High School Math

Probability and Statistics Foundations概率与统计基础

Probability is the mathematics of uncertainty; descriptive statistics is the mathematics of describing a data set you already have. This guide walks the full path from the equally-likely outcome formula $P(A) = |A|/|S|$ through conditional probability and Bayes' rule, the independence test $P(A \cap B) = P(A)P(B)$, expected value and variance of a discrete random variable, the binomial distribution $X \sim B(n, p)$, the normal distribution with the $68$-$95$-$99.7$ rule and $z$-scores, and finally descriptive statistics: mean, median, mode, range, IQR, standard deviation, boxplots, and histograms. The data-management home for this content is Ontario MDM4U; in the US it splits across HSS-CP (conditional probability and rules), HSS-MD (distributions), and HSS-IC (statistical inference). Alberta puts the binomial in Math 30-1 and the bulk of probability + statistics in Math 30-2.概率（probability）是关于不确定性的数学；描述统计（descriptive statistics）则是描述已有数据集的数学。本指南从等可能结果公式 $P(A) = |A|/|S|$ 出发，依次穿过条件概率（conditional probability）与贝叶斯（Bayes）法则、独立性检验 $P(A \cap B) = P(A)P(B)$、离散随机变量（discrete random variable）的期望与方差、二项分布（binomial distribution）$X \sim B(n, p)$、带 $68$-$95$-$99.7$ 法则和 $z$-分数（z-score）的正态分布（normal distribution），最后到描述统计：均值、中位数、众数、极差、四分位距（IQR）、标准差（standard deviation）、箱线图（boxplot）与直方图（histogram）。这部分内容在安大略的"对口课程"是 MDM4U（数据管理）；在美国分散于 HSS-CP（条件概率与规则）、HSS-MD（分布）、HSS-IC（统计推断）。阿尔伯塔把二项分布放在 Math 30-1，把概率与统计的主体放在 Math 30-2。

7 sections7 节内容 US Common Core · ON · BC · ABUS 共同核心 · ON · BC · AB Honors block on Bayes & Variance贝叶斯与方差为荣誉级
Section 1 · Sample spaces and basic probability第 1 节 · 样本空间与基本概率

Sample Spaces, Events, and the Classical Probability Formula样本空间、事件与经典概率公式

Three definitions, one formula.三个定义，一个公式。
  • Sample space $S$.样本空间 $S$。 The set of all possible outcomes of an experiment. Coin: $S = \{H, T\}$. Die: $S = \{1, 2, 3, 4, 5, 6\}$. Two coins: $S = \{HH, HT, TH, TT\}$.实验所有可能结果的集合。掷一枚硬币：$S = \{H, T\}$；掷一颗骰子：$S = \{1, 2, 3, 4, 5, 6\}$；掷两枚硬币：$S = \{HH, HT, TH, TT\}$。
  • Event $A$.事件 $A$。 A subset of $S$. "Roll an even number" is the event $A = \{2, 4, 6\} \subseteq S$. CCSSM HSS-CP.A.1 names this verbatim.$S$ 的子集。"掷出偶数"即事件 $A = \{2, 4, 6\} \subseteq S$。CCSSM HSS-CP.A.1 原文如此。
  • Classical formula经典公式 (equally likely outcomes only):（仅适用等可能结果）：
$$ P(A) \;=\; \frac{|A|}{|S|} \;=\; \frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}. $$
  • Complement rule.补集规则。 $P(A^c) = 1 - P(A)$. Often easier: "at least one" $=$ $1 - $ "none".$P(A^c) = 1 - P(A)$。常用技巧："至少一个" $=$ $1 - $"一个也没有"。
  • Addition rule.加法规则。 $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Subtract the overlap so you don't count it twice. CCSSM HSS-CP.B.7.$P(A \cup B) = P(A) + P(B) - P(A \cap B)$。减去交集避免重复计数。CCSSM HSS-CP.B.7。
  • Range.取值范围。 $0 \le P(A) \le 1$. $P(S) = 1$ (something must happen); $P(\emptyset) = 0$ (impossible event).$0 \le P(A) \le 1$。$P(S) = 1$（必然事件）；$P(\emptyset) = 0$（不可能事件）。
AB Math 30-2 indicator 2.2: "determine if two events are complementary, and explain the reasoning." The complement rule is on the curriculum, not just the formula sheet.AB Math 30-2 指标 2.2："判断两事件是否互补，并说明理由"。补集规则在课纲上，不只是公式表。
Worked Example 1 · Two dice, sum of seven例题 1 · 两颗骰子，点数和为 7

