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Unit D3 · SolutionsUnit D3 · 解析

Probability Distributions — Solutions概率分布 —— 解析

Companion to the IB-Style Practice SetIB 风格练习题的解析配套

EASYMEDIUMHARD Paper 1APaper 1BPaper 2Paper 3 HL ONLY

Syllabus 4.7 – 4.9 SL + AHL 4.11考纲 4.7 – 4.9 SL + AHL 4.11 AA HL

v1.1 · companion to Unit_D3_Probability_Distributions_Practice.html v1.1 · 12 Qs · 91 marks · mark-by-mark withUnit_D3_Probability_Distributions_Practice.html v1.1 配套 · 12 题 · 91 分 · 按 IB 风格逐分给出解析，附 M1 / A1 / R1 callouts分点标注

PART I  ·  PAPER 1 SECTION A — SOLUTIONS第一部分  ·  第一卷 A 节 —— 解析No calculator · 22 marks不可使用计算器 · 22 分

Section A — Worked SolutionsA 节 —— 详细解析

Q1EASYPaper 1A4.7 Discrete RV / $E(X)$[6 marks]

$X$ takes values $1,2,3,4$ with $P = 0.2, 0.3, k, 0.1$. Find $k$, $P(X \le 2)$, $E(X)$.离散型随机变量（discrete random variable）$X$ 的取值为 $1,2,3,4$，对应概率为 $0.2, 0.3, k, 0.1$。求 $k$、$P(X \le 2)$ 与期望（E(X)，expected value）。

Answers:答案：  (a) $k = 0.4$  ·  (b) $0.5$  ·  (c) $E(X) = 2.4$

(a) Distribution sums to $1$ M1·A1

$0.2 + 0.3 + k + 0.1 = 1 \Rightarrow k = 1 - 0.6 = \boxed{0.4}.$

(b) A1

$P(X \le 2) = P(X = 1) + P(X = 2) = 0.2 + 0.3 = \boxed{0.5}.$

(c) Expected value M1·A1·A1

$$ E(X) \;=\; \sum x\,P(X = x) \;=\; 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1) \;=\; 0.2 + 0.6 + 1.2 + 0.4 \;=\; \boxed{2.4}. $$

(a) 概率之和等于 $1$ M1·A1

$0.2 + 0.3 + k + 0.1 = 1 \Rightarrow k = 1 - 0.6 = \boxed{0.4}.$

(b) A1

$P(X \le 2) = P(X = 1) + P(X = 2) = 0.2 + 0.3 = \boxed{0.5}.$

(c) 期望（expected value） M1·A1·A1

$$ E(X) \;=\; \sum x\,P(X = x) \;=\; 1(0.2) + 2(0.3) + 3(0.4) + 4(0.1) \;=\; 0.2 + 0.6 + 1.2 + 0.4 \;=\; \boxed{2.4}. $$
Q2EASYPaper 1A4.7 Fair Game[5 marks]

Heads $+\$5$, tails $-\$3$, fair coin.均匀硬币：正面 $+\$5$、反面 $-\$3$。

Answers:答案：  (a) $\$1$  ·  (b) not fair (favours player)不公平（对玩家有利）  ·  (c) $w = \$3$

(a) Expected profit M1·A1

$$ E(\text{profit}) \;=\; \tfrac{1}{2}(5) + \tfrac{1}{2}(-3) \;=\; 2.5 - 1.5 \;=\; \boxed{\$1}. $$

(b) Fairness R1

The game is not fair: a fair game has $E(\text{profit}) = 0$, but here $E = \$1 \ne 0$. The player has a long-run advantage.

(c) Fair win amount M1·A1

Solve $\tfrac{1}{2}w - \tfrac{1}{2}(3) = 0 \Rightarrow w = \boxed{\$3}.$
Fair game = $E = 0$. "Fair" in the IB sense means symmetric expectation. Many casino games are unfair by design ($E < 0$ for the player); insurance premiums are unfair to the buyer ($E < 0$ for the buyer) but buyers accept the loss in exchange for risk reduction.

(a) 期望收益（expected profit） M1·A1

$$ E(\text{profit}) \;=\; \tfrac{1}{2}(5) + \tfrac{1}{2}(-3) \;=\; 2.5 - 1.5 \;=\; \boxed{\$1}. $$

(b) 公平性 R1

该博弈不是公平博弈（fair game）：公平博弈要求 $E(\text{profit}) = 0$，但此处 $E = \$1 \ne 0$。长期来看玩家占优。

(c) 使博弈公平的赢款 M1·A1

解 $\tfrac{1}{2}w - \tfrac{1}{2}(3) = 0 \Rightarrow w = \boxed{\$3}.$
公平博弈 $\Leftrightarrow E = 0$。IB 语境下的"公平"指期望对称。许多赌场游戏在设计上就不公平（对玩家而言 $E < 0$）；保险费对买家也不公平（$E < 0$），但买家用这笔损失换取风险下降。
Q3MEDIUMPaper 1A4.8 Binomial Exact[6 marks]

$X \sim B(3, 1/3)$. Distribution table; $E(X)$; $P(X \ge 1)$.$X \sim B(3, 1/3)$ 服从二项分布（binomial distribution）。列出概率分布表；求 $E(X)$ 与 $P(X \ge 1)$。

Answers:答案：  (a) table over $27$分母为 $27$ 的分布表  ·  (b) $E(X) = 1$  ·  (c) $\dfrac{19}{27}$

(a) Distribution table M1·A1·A1

Use $P(X = r) = \binom{3}{r}\,p^r\,(1-p)^{3-r}$ with $p = 1/3$:
$x$$0$$1$$2$$3$
$P(X = x)$$\dfrac{8}{27}$$\dfrac{12}{27}$$\dfrac{6}{27}$$\dfrac{1}{27}$
Check sums: $8 + 12 + 6 + 1 = 27$. ✓

(b) $E(X)$ A1

$E(X) = np = 3 \cdot \tfrac{1}{3} = \boxed{1}.$ (Verify from table: $0 \cdot 8/27 + 1 \cdot 12/27 + 2 \cdot 6/27 + 3 \cdot 1/27 = 27/27 = 1$. ✓)

