma$, and any tail/interval before reading off the value.合适时使用图形计算器（GDC）的正态分布命令（normalcdf 与 invNorm）。读出数值前，请先写明 $\mu$、$\sigma$ 以及所求的尾区间。
Q8EASYPaper 24.9 Normal Interval[4 marks]
The mass of a packaged biscuit is modelled by $X \sim N(50,\, 10^2)$ grams.某包装饼干的质量服从正态分布 $X \sim N(50,\, 10^2)$（单位：克）。
(a)Find $P(45 < X < 55)$, to 3 s.f.求 $P(45 < X < 55)$，保留 3 位有效数字。[3]
(b)Interpret this probability in context, in one sentence.结合题境，用一句话解释该概率的实际意义。[1]
Q9MEDIUMPaper 24.9 Inverse Normal[5 marks]
Adult male heights in a population are modelled by $X \sim N(170,\, 7^2)$ cm.某人群成年男性身高服从 $X \sim N(170,\, 7^2)$（单位：cm）。
(b)Find the height exceeded by exactly $10\%$ of this population, to 1 d.p. (cm).求该人群中恰有 $10\%$ 的人超过的身高（精确到小数点后 1 位，单位 cm）。[3]
Q10HARDPaper 24.9 Solve for $\sigma$[6 marks]
A random variable $X$ is normally distributed with mean $60$ and unknown standard deviation $\sigma$. It is given that $P(X > 70) = 0.10$.随机变量（random variable）$X$ 服从均值为 $60$、标准差（standard deviation，SD）$\sigma$ 未知的正态分布。已知 $P(X > 70) = 0.10$。
(a)Find $\sigma$, to 3 s.f.求 $\sigma$，保留 3 位有效数字。[4]
(b)For this $\sigma$, find $P(X < 50)$, to 3 s.f.使用所求的 $\sigma$，求 $P(X < 50)$，保留 3 位有效数字。[2]
PART IV · PAPER 3 (HL ONLY)第四部分 · 第三卷（仅 HL）Calculator · 31 marks可使用计算器 · 31 分
Section D — Paper 3 (HL Extended Exploration)D 节 —— 第三卷（HL 长题探究）
A multi-part exploration on linear transformations of random variables and sums of independent normals. HL-only material (AHL 4.11).关于随机变量线性变换与独立正态分布之和的多部分探究题。仅 HL 内容（AHL 4.11）。
Q11HARDPaper 3AHL 4.11 Linear Transforms / Sums[15 marks]HL
Let $X \sim N(100,\, 15^2)$ and $Y \sim N(80,\, 20^2)$ be independent random variables.设 $X \sim N(100,\, 15^2)$ 与 $Y \sim N(80,\, 20^2)$ 为独立的随机变量。
(e)Find $c$ such that $P(X + Y > c) = 0.05$, to 1 d.p.求满足 $P(X + Y > c) = 0.05$ 的 $c$，保留 1 位小数。[3]
(f)Explain, in one sentence, why $\text{Var}(X - Y) = \text{Var}(X + Y)$ even though $E(X - Y) \ne E(X + Y)$.用一句话说明：为何尽管 $E(X - Y) \ne E(X + Y)$，仍有 $\text{Var}(X - Y) = \text{Var}(X + Y)$。[2]
Q12HARDPaper 3AHL 4.12 Sample Mean / Standard Error[16 marks]HL
An espresso machine pours individual shots with volume $X \sim N(30,\,1.5^{2})$ ml. Quality control draws a random sample of $n$ independent shots and computes the sample mean $\bar X$.某浓缩咖啡机单杯出品的体积 $X \sim N(30,\,1.5^{2})$（单位：ml）。质检部抽取 $n$ 杯相互独立的样本，计算样本均值（sample mean）$\bar X$。
(a)State the distribution of $\bar X$ for a sample of size $n$, giving its mean and variance in terms of $\mu = 30$ and $\sigma = 1.5$.写出样本量为 $n$ 时 $\bar X$ 的抽样分布（sampling distribution），并用 $\mu = 30$ 与 $\sigma = 1.5$ 表示其均值与方差。[2]
(b)For $n = 9$, find $P(|\bar X - 30| > 0.5)$, to 3 s.f.当 $n = 9$ 时，求 $P(|\bar X - 30| > 0.5)$，保留 3 位有效数字。[3]
(c)Find the smallest sample size $n$ such that $P(|\bar X - 30| > 0.5) \le 0.01$. Show the inequality on $z$ you set up before iterating on the GDC.求使 $P(|\bar X - 30| > 0.5) \le 0.01$ 成立的最小样本量 $n$。在用 GDC 迭代之前，先写出所建立的关于 $z$ 值（z-score）的不等式。[3]
(d)A bartender pours a "double" by combining two independent shots. Let $T = X_{1} + X_{2}$. State the distribution of $T$, then find $P(T < 58)$, to 3 s.f.调酒师将两杯独立的浓缩合为"双份"，设 $T = X_{1} + X_{2}$。写出 $T$ 的分布，并求 $P(T < 58)$，保留 3 位有效数字。[3]
(e)Explain why $\text{Var}(\bar X) = \sigma^{2}/n$ falls off as $1/n$, but the standard error $\text{SE}(\bar X) = \sigma/\sqrt{n}$ only falls off as $1/\sqrt{n}$. State, as a practical consequence, the factor by which $n$ must grow to halve the standard error.说明为何 $\text{Var}(\bar X) = \sigma^{2}/n$ 以 $1/n$ 的速率减小，而标准误（standard error，SE）$\text{SE}(\bar X) = \sigma/\sqrt{n}$ 仅以 $1/\sqrt{n}$ 的速率减小。作为实际后果，写出要使标准误减半，$n$ 必须扩大的倍数。[2]
(f)Now suppose individual shot volumes are not normal — say uniformly distributed on $[28,\,32]$ ml. State, without proof, what the distribution of $\bar X$ tends to as $n \to \infty$, including the name of the theorem. Give a one-sentence intuition for why averaging "normalises".现假设单杯体积不再服从正态，而是在 $[28,\,32]$ ml 上均匀分布。无需证明，写出当 $n \to \infty$ 时 $\bar X$ 趋于何种分布，并写出对应定理的名称（中心极限定理（central limit theorem，CLT））。用一句话直观说明为何"求平均"会让分布趋于正态。[3]