Two fair six-sided dice are rolled. Find the probability that the sum is $7$.掷两颗均匀六面骰子。求点数和为 $7$ 的概率。

List the sample space.列出样本空间。 All ordered pairs $(d_1, d_2)$ with each $d_i \in \{1, 2, 3, 4, 5, 6\}$. Size $|S| = 6 \times 6 = 36$, and all $36$ pairs are equally likely.所有有序对 $(d_1, d_2)$，其中 $d_i \in \{1, 2, 3, 4, 5, 6\}$。$|S| = 6 \times 6 = 36$，且 $36$ 对等可能。

List the event.列出事件。 $A = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$. $|A| = 6$.$A = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$。$|A| = 6$。

$$ P(A) = \frac{|A|}{|S|} = \frac{6}{36} = \frac{1}{6} \approx 0.167. $$

Sanity-check.合理性核验。 Sums range $2$-$12$; $7$ is the centre and most-likely sum. The complement "sum $\ne 7$" has probability $1 - 1/6 = 5/6 \approx 0.833$ , confirms most outcomes are not a 7.两骰之和范围 $2$-$12$，$7$ 居中是最可能值。补事件"和 $\ne 7$"概率 $1 - 1/6 = 5/6 \approx 0.833$ , 多数结果非 7，吻合。

A card is drawn at random from a standard $52$-card deck. Find $P($ face card or heart $)$. (Face cards: J, Q, K of each suit.)从标准 $52$ 张牌中随机抽一张。求 $P($ 人头牌或红心 $)$。（人头牌：每花色的 J、Q、K。）
§1 · Q1
$25/52$
$22/52 = 11/26$
$15/52$
$1/2$
Addition rule: $P($ face $) + P($ heart $) - P($ face $\cap$ heart $) = 12/52 + 13/52 - 3/52 = 22/52 = 11/26$. (Three face hearts: J$\heartsuit$, Q$\heartsuit$, K$\heartsuit$.)加法规则：$P($ 人头 $) + P($ 红心 $) - P($ 人头 $\cap$ 红心 $) = 12/52 + 13/52 - 3/52 = 22/52 = 11/26$。（三张人头红心：J$\heartsuit$、Q$\heartsuit$、K$\heartsuit$。）
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$. Don't forget to subtract the three face hearts you counted twice.$P(A \cup B) = P(A) + P(B) - P(A \cap B)$。别忘了减去被重复计数的三张人头红心。
A fair coin is flipped $4$ times. Find $P($ at least one head $)$.均匀硬币掷 $4$ 次。求 $P($ 至少出现一次正面 $)$。
§1 · Q2
$1/2$
$1/4$
$15/16$
$1/16$
Complement trick: $P($ at least one head $) = 1 - P($ no heads $) = 1 - (1/2)^4 = 1 - 1/16 = 15/16$.补集技巧：$P($ 至少一次正面 $) = 1 - P($ 无正面 $) = 1 - (1/2)^4 = 1 - 1/16 = 15/16$。
"At least one" = $1 - $ "none". For independent flips, $P($ no heads $) = (1/2)^4$."至少一次" $= 1 - $ "一次也没有"。独立投掷下 $P($ 无正面 $) = (1/2)^4$。
Section 2 · Conditional probability and Bayes第 2 节 · 条件概率与贝叶斯

Conditional Probability, Total Probability, and Bayes' Rule Honors — US A2 / AB 30-2条件概率、全概率与贝叶斯法则 荣誉 — US A2 / AB 30-2

Curriculum note.课纲提示。 Conditional probability is core in CCSSM HSS-CP.A.3, .A.5, .B.6 and in ON MDM4U. Bayes' rule is named in the AP Stats curriculum and in IB Math AI / AA HL; it is honors-level in US Algebra II (often skipped) and in AB Math 30-2 (which covers conditional probability but does not derive Bayes explicitly , indicator 3.2 expects "the probability of an event, given the occurrence of a previous event").条件概率在 CCSSM HSS-CP.A.3、.A.5、.B.6 与 ON MDM4U 中为核心；贝叶斯法则在 AP Stats、IB Math AI / AA HL 中被点名；在美国代数 II（常被跳过）与 AB Math 30-2（涵盖条件概率但不显式推贝叶斯 , 指标 3.2 仅要求"已知前一事件发生时某事件的概率"）中为荣誉级。
Three formulas you should be able to write without notes.三条应能脱稿写出的公式。