(c) M1·A1

$P(X \ge 1) = 1 - P(X = 0) = 1 - \dfrac{8}{27} = \boxed{\dfrac{19}{27}}.$

(a) 分布表 M1·A1·A1

令 $p = 1/3$，使用公式 $P(X = r) = \binom{3}{r}\,p^r\,(1-p)^{3-r}$：
$x$$0$$1$$2$$3$
$P(X = x)$$\dfrac{8}{27}$$\dfrac{12}{27}$$\dfrac{6}{27}$$\dfrac{1}{27}$
验证之和：$8 + 12 + 6 + 1 = 27$。 ✓

(b) $E(X)$ A1

$E(X) = np = 3 \cdot \tfrac{1}{3} = \boxed{1}.$（用表格验证：$0 \cdot 8/27 + 1 \cdot 12/27 + 2 \cdot 6/27 + 3 \cdot 1/27 = 27/27 = 1$。 ✓）

(c) M1·A1

$P(X \ge 1) = 1 - P(X = 0) = 1 - \dfrac{8}{27} = \boxed{\dfrac{19}{27}}.$
Q4MEDIUMPaper 1A4.9 Normal with tabulated $z$[5 marks]

$X \sim N(50, 100)$; given $P(Z < 1) = 0.8413$.$X$ 服从正态分布（normal distribution，N(μ, σ²)）$X \sim N(50, 100)$；已知 $P(Z < 1) = 0.8413$。

Answers:答案：  (a) $z = 1$  ·  (b) $P(X < 60) = 0.8413$  ·  (c) $P(40 < X < 60) = 0.6826$

(a) Standardisation A1

$\sigma = \sqrt{100} = 10$. $z = (x - \mu)/\sigma = (60 - 50)/10 = \boxed{1}.$

(b) A1

$P(X < 60) = P(Z < 1) = \boxed{0.8413}.$

(c) Symmetry M1·M1·A1

Symmetric interval $\pm 1$ SD about the mean: $$ P(40 < X < 60) \;=\; P(-1 < Z < 1) \;=\; 2 \cdot P(0 < Z < 1) \;=\; 2 \cdot (0.8413 - 0.5) \;=\; 2 \cdot 0.3413 \;=\; \boxed{0.6826}. $$
The $68$–$95$–$99.7$ rule. About $68\%$ of a normal lies within $\pm 1\sigma$, $95\%$ within $\pm 2\sigma$, $99.7\%$ within $\pm 3\sigma$. Memorise these; they're the no-calc shortcut for symmetric intervals.

(a) 标准化（z-score） A1

$\sigma = \sqrt{100} = 10$。 $z = (x - \mu)/\sigma = (60 - 50)/10 = \boxed{1}.$

(b) A1

$P(X < 60) = P(Z < 1) = \boxed{0.8413}.$

(c) 对称性 M1·M1·A1

以均值为中心、宽度 $\pm 1$ 个标准差（SD）的对称区间： $$ P(40 < X < 60) \;=\; P(-1 < Z < 1) \;=\; 2 \cdot P(0 < Z < 1) \;=\; 2 \cdot (0.8413 - 0.5) \;=\; 2 \cdot 0.3413 \;=\; \boxed{0.6826}. $$
$68$–$95$–$99.7$ 法则。正态分布约有 $68\%$ 落在 $\pm 1\sigma$ 内、$95\%$ 落在 $\pm 2\sigma$ 内、$99.7\%$ 落在 $\pm 3\sigma$ 内。务必记牢；它们是无计算器条件下处理对称区间的捷径。
PART II  ·  PAPER 1 SECTION B — SOLUTIONS第二部分  ·  第一卷 B 节 —— 解析No calculator · 23 marks不可使用计算器 · 23 分

Section B — Extended SolutionsB 节 —— 长题解析

Q5MEDIUMPaper 1B4.8 Binomial Cumulative[7 marks]

$X \sim B(5, 1/2)$.$X \sim B(5, 1/2)$（二项分布）。

Answers:答案：  (a) $\dfrac{10}{32} = \dfrac{5}{16}$  ·  (b) $\dfrac{6}{32} = \dfrac{3}{16}$  ·  (c) $\dfrac{16}{32} = \dfrac{1}{2}$  ·  (d) symmetry verified对称性已验证

(a) $P(X = 2)$ M1·A1

$$ P(X = 2) \;=\; \binom{5}{2}\left(\tfrac{1}{2}\right)^2\left(\tfrac{1}{2}\right)^3 \;=\; 10 \cdot \tfrac{1}{32} \;=\; \boxed{\dfrac{10}{32} = \dfrac{5}{16}}. $$

(b) $P(X \le 1)$ M1·A1

$$ P(X \le 1) \;=\; P(X=0) + P(X=1) \;=\; \tfrac{1}{32} + \tfrac{5}{32} \;=\; \boxed{\dfrac{6}{32} = \dfrac{3}{16}}. $$

(c) $P(X \ge 3)$ M1·A1

$$ P(X \ge 3) \;=\; P(X=3) + P(X=4) + P(X=5) \;=\; \tfrac{10}{32} + \tfrac{5}{32} + \tfrac{1}{32} \;=\; \boxed{\dfrac{16}{32} = \dfrac{1}{2}}. $$

(d) Symmetry check R1

$B(5, 1/2)$ is symmetric about $E(X) = 2.5$. So $P(X \ge 3) = P(X \le 2)$ by reflecting outcomes (replace each $X = r$ outcome with $5 - r$). Check: $P(X \le 2) = \tfrac{1}{32} + \tfrac{5}{32} + \tfrac{10}{32} = \tfrac{16}{32} = \tfrac{1}{2}$. ✓
$B(n, 1/2)$ is symmetric. Only the binomial with $p = 1/2$ is symmetric (and only then). For other $p$, $P(X = r) \ne P(X = n-r)$ in general. The Pascal-triangle row is the no-calc shortcut for $B(n, 1/2)$: divide row $n$ by $2^n$.

(a) $P(X = 2)$ M1·A1

$$ P(X = 2) \;=\; \binom{5}{2}\left(\tfrac{1}{2}\right)^2\left(\tfrac{1}{2}\right)^3 \;=\; 10 \cdot \tfrac{1}{32} \;=\; \boxed{\dfrac{10}{32} = \dfrac{5}{16}}. $$

(b) $P(X \le 1)$ M1·A1

$$ P(X \le 1) \;=\; P(X=0) + P(X=1) \;=\; \tfrac{1}{32} + \tfrac{5}{32} \;=\; \boxed{\dfrac{6}{32} = \dfrac{3}{16}}. $$

(c) $P(X \ge 3)$ M1·A1

$$ P(X \ge 3) \;=\; P(X=3) + P(X=4) + P(X=5) \;=\; \tfrac{10}{32} + \tfrac{5}{32} + \tfrac{1}{32} \;=\; \boxed{\dfrac{16}{32} = \dfrac{1}{2}}. $$

(d) 对称性验证 R1

$B(5, 1/2)$ 关于 $E(X) = 2.5$ 对称，因此通过将结果映射 $X = r \mapsto 5 - r$ 即得 $P(X \ge 3) = P(X \le 2)$。验证：$P(X \le 2) = \tfrac{1}{32} + \tfrac{5}{32} + \tfrac{10}{32} = \tfrac{16}{32} = \tfrac{1}{2}$。 ✓
$B(n, 1/2)$ 是对称分布。只有 $p = 1/2$ 的二项分布对称（且仅此情况）。对其他 $p$，一般而言 $P(X = r) \ne P(X = n-r)$。帕斯卡三角的第 $n$ 行除以 $2^n$，就是 $B(n, 1/2)$ 的无计算器捷径。
Q6HARDPaper 1BAHL 4.11 Variance[8 marks]HL

$X \in \{-1, 0, 1, 2\}$ with $P = 1/8, 3/8, 3/8, 1/8$. Find $E(X)$, $E(X^2)$, $\text{Var}(X)$, $\text{Var}(2X-1)$.$X \in \{-1, 0, 1, 2\}$，对应概率为 $1/8, 3/8, 3/8, 1/8$。求 $E(X)$、$E(X^2)$、方差（Var(X)）、$\text{Var}(2X-1)$。

Answers:答案：  (b) $E(X) = \tfrac{1}{2}$  ·  (c) $E(X^2) = 1$, $\text{Var}(X) = \tfrac{3}{4}$  ·  (d) $\text{Var}(2X-1) = 3$

(a) Sum check A1

$\tfrac{1}{8} + \tfrac{3}{8} + \tfrac{3}{8} + \tfrac{1}{8} = \tfrac{8}{8} = 1$. ✓

(b) $E(X)$ M1·A1

$$ E(X) \;=\; (-1)\tfrac{1}{8} + 0 \cdot \tfrac{3}{8} + 1 \cdot \tfrac{3}{8} + 2 \cdot \tfrac{1}{8} \;=\; -\tfrac{1}{8} + \tfrac{3}{8} + \tfrac{2}{8} \;=\; \dfrac{4}{8} \;=\; \boxed{\dfrac{1}{2}}. $$

(c) $E(X^2)$ and $\text{Var}(X)$ M1·A1·A1

$$ E(X^2) \;=\; (-1)^2 \cdot \tfrac{1}{8} + 0^2 \cdot \tfrac{3}{8} + 1^2 \cdot \tfrac{3}{8} + 2^2 \cdot \tfrac{1}{8} \;=\; \tfrac{1}{8} + \tfrac{3}{8} + \tfrac{4}{8} \;=\; \dfrac{8}{8} \;=\; 1. $$ $$ \text{Var}(X) \;=\; E(X^2) - [E(X)]^2 \;=\; 1 - \left(\tfrac{1}{2}\right)^2 \;=\; 1 - \tfrac{1}{4} \;=\; \boxed{\dfrac{3}{4}}. $$

(d) $\text{Var}(2X - 1)$ M1·A1

Lemma: $\text{Var}(aX + b) = a^2\,\text{Var}(X)$ (variance unaffected by shifts; scales by $a^2$). $$ \text{Var}(2X - 1) \;=\; 2^2 \cdot \text{Var}(X) \;=\; 4 \cdot \tfrac{3}{4} \;=\; \boxed{3}. $$
Variance vs SD scaling. $\text{Var}(aX + b) = a^2\,\text{Var}(X)$, so $\sigma(aX + b) = |a|\,\sigma(X)$. The square in variance is what produces the $a^2$ instead of $|a|$.

(a) 概率之和验证 A1

$\tfrac{1}{8} + \tfrac{3}{8} + \tfrac{3}{8} + \tfrac{1}{8} = \tfrac{8}{8} = 1$。 ✓

(b) $E(X)$ M1·A1

$$ E(X) \;=\; (-1)\tfrac{1}{8} + 0 \cdot \tfrac{3}{8} + 1 \cdot \tfrac{3}{8} + 2 \cdot \tfrac{1}{8} \;=\; -\tfrac{1}{8} + \tfrac{3}{8} + \tfrac{2}{8} \;=\; \dfrac{4}{8} \;=\; \boxed{\dfrac{1}{2}}. $$

(c) $E(X^2)$ 与方差 $\text{Var}(X)$ M1·A1·A1

$$ E(X^2) \;=\; (-1)^2 \cdot \tfrac{1}{8} + 0^2 \cdot \tfrac{3}{8} + 1^2 \cdot \tfrac{3}{8} + 2^2 \cdot \tfrac{1}{8} \;=\; \tfrac{1}{8} + \tfrac{3}{8} + \tfrac{4}{8} \;=\; \dfrac{8}{8} \;=\; 1. $$ $$ \text{Var}(X) \;=\; E(X^2) - [E(X)]^2 \;=\; 1 - \left(\tfrac{1}{2}\right)^2 \;=\; 1 - \tfrac{1}{4} \;=\; \boxed{\dfrac{3}{4}}. $$