Conditional probability:条件概率：

$$ P(A \mid B) \;=\; \frac{P(A \cap B)}{P(B)}, \qquad P(B) > 0. $$

Multiplication rule:乘法规则：

$$ P(A \cap B) \;=\; P(A \mid B) \, P(B) \;=\; P(B \mid A) \, P(A). $$

Bayes' rule (the conditional probabilities $P(A | B)$ and $P(B | A)$ are exchanged):贝叶斯法则（交换条件概率 $P(A | B)$ 与 $P(B | A)$）：

$$ P(A \mid B) \;=\; \frac{P(B \mid A) \, P(A)}{P(B)}. $$
  • Law of total probability.全概率公式。 If $A_1, \ldots, A_n$ partition $S$ then $P(B) = \sum_i P(B | A_i) P(A_i)$. This is the denominator of Bayes' rule when you have several "hypotheses" $A_i$.若 $A_1, \ldots, A_n$ 是 $S$ 的一个划分，则 $P(B) = \sum_i P(B | A_i) P(A_i)$。当有多个"假设" $A_i$ 时，这是贝叶斯法则的分母。
  • Two-way table reading.双向表读取。 $P(A | B)$ is the row (or column) relative frequency , restrict to the $B$ slice, then ask what fraction is also in $A$. CCSSM HSS-CP.B.6 writes this method verbatim.$P(A | B)$ 即"行（或列）相对频率", 先限制在 $B$ 这一切片内，再问其中有多少属于 $A$。CCSSM HSS-CP.B.6 原文表述此法。
Worked Example 2 · Medical test (Bayes inversion)例题 2 · 医学检测（贝叶斯反演）

A disease affects $1\%$ of a population ($P(D) = 0.01$). A diagnostic test has sensitivity $99\%$ ($P(+ | D) = 0.99$) and false-positive rate $5\%$ ($P(+ | D^c) = 0.05$). A person tests positive. What is the probability they have the disease, $P(D | +)$?某疾病在人群中患病率 $1\%$（$P(D) = 0.01$）。某诊断检测灵敏度 $99\%$（$P(+ | D) = 0.99$），假阳性率 $5\%$（$P(+ | D^c) = 0.05$）。某人检测呈阳性，问其患病概率 $P(D | +)$ 为多少？

Apply Bayes.套用贝叶斯。

$$ P(D \mid +) = \frac{P(+ \mid D) \, P(D)}{P(+)}. $$

Compute the denominator via total probability.用全概率公式算分母。

$$ P(+) = P(+ \mid D) P(D) + P(+ \mid D^c) P(D^c) = 0.99 \cdot 0.01 + 0.05 \cdot 0.99 = 0.0099 + 0.0495 = 0.0594. $$

Plug in.代入。

$$ P(D \mid +) = \frac{0.99 \cdot 0.01}{0.0594} = \frac{0.0099}{0.0594} \approx 0.167. $$

Interpret.解读。 Despite a $99\%$-sensitive test, a positive result only gives about a $17\%$ chance of having the disease , because the disease is rare ($1\%$), false positives ($5\%$ of $99\%$ healthy) dominate true positives. This is the classic base-rate fallacy that Bayes corrects.尽管灵敏度高达 $99\%$，阳性结果仅意味着约 $17\%$ 的患病概率 , 因为疾病稀有（$1\%$），假阳性（$99\%$ 健康人中的 $5\%$）压倒真阳性。这就是贝叶斯所纠正的"基础概率谬误"。

A bag has $5$ red and $3$ blue marbles. Two are drawn without replacement. Find $P($ second is red $|$ first is red $)$.袋中 $5$ 红 $3$ 蓝弹珠。无放回抽两个。求 $P($ 第二次为红 $|$ 第一次为红 $)$。
§2 · Q1
$5/8$
$5/7$
$4/7$
$4/8 = 1/2$
After removing one red, $4$ reds remain out of $7$ marbles. So $P($ 2nd red $|$ 1st red $) = 4/7$. (Without replacement makes the events dependent , conditional probability is essential.)取走一红后，剩 $4$ 红共 $7$ 个。故 $P($ 2nd 红 $|$ 1st 红 $) = 4/7$。（无放回使事件相关 , 必须用条件概率。）
After the first red is removed, recount reds and total. The conditional probability changes because the sample space shrinks.取走第一颗红后重新数红色与总数。条件概率会变化，因为样本空间缩小。
Two events satisfy $P(A) = 0.4$, $P(B) = 0.5$, $P(A \cap B) = 0.2$. Find $P(A | B)$.两事件 $P(A) = 0.4$、$P(B) = 0.5$、$P(A \cap B) = 0.2$。求 $P(A | B)$。
§2 · Q2
$0.20$
$0.40$
$0.50$
$0.80$
$P(A | B) = P(A \cap B) / P(B) = 0.2 / 0.5 = 0.4$. Notice $P(A | B) = P(A) = 0.4$ here , that means $A$ and $B$ are independent (foreshadows §3).$P(A | B) = P(A \cap B) / P(B) = 0.2 / 0.5 = 0.4$。注意此处 $P(A | B) = P(A) = 0.4$ , 即 $A$ 与 $B$ 独立（§3 预告）。
Divide the intersection by the conditioning event: $P(A \cap B) / P(B)$.用交集除以条件事件：$P(A \cap B) / P(B)$。
Section 3 · Independent vs dependent events第 3 节 · 独立与非独立事件

Independence, Dependence, and Tree Diagrams独立性、相依性与树图

The independence test in one line.一行的独立性检验。 $$ A \text{ and } B \text{ are independent} \iff P(A \cap B) = P(A) \, P(B). $$
  • Equivalent forms.等价形式。 $P(A | B) = P(A)$ and $P(B | A) = P(B)$ , "knowing $B$ tells you nothing extra about $A$." CCSSM HSS-CP.A.2, .A.3.$P(A | B) = P(A)$ 与 $P(B | A) = P(B)$ , "知道 $B$ 不增添关于 $A$ 的信息"。CCSSM HSS-CP.A.2、.A.3。
  • Independent vs mutually exclusive.独立 与 互斥。 Different concepts. Mutually exclusive: $A \cap B = \emptyset$, so $P(A \cap B) = 0$. Independent: $P(A \cap B) = P(A) P(B)$, generally nonzero. Two events with positive probability cannot be both.两个不同的概念。互斥：$A \cap B = \emptyset$，故 $P(A \cap B) = 0$；独立：$P(A \cap B) = P(A) P(B)$，通常不为零。两个概率均为正的事件不可能既互斥又独立。
  • Sequential / dependent.序列 / 相依。 For without-replacement draws, the second probability depends on the first , use the multiplication rule with the conditional: $P(A_1 \cap A_2) = P(A_1) \, P(A_2 | A_1)$.无放回抽取时，第二次概率依赖第一次 , 用带条件的乘法规则：$P(A_1 \cap A_2) = P(A_1) \, P(A_2 | A_1)$。
  • Tree diagrams.树图。 Multiply along a branch (intersections), add across branches (unions). AB Math 30-2 indicator 3.4 expects "determining the probability of dependent or independent events" via this kind of organizer.沿一条枝相乘（交集），跨枝相加（并集）。AB Math 30-2 指标 3.4 要求"判定相依或独立事件的概率"，正是用此类图示。
AB Math 30-2 indicator 3.1: "compare, using examples, dependent and independent events" , the comparison itself is on the curriculum.AB Math 30-2 指标 3.1："用例子比较相依与独立事件" , "对比"本身就是课纲要求。
Worked Example 3 · Two-card draw, with and without replacement例题 3 · 抽两张牌（有 / 无放回）

From a standard $52$-card deck, draw two cards. Find $P($ both kings $)$ in two scenarios: (a) with replacement (return and reshuffle after card 1); (b) without replacement.从标准 $52$ 张牌中抽两张。在两种情形下求 $P($ 两张都是 K $)$：(a) 有放回（抽完第一张放回洗匀）；(b) 无放回。

(a) With replacement , independent.(a) 有放回 , 独立。 $P($ K on draw 1 $) = 4/52 = 1/13$. After replacement, deck is restored, so $P($ K on draw 2 $) = 1/13$ again.$P($ 第 1 张为 K $) = 4/52 = 1/13$。放回后牌组恢复，故 $P($ 第 2 张为 K $) = 1/13$ 不变。

$$ P(\text{both K, replacement}) = \frac{1}{13} \cdot \frac{1}{13} = \frac{1}{169} \approx 0.00592. $$

(b) Without replacement , dependent.(b) 无放回 , 相依。 $P($ K on 1 $) = 4/52 = 1/13$. Given a K was drawn, $3$ kings remain in $51$ cards, so $P($ K on 2 $| $ K on 1 $) = 3/51 = 1/17$.$P($ 第 1 张为 K $) = 4/52 = 1/13$。在第一张是 K 的条件下，$51$ 张中剩 $3$ 张 K，故 $P($ 第 2 张为 K $| $ 第 1 张为 K $) = 3/51 = 1/17$。

$$ P(\text{both K, no replacement}) = \frac{1}{13} \cdot \frac{1}{17} = \frac{1}{221} \approx 0.00452. $$