(d) $\text{Var}(2X - 1)$ M1·A1

引理：$\text{Var}(aX + b) = a^2\,\text{Var}(X)$（平移不改变方差，缩放因子按 $a^2$ 计入）。 $$ \text{Var}(2X - 1) \;=\; 2^2 \cdot \text{Var}(X) \;=\; 4 \cdot \tfrac{3}{4} \;=\; \boxed{3}. $$
方差与标准差的缩放。$\text{Var}(aX + b) = a^2\,\text{Var}(X)$，因此 $\sigma(aX + b) = |a|\,\sigma(X)$。正是方差中的平方项把因子做成了 $a^2$，而不是 $|a|$。
Q7HARDPaper 1BAHL 4.11 Continuous PDF[8 marks]HL

$f(x) = kx$ on $[0, 2]$. Find $k$, $P(X > 1)$, median $m$.概率密度函数（probability density function，pdf）$f(x) = kx$，定义域 $[0, 2]$。求 $k$、$P(X > 1)$、中位数 $m$。

Answers:答案：  (a) $k = \dfrac{1}{2}$  ·  (b) $\dfrac{3}{4}$  ·  (c) $m = \sqrt{2}$

(a) Normalising constant M1·M1·A1

Continuous-PDF axiom: $\displaystyle \int_{-\infty}^{\infty} f(x)\,dx = 1$. Since $f \equiv 0$ outside $[0, 2]$: $$ \int_0^2 kx\,dx \;=\; k \cdot \left[\dfrac{x^2}{2}\right]_0^2 \;=\; k \cdot 2 \;=\; 1 \;\;\Longrightarrow\;\; \boxed{k = \dfrac{1}{2}}. $$

(b) $P(X > 1)$ M1·A1

$$ P(X > 1) \;=\; \int_1^2 \tfrac{1}{2}x\,dx \;=\; \tfrac{1}{4}\,x^2\bigg|_1^2 \;=\; \tfrac{1}{4}(4 - 1) \;=\; \boxed{\dfrac{3}{4}}. $$

(c) Median M1·M1·A1

Median $m$ satisfies $\int_0^m f(x)\,dx = 1/2$: $$ \int_0^m \tfrac{1}{2}x\,dx \;=\; \tfrac{1}{4}\,m^2 \;=\; \tfrac{1}{2} \;\;\Longrightarrow\;\; m^2 = 2 \;\;\Longrightarrow\;\; \boxed{m = \sqrt{2}}. $$ (Sanity check: $\sqrt{2} \approx 1.414$, comfortably inside $[0, 2]$. ✓)
Continuous distribution checklist. Normalising constant from $\int f = 1$; probability from $\int_a^b f$; median from $\int_0^m f = 1/2$; mode from where $f$ is maximised; mean from $\int x f(x)\,dx$. Each integral lives in one line on the page — but you must set them up correctly.

(a) 归一化常数 M1·M1·A1

连续 pdf 的归一化公理：$\displaystyle \int_{-\infty}^{\infty} f(x)\,dx = 1$。由于 $f$ 在 $[0, 2]$ 之外恒为 $0$： $$ \int_0^2 kx\,dx \;=\; k \cdot \left[\dfrac{x^2}{2}\right]_0^2 \;=\; k \cdot 2 \;=\; 1 \;\;\Longrightarrow\;\; \boxed{k = \dfrac{1}{2}}. $$

(b) $P(X > 1)$ M1·A1

$$ P(X > 1) \;=\; \int_1^2 \tfrac{1}{2}x\,dx \;=\; \tfrac{1}{4}\,x^2\bigg|_1^2 \;=\; \tfrac{1}{4}(4 - 1) \;=\; \boxed{\dfrac{3}{4}}. $$

(c) 中位数（median） M1·M1·A1

中位数 $m$ 满足 $\int_0^m f(x)\,dx = 1/2$： $$ \int_0^m \tfrac{1}{2}x\,dx \;=\; \tfrac{1}{4}\,m^2 \;=\; \tfrac{1}{2} \;\;\Longrightarrow\;\; m^2 = 2 \;\;\Longrightarrow\;\; \boxed{m = \sqrt{2}}. $$ （合理性检查：$\sqrt{2} \approx 1.414$，确实在 $[0, 2]$ 内。 ✓）
连续型分布工具清单。归一化常数：由 $\int f = 1$；概率：由 $\int_a^b f$；中位数：由 $\int_0^m f = 1/2$；众数（mode）：$f$ 取得最大值之处；均值：$\int x f(x)\,dx$。每个积分都只需写一行 —— 但你必须把它们的形式写对。
PART III  ·  PAPER 2 — SOLUTIONS第三部分  ·  第二卷 —— 解析Calculator · 15 marks可使用计算器 · 15 分

Section C — Paper 2 (Calculator) SolutionsC 节 —— 第二卷（可用计算器）解析

Q8EASYPaper 24.9 Normal Interval[4 marks]

Biscuit mass $X \sim N(50, 10^2)$ g.饼干质量 $X \sim N(50, 10^2)$ 克。

Answers:答案：  (a) $P(45 < X < 55) \approx 0.383$  ·  (b) ~$38\%$ of biscuits weigh between $45$ g and $55$ g约 $38\%$ 的饼干质量介于 $45$ g 与 $55$ g 之间

(a) GDC normalcdf M1·M1·A1

Use normalcdf(lower=45, upper=55, μ=50, σ=10): $$ P(45 < X < 55) \;=\; \boxed{0.383}\ (3\text{ s.f.}). $$ Equivalent via standardisation: $z_1 = -0.5$, $z_2 = 0.5$; $P = 2(0.6915 - 0.5) = 0.383$.

(b) Interpretation A1

About $38.3\%$ of biscuits — roughly two in five — have a mass between $45$ g and $55$ g.
State $\mu$ and $\sigma$ explicitly. IB markers want to see you identified the distribution before reading the GDC output. "$P(45 < X < 55) = 0.383$ via normalcdf" is fine; bare $0.383$ is not.

(a) GDC 调用 normalcdf M1·M1·A1

使用 normalcdf(lower=45, upper=55, μ=50, σ=10)： $$ P(45 < X < 55) \;=\; \boxed{0.383}\ (\text{3 位有效数字})。 $$ 也可用标准化（z-score）法：$z_1 = -0.5$，$z_2 = 0.5$；$P = 2(0.6915 - 0.5) = 0.383$。

(b) 解释 A1

约有 $38.3\%$ 的饼干 —— 大致每五块里有两块 —— 质量介于 $45$ g 与 $55$ g 之间。
务必显式声明 $\mu$ 与 $\sigma$。IB 评卷想看到你在读 GDC 输出之前就确定了分布。写"$P(45 < X < 55) = 0.383$，由 normalcdf 得"可以接受；只写一个 $0.383$ 则不行。
Q9MEDIUMPaper 24.9 Inverse Normal[5 marks]

Adult male heights $X \sim N(170, 7^2)$ cm.成年男性身高 $X \sim N(170, 7^2)$ cm。

Answers:答案：  (a) $P(X > 175) \approx 0.238$  ·  (b) $\approx 179.0$ cm

(a) M1·A1

$z = (175 - 170)/7 = 5/7 \approx 0.714$. GDC: $P(X > 175) = 1 - P(X \le 175) \approx 1 - 0.762 = \boxed{0.238}.$

(b) Inverse normal M1·M1·A1

We need $x_0$ with $P(X > x_0) = 0.10$, i.e. $P(X \le x_0) = 0.90$. Using invNorm(area=0.90, μ=170, σ=7): $$ x_0 \;\approx\; 170 + 7 \cdot 1.2816 \;\approx\; 170 + 8.97 \;\approx\; \boxed{179.0\ \text{cm}}. $$ (Equivalent: $z_0 = \text{invNorm}(0.90) \approx 1.28$; then $x_0 = \mu + z_0\sigma$.)
Direction of the tail. invNorm returns the value with the given area to the left. For "top $10\%$" you need area-to-left $= 0.90$, not $0.10$. Read this carefully on the calculator.

(a) M1·A1

$z = (175 - 170)/7 = 5/7 \approx 0.714$。 GDC：$P(X > 175) = 1 - P(X \le 175) \approx 1 - 0.762 = \boxed{0.238}.$

(b) 逆正态（invNorm） M1·M1·A1

需要找 $x_0$ 使 $P(X > x_0) = 0.10$，即 $P(X \le x_0) = 0.90$。 使用 invNorm(area=0.90, μ=170, σ=7)： $$ x_0 \;\approx\; 170 + 7 \cdot 1.2816 \;\approx\; 170 + 8.97 \;\approx\; \boxed{179.0\ \text{cm}}. $$ （等价做法：$z_0 = \text{invNorm}(0.90) \approx 1.28$；再由 $x_0 = \mu + z_0\sigma$ 求得。）
注意尾部方向。invNorm 返回的是给定左侧累积面积对应的值。求"前 $10\%$"对应的身高，应使用左侧面积 $= 0.90$，而非 $0.10$。在计算器上务必看清。
Q10HARDPaper 24.9 Solve for $\sigma$[6 marks]

$X \sim N(60, \sigma^2)$, $P(X > 70) = 0.10$.$X \sim N(60, \sigma^2)$，$P(X > 70) = 0.10$。

Answers:答案：  (a) $\sigma \approx 7.80$  ·  (b) $P(X < 50) \approx 0.100$

(a) Solve for $\sigma$ M1·M1·M1·A1

Standardise: $\dfrac{70 - 60}{\sigma} = z_0$ where $P(Z > z_0) = 0.10$, i.e. $z_0 = \text{invNorm}(0.90) \approx 1.2816$. $$ \dfrac{10}{\sigma} \;=\; 1.2816 \;\;\Longrightarrow\;\; \sigma \;=\; \dfrac{10}{1.2816} \;\approx\; \boxed{7.80}. $$

(b) M1·A1

By symmetry, $50$ is the mirror image of $70$ about $\mu = 60$, so $$ P(X < 50) \;=\; P(X > 70) \;=\; \boxed{0.100}. $$ Confirm with GDC: normalcdf(-∞, 50, 60, 7.80) $\approx 0.100$. ✓
Symmetry shortcut. If $a$ and $b$ are equidistant from $\mu$, then $P(X < a)$ on one tail equals $P(X > b)$ on the other. Faster than re-running the GDC; useful as a check.

(a) 求 $\sigma$ M1·M1·M1·A1

标准化：$\dfrac{70 - 60}{\sigma} = z_0$，其中 $P(Z > z_0) = 0.10$，即 $z_0 = \text{invNorm}(0.90) \approx 1.2816$。 $$ \dfrac{10}{\sigma} \;=\; 1.2816 \;\;\Longrightarrow\;\; \sigma \;=\; \dfrac{10}{1.2816} \;\approx\; \boxed{7.80}. $$

(b) M1·A1

由对称性，$50$ 是 $70$ 关于均值 $\mu = 60$ 的镜像，因此 $$ P(X < 50) \;=\; P(X > 70) \;=\; \boxed{0.100}. $$ 用 GDC 验证：normalcdf(-∞, 50, 60, 7.80) $\approx 0.100$。 ✓
对称性捷径。若 $a$ 与 $b$ 到均值 $\mu$ 的距离相等，则一侧尾概率 $P(X < a)$ 等于另一侧尾概率 $P(X > b)$。比重新调用 GDC 更快，也便于核对。
PART IV  ·  PAPER 3 (HL) — SOLUTIONS第四部分  ·  第三卷（HL）—— 解析Calculator · 31 marks可使用计算器 · 31 分

Section D — Paper 3 (HL Extended) SolutionsD 节 —— 第三卷（HL 长题）解析

Q11HARDPaper 3AHL 4.11 Linear Transforms / Sums[15 marks]HL

$X \sim N(100, 15^2)$, $Y \sim N(80, 20^2)$, independent.$X \sim N(100, 15^2)$，$Y \sim N(80, 20^2)$，相互独立。

Answers:答案：  (a) $E = 203$, $\text{Var} = 900$  ·  (b) $E = 180$, $\text{Var} = 625$  ·  (c) $P \approx 0.212$  ·  (d) $E = 20$, $\text{Var} = 625$, $P(X > Y) \approx 0.788$  ·  (e) $c \approx 221.1$  ·  (f) independent variances add独立时方差相加

Lemmas (used throughout) R1

For RVs $X, Y$ and constants $a, b, c$:
  • $E(aX + b) = aE(X) + b$, $\;\text{Var}(aX + b) = a^2\,\text{Var}(X)$
  • $E(X \pm Y) = E(X) \pm E(Y)$ (always)
  • $\text{Var}(X \pm Y) = \text{Var}(X) + \text{Var}(Y)$ — when $X, Y$ are independent (the sign on $Y$ squares away)
  • Sum/difference of independent normals is normal: $X \pm Y \sim N(E(X) \pm E(Y),\, \text{Var}(X) + \text{Var}(Y))$

(a) A1·A1

$E(2X + 3) = 2(100) + 3 = \boxed{203}$. $\text{Var}(2X + 3) = 4 \cdot 225 = \boxed{900}$.