Compare.对比。 Without replacement is less likely , once a king is removed, the chance of a second king drops. The difference $(1/169) - (1/221) \approx 0.0014$ measures the dependence.无放回概率更小 , 取走一张 K 后，第二次抽 K 的机会下降。差 $(1/169) - (1/221) \approx 0.0014$ 衡量了相依程度。

$P(A) = 0.3$, $P(B) = 0.4$, $P(A \cap B) = 0.12$. Are $A$ and $B$ independent?$P(A) = 0.3$、$P(B) = 0.4$、$P(A \cap B) = 0.12$。$A$、$B$ 独立吗？
§3 · Q1
No, because $P(A) \ne P(B)$否，因为 $P(A) \ne P(B)$
No, because $P(A) + P(B) \ne 1$否，因为 $P(A) + P(B) \ne 1$
Yes, because $P(A)P(B) = 0.12 = P(A \cap B)$是，因为 $P(A)P(B) = 0.12 = P(A \cap B)$
Cannot be determined无法判定
$P(A) P(B) = 0.3 \cdot 0.4 = 0.12$, which equals $P(A \cap B)$. So independent by definition. Equivalently, $P(A | B) = 0.12 / 0.4 = 0.3 = P(A)$ , $B$ tells us nothing about $A$.$P(A) P(B) = 0.3 \cdot 0.4 = 0.12 = P(A \cap B)$，按定义独立。等价地，$P(A | B) = 0.12 / 0.4 = 0.3 = P(A)$ , $B$ 不提供关于 $A$ 的信息。
Apply the test $P(A \cap B) = P(A) P(B)$ literally. If equal, independent; if not, dependent.直接套用 $P(A \cap B) = P(A) P(B)$。相等则独立；否则相依。
Three components in a series circuit each work independently with probability $0.95$. The circuit works iff all three work. Find $P($ circuit works $)$ to four decimals.串联电路中三个元件各自独立工作的概率均为 $0.95$。电路正常当且仅当三者全部工作。求 $P($ 电路工作 $)$，保留四位小数。
§3 · Q2
$0.9500$
$0.8574$
$0.1426$
$2.8500$
Independence $\Rightarrow$ multiply: $P($ all work $) = 0.95^3 = 0.857375 \approx 0.8574$. (Add $0.95 \cdot 3$ is wrong , addition is for unions of mutually exclusive events.)独立 $\Rightarrow$ 相乘：$P($ 全部工作 $) = 0.95^3 = 0.857375 \approx 0.8574$。（$0.95 \cdot 3$ 错 , 加法用于互斥事件之并。）
Independent intersections multiply. $(0.95)^3 \ne 3 \cdot 0.95$.独立事件的交集要相乘。$(0.95)^3 \ne 3 \cdot 0.95$。
Section 4 · Discrete probability distributions第 4 节 · 离散概率分布

Random Variables, Expected Value, and Variance Honors — US A2 / AB 30-2随机变量、期望与方差 荣誉 — US A2 / AB 30-2

Curriculum note.课纲提示。 CCSSM puts random variables and expected value in the (+) cluster HSS-MD.A.1-.A.4 , honors / AP-feeder. ON MDM4U covers them in core under the Probability Distributions strand. AB Math 30-2 names expected outcomes informally but does not develop a full variance formula by hand. Treat §4 as required for ON MDM4U and AP Stats prep; honors for US Algebra II and AB Math 30-2.CCSSM 把随机变量与期望放在 (+) 簇 HSS-MD.A.1-.A.4 , 荣誉 / AP 衔接。ON MDM4U 在"概率分布"单元中作为核心覆盖。AB Math 30-2 非形式化提及期望结果，但不手算完整方差公式。本节在 ON MDM4U 与 AP Stats 备考中为必学；在美国代数 II 与 AB Math 30-2 中为荣誉级。
The PMF table is the object.概率质量函数表就是对象本身。

A discrete random variable $X$ takes finitely many values $x_1, x_2, \ldots, x_n$ with probabilities $p_i = P(X = x_i)$. The pair-list $\{(x_i, p_i)\}$ is the probability mass function (PMF). Required: $p_i \ge 0$ and $\sum p_i = 1$.离散随机变量 $X$ 取有限多个值 $x_1, x_2, \ldots, x_n$，对应概率 $p_i = P(X = x_i)$。对子列表 $\{(x_i, p_i)\}$ 即概率质量函数（PMF）。条件：$p_i \ge 0$ 且 $\sum p_i = 1$。

Expected value (mean):期望（均值）：

$$ E(X) \;=\; \mu \;=\; \sum_i x_i \, P(X = x_i) \;=\; \sum_i x_i \, p_i. $$

Variance and standard deviation:方差与标准差：

$$ \mathrm{Var}(X) \;=\; \sigma^2 \;=\; \sum_i (x_i - \mu)^2 \, p_i, \qquad \sigma \;=\; \sqrt{\mathrm{Var}(X)}. $$

Computational shortcut:计算捷径：

$$ \mathrm{Var}(X) \;=\; E(X^2) - [E(X)]^2 \;=\; \sum_i x_i^2 \, p_i - \mu^2. $$
  • $E$ is a weighted average.$E$ 是加权平均。 Each value times its probability. If outcomes are equally likely $p_i = 1/n$, then $E(X)$ reduces to the ordinary mean.每个值乘以其概率。等可能时 $p_i = 1/n$，$E(X)$ 退化为普通平均。
  • Variance has squared units.方差单位带平方。 If $X$ is in metres, $\mathrm{Var}(X)$ is in metres-squared; $\sigma$ restores the original units. Standard deviation is usually the reportable spread.若 $X$ 单位为米，$\mathrm{Var}(X)$ 单位为米-平方；$\sigma$ 回到原单位。标准差通常是可报告的离散度。
  • Linearity.线性性。 $E(aX + b) = a E(X) + b$ and $\mathrm{Var}(aX + b) = a^2 \mathrm{Var}(X)$ , constants come out of $E$, square out of $\mathrm{Var}$.$E(aX + b) = a E(X) + b$，$\mathrm{Var}(aX + b) = a^2 \mathrm{Var}(X)$ , 常数可从 $E$ 中提出，方差中需平方。
Worked Example 4 · Expected value and variance of a die roll例题 4 · 骰子掷一次的期望与方差

Let $X$ be the result of rolling one fair six-sided die. Find $E(X)$, $\mathrm{Var}(X)$, and $\sigma$.设 $X$ 为掷一颗均匀六面骰子的结果。求 $E(X)$、$\mathrm{Var}(X)$、$\sigma$。

PMF.PMF。 $P(X = k) = 1/6$ for $k = 1, 2, 3, 4, 5, 6$.$P(X = k) = 1/6$，$k = 1, 2, 3, 4, 5, 6$。

Expected value.期望。

$$ E(X) = \frac{1 + 2 + 3 + 4 + 5 + 6}{6} = \frac{21}{6} = 3.5. $$

Variance via $E(X^2) - \mu^2$.用 $E(X^2) - \mu^2$ 求方差。

$$ E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6} \approx 15.167. $$ $$ \mathrm{Var}(X) = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12} \approx 2.917. $$ $$ \sigma = \sqrt{35/12} \approx 1.708. $$

Interpret.解读。 $E(X) = 3.5$ sits at the centre (you cannot actually roll $3.5$). $\sigma \approx 1.7$ means typical rolls deviate from the mean by about $1.7$ , consistent with the $1$-$6$ range.$E(X) = 3.5$ 居中（实际掷不出 $3.5$）。$\sigma \approx 1.7$ 表示典型偏离均值约 $1.7$ , 与 $1$-$6$ 范围吻合。