(b) A1·A1

$E(X + Y) = 100 + 80 = \boxed{180}$. $\text{Var}(X + Y) = 225 + 400 = \boxed{625}$. So $\sigma(X + Y) = 25$.

(c) M1·M1·A1

$X + Y \sim N(180, 625)$ with $\sigma = 25$. Standardise: $$ P(X + Y > 200) \;=\; P\!\left(Z > \dfrac{200 - 180}{25}\right) \;=\; P(Z > 0.8) \;\approx\; 1 - 0.7881 \;=\; \boxed{0.212}. $$

(d) Difference of independent normals M1·A1·M1

$E(X - Y) = 100 - 80 = \boxed{20}$. $\text{Var}(X - Y) = 225 + 400 = \boxed{625}$ (variances ADD for both sum and difference of independent RVs). $$ P(X > Y) \;=\; P(X - Y > 0) \;=\; P\!\left(Z > \dfrac{0 - 20}{25}\right) \;=\; P(Z > -0.8) \;=\; P(Z < 0.8) \;\approx\; \boxed{0.788}. $$

(e) Inverse normal on $X + Y$ M1·M1·A1

Need $P(X + Y > c) = 0.05 \Rightarrow P(X + Y \le c) = 0.95$. $z_{0.95} = \text{invNorm}(0.95) \approx 1.6449$. $$ c \;=\; 180 + 25 \cdot 1.6449 \;\approx\; 180 + 41.12 \;\approx\; \boxed{221.1}. $$

(f) Why $\text{Var}(X - Y) = \text{Var}(X + Y)$ R1·R1

Subtracting $Y$ is the same as adding $-Y$. For variance, $\text{Var}(-Y) = (-1)^2\,\text{Var}(Y) = \text{Var}(Y)$, so flipping the sign on $Y$ doesn't change its contribution to total variance. Under independence the two variances add (with no covariance term to subtract), so both sum and difference get the same total variance.
The independence assumption is doing the work. If $X$ and $Y$ were correlated, $\text{Var}(X \pm Y) = \text{Var}(X) + \text{Var}(Y) \pm 2\,\text{Cov}(X, Y)$, and sum vs difference would give different variances. AA HL only handles the independent case; the covariance-aware version sits in first-year university probability.

引理（贯穿全题使用） R1

对随机变量 $X, Y$ 与常数 $a, b, c$：
  • $E(aX + b) = aE(X) + b$；$\;\text{Var}(aX + b) = a^2\,\text{Var}(X)$（线性变换（linear transformation））
  • $E(X \pm Y) = E(X) \pm E(Y)$（无条件成立）
  • $\text{Var}(X \pm Y) = \text{Var}(X) + \text{Var}(Y)$ —— 仅当 $X, Y$ 独立时成立（$Y$ 的符号在平方下被消去）
  • 独立正态分布之和（差）仍为正态：$X \pm Y \sim N(E(X) \pm E(Y),\, \text{Var}(X) + \text{Var}(Y))$

(a) A1·A1

$E(2X + 3) = 2(100) + 3 = \boxed{203}$。 $\text{Var}(2X + 3) = 4 \cdot 225 = \boxed{900}$。

(b) A1·A1

$E(X + Y) = 100 + 80 = \boxed{180}$。 $\text{Var}(X + Y) = 225 + 400 = \boxed{625}$。 所以 $\sigma(X + Y) = 25$。

(c) M1·M1·A1

$X + Y \sim N(180, 625)$，标准差 $\sigma = 25$。标准化： $$ P(X + Y > 200) \;=\; P\!\left(Z > \dfrac{200 - 180}{25}\right) \;=\; P(Z > 0.8) \;\approx\; 1 - 0.7881 \;=\; \boxed{0.212}. $$

(d) 独立正态分布之差 M1·A1·M1

$E(X - Y) = 100 - 80 = \boxed{20}$。 $\text{Var}(X - Y) = 225 + 400 = \boxed{625}$（独立随机变量的和与差，方差均为相加）。 $$ P(X > Y) \;=\; P(X - Y > 0) \;=\; P\!\left(Z > \dfrac{0 - 20}{25}\right) \;=\; P(Z > -0.8) \;=\; P(Z < 0.8) \;\approx\; \boxed{0.788}. $$

(e) 对 $X + Y$ 做逆正态 M1·M1·A1

需要 $P(X + Y > c) = 0.05 \Rightarrow P(X + Y \le c) = 0.95$。 $z_{0.95} = \text{invNorm}(0.95) \approx 1.6449$。 $$ c \;=\; 180 + 25 \cdot 1.6449 \;\approx\; 180 + 41.12 \;\approx\; \boxed{221.1}. $$

(f) 为何 $\text{Var}(X - Y) = \text{Var}(X + Y)$ R1·R1

减去 $Y$ 等价于加上 $-Y$。对方差而言，$\text{Var}(-Y) = (-1)^2\,\text{Var}(Y) = \text{Var}(Y)$，即翻转 $Y$ 的符号并不改变它对总方差的贡献。在独立条件下两项方差相加（不存在协方差项），因此和与差具有相同的总方差。
独立性假设是关键。若 $X$ 与 $Y$ 相关，则 $\text{Var}(X \pm Y) = \text{Var}(X) + \text{Var}(Y) \pm 2\,\text{Cov}(X, Y)$，和与差的方差将不再相等。AA HL 仅处理独立情形；含协方差的版本要等到大学一年级概率课程才学。
Q12HARDPaper 3AHL 4.12 Sample Mean / Standard Error[16 marks]HL