A lottery ticket costs $\$2$. The prize is $\$500$, won with probability $1/1000$. Otherwise you lose your $\$2$. Find the expected profit per ticket.彩票每张 $\$2$。奖金 $\$500$，中奖概率 $1/1000$；否则损失 $\$2$。求每张票的期望盈亏。
§4 · Q1
$+\$0.50$
$-\$1.50$
$-\$2.00$
$+\$500$
Profit $= +498$ with probability $0.001$, $-2$ with probability $0.999$. $E(\text{profit}) = 498 \cdot 0.001 + (-2) \cdot 0.999 = 0.498 - 1.998 = -1.50$. The expected loss per ticket is $\$1.50$.盈亏 $= +498$ 概率 $0.001$；$-2$ 概率 $0.999$。$E($ 盈亏 $) = 498 \cdot 0.001 + (-2) \cdot 0.999 = 0.498 - 1.998 = -1.50$。每张票期望损失 $\$1.50$。
Profit on a win is $\$500 - \$2 = \$498$, not $\$500$. Loss otherwise is $\$2$. Multiply each by its probability and sum.中奖盈利 $\$500 - \$2 = \$498$（不是 $\$500$），否则损失 $\$2$。各乘对应概率再求和。
$X$ has PMF $P(X=0) = 0.2$, $P(X=1) = 0.5$, $P(X=2) = 0.3$. Find $E(X)$ and $\mathrm{Var}(X)$.$X$ 的 PMF：$P(X=0) = 0.2$、$P(X=1) = 0.5$、$P(X=2) = 0.3$。求 $E(X)$ 与 $\mathrm{Var}(X)$。
§4 · Q2
$E(X) = 1.1$, $\mathrm{Var}(X) = 0.49$
$E(X) = 1.0$, $\mathrm{Var}(X) = 0.50$
$E(X) = 1.1$, $\mathrm{Var}(X) = 1.70$
$E(X) = 1.5$, $\mathrm{Var}(X) = 0.25$
$E(X) = 0 \cdot 0.2 + 1 \cdot 0.5 + 2 \cdot 0.3 = 1.1$. $E(X^2) = 0 + 1 \cdot 0.5 + 4 \cdot 0.3 = 1.7$. $\mathrm{Var}(X) = 1.7 - (1.1)^2 = 1.7 - 1.21 = 0.49$.$E(X) = 0 \cdot 0.2 + 1 \cdot 0.5 + 2 \cdot 0.3 = 1.1$。$E(X^2) = 0 + 1 \cdot 0.5 + 4 \cdot 0.3 = 1.7$。$\mathrm{Var}(X) = 1.7 - (1.1)^2 = 1.7 - 1.21 = 0.49$。
Use the shortcut $\mathrm{Var}(X) = E(X^2) - \mu^2$. Compute $E(X^2)$ by squaring each value before multiplying by its probability.用捷径 $\mathrm{Var}(X) = E(X^2) - \mu^2$。先把每个值平方再乘概率，求 $E(X^2)$。
Section 5 · The binomial distribution第 5 节 · 二项分布

The Binomial Distribution: $X \sim B(n, p)$二项分布：$X \sim B(n, p)$

Cross-reference.交叉引用。 The binomial coefficient $\binom{n}{k}$ used here is developed in our HS Math guide on Combinatorics and the Binomial Theorem , see §3 (combinations) and §5 (Binomial Theorem). If $\binom{n}{k} = \tfrac{n!}{k!(n-k)!}$ does not look familiar, read that unit first.本节用到的二项系数 $\binom{n}{k}$ 在 HS Math 组合学与二项式定理一文中已建立 , 参见 §3（组合）与 §5（二项式定理）。若 $\binom{n}{k} = \tfrac{n!}{k!(n-k)!}$ 不熟悉，请先读那一单元。
Four conditions, one formula.四个条件，一个公式。

$X \sim B(n, p)$ means $X$ counts successes in $n$ trials where:$X \sim B(n, p)$ 表示 $X$ 计数 $n$ 次试验中的"成功"次数，且满足：

  1. $n$ trials,$n$ 次试验， fixed in advance.事先固定。
  2. Each trial is "success" or "failure"每次试验只有"成功"或"失败" (Bernoulli).（伯努利试验）。
  3. $P(\text{success}) = p$ is the same on every trial.每次 $P($ 成功 $) = p$ 不变。
  4. Trials are independent.各次试验独立。

PMF:概率质量函数：

$$ P(X = k) \;=\; \binom{n}{k} \, p^k \, (1-p)^{n-k}, \qquad k = 0, 1, 2, \ldots, n. $$