$X \sim N(30, 1.5^2)$ ml espresso shots; sample-mean distribution, sample-size threshold, sum of two shots, then a sentence on the CLT.浓缩咖啡单杯体积 $X \sim N(30, 1.5^2)$ ml；样本均值的抽样分布（sampling distribution）、样本量阈值、两杯之和，最后用一句话陈述中心极限定理（central limit theorem，CLT）。

Answers:答案：  (a) $\bar X \sim N(30,\, 2.25/n)$  ·  (b) $\approx 0.317$  ·  (c) $n = 60$  ·  (d) $T \sim N(60,\,4.5)$, $P(T < 58) \approx 0.173$  ·  (e) halving SE requires $4\times$ the sample size使 SE 减半需要样本量扩大 $4$ 倍  ·  (f) Central Limit Theorem中心极限定理

Lemma (used throughout) R1

For $n$ independent draws of $X \sim N(\mu, \sigma^{2})$: $$ \bar X \;=\; \frac{1}{n}\sum_{i=1}^{n} X_{i} \;\sim\; N\!\left(\mu,\; \frac{\sigma^{2}}{n}\right), \qquad \text{SE}(\bar X) \;=\; \frac{\sigma}{\sqrt{n}}. $$ Sum is also normal: $\sum_{i=1}^{n} X_{i} \sim N(n\mu, n\sigma^{2})$.

(a) Distribution of $\bar X$ A1·A1

$$ \bar X \;\sim\; N\!\left(30,\; \frac{1.5^{2}}{n}\right) \;=\; \boxed{N\!\left(30,\; \frac{2.25}{n}\right)}. $$ Mean $30$, variance $2.25/n$, standard error $1.5/\sqrt{n}$.

(b) $n = 9$ tail probability M1·M1·A1

$\sigma_{\bar X} = 1.5/\sqrt{9} = 0.5$. Standardise: $$ P(|\bar X - 30| > 0.5) \;=\; P\!\left(|Z| > \dfrac{0.5}{0.5}\right) \;=\; P(|Z| > 1) \;=\; 2\bigl(1 - \Phi(1)\bigr) \;=\; 2(1 - 0.8413) \;\approx\; \boxed{0.317}. $$ So nearly $1$ in $3$ samples of $9$ shots will already mis-estimate the mean by more than $0.5$ ml — calling for a bigger sample (part c).

(c) Smallest $n$ for tail $\le 0.01$ M1·M1·A1

$\sigma_{\bar X} = 1.5/\sqrt{n}$, so $$ P(|\bar X - 30| > 0.5) \;=\; 2\!\left(1 - \Phi\!\left(\dfrac{0.5\sqrt{n}}{1.5}\right)\right) \;=\; 2\!\left(1 - \Phi\!\left(\dfrac{\sqrt{n}}{3}\right)\right) \;\le\; 0.01. $$ Need $\Phi(\sqrt{n}/3) \ge 0.995 \;\Leftrightarrow\; \sqrt{n}/3 \ge z_{0.995} \approx 2.5758$, so $$ \sqrt{n} \;\ge\; 3 \cdot 2.5758 \;\approx\; 7.7274 \;\Longrightarrow\; n \;\ge\; 59.71. $$ GDC iteration:
$n$$\sigma_{\bar X}$$z = 0.5/\sigma_{\bar X}$tail $= 2(1-\Phi(z))$
$59$$0.1953$$2.560$$\approx 0.0105$ (fails)
$60$$0.1936$$2.582$$\approx 0.0098$ (succeeds)
Smallest $n = \boxed{60}$.

(d) Double-shot $T = X_{1} + X_{2}$ A1·M1·A1

Independent normals sum to normal: $T \sim N(30 + 30,\; 1.5^{2} + 1.5^{2}) = \boxed{N(60,\, 4.5)}$, so $\sigma_{T} = \sqrt{4.5} \approx 2.121$. $$ P(T < 58) \;=\; P\!\left(Z < \dfrac{58 - 60}{2.121}\right) \;=\; P(Z < -0.9428) \;=\; 1 - \Phi(0.9428) \;\approx\; \boxed{0.173}. $$

(e) $1/n$ vs $1/\sqrt{n}$ scaling R1·A1

Variance is a sum of squared deviations divided by $n$; that division produces the $1/n$ factor in $\text{Var}(\bar X)$. The standard error is the square root of the variance, which exchanges $1/n$ for $1/\sqrt{n}$. Practical consequence: to halve $\text{SE}(\bar X) = \sigma/\sqrt{n}$, you need $\sqrt{n}$ to double, so $n$ must quadruple. Precision is expensive — the diminishing-returns curve in sample-size planning starts here.

(f) Central Limit Theorem A1·R1·R1

For any distribution with finite mean $\mu$ and finite variance $\sigma^{2}$, the sample mean of $n$ i.i.d. draws satisfies $$ \bar X \;\xrightarrow{d}\; N\!\left(\mu,\; \frac{\sigma^{2}}{n}\right) \qquad \text{as } n \to \infty. $$ This is the Central Limit Theorem. For uniform$[28, 32]$ shots: $\mu = 30$, $\sigma^{2} = 4^{2}/12 = 4/3$, so $\bar X \approx N(30,\; (4/3)/n)$ for large $n$. Intuition: the original distribution's odd shape is built from many small independent contributions whose deviations cancel in proportions controlled only by the variance; the shape information washes out, and the only thing the limit "remembers" about each individual is its variance — hence the same Gaussian limit regardless of starting distribution.
Why this matters beyond bottle-fill. The CLT is what lets confidence intervals like $\bar X \pm 1.96\,\text{SE}$ work in practice even when $X$'s distribution is unknown — provided $n$ is "large enough" (rule of thumb: $n \ge 30$, much smaller if the underlying distribution is close to symmetric). The price you pay is that precision grows only as $\sqrt{n}$: in (c), going from a tail probability of $0.317$ at $n = 9$ down to $0.01$ at $n = 60$ took a $\sim 7\times$ increase in sample size for a $\sim 30\times$ tightening in the tail, exactly because $z = 0.5/(\sigma/\sqrt{n})$ scales with $\sqrt{n}$, not $n$.

引理（贯穿全题使用） R1

对 $X \sim N(\mu, \sigma^{2})$ 进行 $n$ 次独立抽样： $$ \bar X \;=\; \frac{1}{n}\sum_{i=1}^{n} X_{i} \;\sim\; N\!\left(\mu,\; \frac{\sigma^{2}}{n}\right), \qquad \text{SE}(\bar X) \;=\; \frac{\sigma}{\sqrt{n}}. $$ 和也是正态的：$\sum_{i=1}^{n} X_{i} \sim N(n\mu, n\sigma^{2})$。

(a) $\bar X$ 的分布 A1·A1

$$ \bar X \;\sim\; N\!\left(30,\; \frac{1.5^{2}}{n}\right) \;=\; \boxed{N\!\left(30,\; \frac{2.25}{n}\right)}. $$ 均值 $30$、方差 $2.25/n$、标准误（standard error，SE）$1.5/\sqrt{n}$。

(b) $n = 9$ 时的尾概率 M1·M1·A1

$\sigma_{\bar X} = 1.5/\sqrt{9} = 0.5$。标准化： $$ P(|\bar X - 30| > 0.5) \;=\; P\!\left(|Z| > \dfrac{0.5}{0.5}\right) \;=\; P(|Z| > 1) \;=\; 2\bigl(1 - \Phi(1)\bigr) \;=\; 2(1 - 0.8413) \;\approx\; \boxed{0.317}. $$ 也就是说，$9$ 杯的样本中近 $1/3$ 的次数对均值的估计就会偏离 $0.5$ ml 以上 —— 这就需要更大的样本（见 (c)）。

(c) 使尾概率 $\le 0.01$ 的最小 $n$ M1·M1·A1

$\sigma_{\bar X} = 1.5/\sqrt{n}$，所以 $$ P(|\bar X - 30| > 0.5) \;=\; 2\!\left(1 - \Phi\!\left(\dfrac{0.5\sqrt{n}}{1.5}\right)\right) \;=\; 2\!\left(1 - \Phi\!\left(\dfrac{\sqrt{n}}{3}\right)\right) \;\le\; 0.01. $$ 要求 $\Phi(\sqrt{n}/3) \ge 0.995 \;\Leftrightarrow\; \sqrt{n}/3 \ge z_{0.995} \approx 2.5758$，所以 $$ \sqrt{n} \;\ge\; 3 \cdot 2.5758 \;\approx\; 7.7274 \;\Longrightarrow\; n \;\ge\; 59.71. $$ GDC 迭代：
$n$$\sigma_{\bar X}$$z = 0.5/\sigma_{\bar X}$tail $= 2(1-\Phi(z))$
$59$$0.1953$$2.560$$\approx 0.0105$ （不满足）
$60$$0.1936$$2.582$$\approx 0.0098$ （满足）
最小的 $n = \boxed{60}$。

(d) 双份 $T = X_{1} + X_{2}$ A1·M1·A1

独立正态分布之和仍为正态：$T \sim N(30 + 30,\; 1.5^{2} + 1.5^{2}) = \boxed{N(60,\, 4.5)}$，因此 $\sigma_{T} = \sqrt{4.5} \approx 2.121$。 $$ P(T < 58) \;=\; P\!\left(Z < \dfrac{58 - 60}{2.121}\right) \;=\; P(Z < -0.9428) \;=\; 1 - \Phi(0.9428) \;\approx\; \boxed{0.173}. $$

(e) $1/n$ 与 $1/\sqrt{n}$ 的衰减对比 R1·A1

方差是偏差平方之和再除以 $n$，这个除法直接给 $\text{Var}(\bar X)$ 带来 $1/n$ 的因子。标准误是方差的算术平方根，于是把 $1/n$ 换成了 $1/\sqrt{n}$。实际后果：要把 $\text{SE}(\bar X) = \sigma/\sqrt{n}$ 减半，需要 $\sqrt{n}$ 翻倍，即 $n$ 必须扩大 $4$ 倍。精度是昂贵的 —— 样本量规划中的"边际效益递减"曲线就从这里开始。

(f) 中心极限定理 A1·R1·R1

对均值 $\mu$ 与方差 $\sigma^{2}$ 都有限的任意分布，$n$ 个独立同分布样本的样本均值满足 $$ \bar X \;\xrightarrow{d}\; N\!\left(\mu,\; \frac{\sigma^{2}}{n}\right) \qquad \text{当 } n \to \infty。 $$ 这就是中心极限定理。对 $[28, 32]$ 上均匀分布的单杯：$\mu = 30$，$\sigma^{2} = 4^{2}/12 = 4/3$，所以大 $n$ 时 $\bar X \approx N(30,\; (4/3)/n)$。直观理解：原分布奇特的形状由许多小而独立的贡献叠加而成，这些贡献的偏差只按方差控制的比例相互抵消；形状信息被"洗掉"，极限只"记住"每个个体的方差 —— 因此无论起始分布形态如何，极限都是同一族高斯分布。
这件事在牛奶灌装之外为何也重要。正是中心极限定理（CLT）让形如 $\bar X \pm 1.96\,\text{SE}$ 的置信区间（confidence interval，CI）在实际中可用 —— 即便不知道 $X$ 的分布形态，只要 $n$ "足够大"（经验法则：$n \ge 30$，若原分布接近对称则更小即可）。代价是精度仅按 $\sqrt{n}$ 增长：在 (c) 中将尾概率从 $n = 9$ 时的 $0.317$ 压到 $n = 60$ 时的 $0.01$，样本量约扩 $7$ 倍而尾概率约缩 $30$ 倍 —— 正因为 $z = 0.5/(\sigma/\sqrt{n})$ 与 $\sqrt{n}$ 同阶，而非与 $n$ 同阶。