Mean and variance:均值与方差：

$$ E(X) \;=\; n p, \qquad \mathrm{Var}(X) \;=\; n p (1 - p), \qquad \sigma \;=\; \sqrt{n p (1 - p)}. $$
  • "At least", "at most" require sums."至少"、"至多"需要求和。 $P(X \ge k) = \sum_{j=k}^{n} P(X = j)$. Often easier via the complement: $P(X \ge 1) = 1 - P(X = 0) = 1 - (1-p)^n$.$P(X \ge k) = \sum_{j=k}^{n} P(X = j)$。常通过补集更省事：$P(X \ge 1) = 1 - P(X = 0) = 1 - (1-p)^n$。
  • Why $E(X) = np$.$E(X) = np$ 的来由。 Write $X = X_1 + X_2 + \cdots + X_n$ where $X_i$ indicates "success on trial $i$" ($X_i = 1$ or $0$). Then $E(X_i) = p$ and $E(X) = \sum E(X_i) = np$ by linearity. Independence is not needed for the mean , it is needed for the variance.写 $X = X_1 + X_2 + \cdots + X_n$，其中 $X_i$ 指示"第 $i$ 次成功"（$X_i = 1$ 或 $0$）。则 $E(X_i) = p$，由线性性 $E(X) = \sum E(X_i) = np$。均值不需独立性；方差需要。
ON MDM4U Probability Distributions strand develops this distribution from first principles and uses $E(X) = np$, $\mathrm{Var}(X) = np(1-p)$ as required formulas. AB Math 30-1 builds the binomial coefficient under Permutations, Combinations and Binomial Theorem (indicators 3.1-3.6, 4.1-4.4) , the application to the binomial distribution is natural but is not explicitly in the AB 30-1 curriculum (which stays inside combinatorics).ON MDM4U "概率分布"单元从第一性原理推导本分布，并把 $E(X) = np$、$\mathrm{Var}(X) = np(1-p)$ 作为必背公式。AB Math 30-1 在"排列、组合与二项式定理"中建立二项系数（指标 3.1-3.6、4.1-4.4）, 应用到二项分布是自然延伸，但 AB 30-1 课纲本身停在组合学层面。
Worked Example 5 · Exactly 3 of 5 free throws例题 5 · 5 罚 3 中

A basketball player makes free throws with probability $p = 0.7$, independent across attempts. They take $n = 5$ shots. Find $P($ exactly 3 made $)$ and $P($ at least 3 made $)$ to four decimals.某篮球运动员每次罚球独立命中概率 $p = 0.7$。共投 $n = 5$ 次。求 $P($ 恰中 3 球 $)$ 与 $P($ 至少中 3 球 $)$，保留四位小数。

Set up $X \sim B(5, 0.7)$.建立 $X \sim B(5, 0.7)$。

Exactly $3$.恰好 $3$ 次。

$$ P(X = 3) = \binom{5}{3} (0.7)^3 (0.3)^2 = 10 \cdot 0.343 \cdot 0.09 = 0.3087. $$

At least $3$ , sum $k = 3, 4, 5$.至少 $3$ 次 , 求和 $k = 3, 4, 5$。

$$ P(X = 4) = \binom{5}{4} (0.7)^4 (0.3)^1 = 5 \cdot 0.2401 \cdot 0.3 = 0.36015. $$ $$ P(X = 5) = \binom{5}{5} (0.7)^5 (0.3)^0 = 1 \cdot 0.16807 \cdot 1 = 0.16807. $$ $$ P(X \ge 3) = 0.3087 + 0.36015 + 0.16807 \approx 0.8369. $$

Sanity-check via mean.用均值核验。 $E(X) = 5 \cdot 0.7 = 3.5$. Most likely outcomes cluster near $3.5$ , consistent with the bulk of the mass landing on $k = 3, 4, 5$.$E(X) = 5 \cdot 0.7 = 3.5$。最可能结果集中在 $3.5$ 附近 , 与 $k = 3, 4, 5$ 处概率质量集中一致。

A fair coin is tossed $10$ times. Find $P($ exactly $6$ heads $)$ to four decimals.均匀硬币掷 $10$ 次。求 $P($ 恰好 $6$ 次正面 $)$，保留四位小数。
§5 · Q1
$0.2051$
$0.6000$
$0.0098$
$0.5000$
$X \sim B(10, 0.5)$. $P(X = 6) = \binom{10}{6} (0.5)^6 (0.5)^4 = 210 / 1024 \approx 0.2051$.$X \sim B(10, 0.5)$。$P(X = 6) = \binom{10}{6} (0.5)^6 (0.5)^4 = 210 / 1024 \approx 0.2051$。
Plug into $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ with $n = 10$, $k = 6$, $p = 1/2$. $\binom{10}{6} = 210$.代入 $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$，$n = 10$、$k = 6$、$p = 1/2$。$\binom{10}{6} = 210$。
A defective-rate is $p = 0.02$. A batch of $n = 200$ items is shipped, with $X = $ number defective. Find $P(X \ge 1)$, the probability the batch contains at least one defective item, to four decimals.次品率 $p = 0.02$。出货 $n = 200$ 件，$X = $ 次品数。求 $P(X \ge 1)$，即该批至少含一件次品的概率，保留四位小数。
§5 · Q2
$0.0176$
$0.0400$
$0.9824$
$0.9800$
Use the complement: $P(X \ge 1) = 1 - P(X = 0) = 1 - \binom{200}{0}(0.02)^0(0.98)^{200} = 1 - (0.98)^{200} \approx 1 - 0.0176 = 0.9824$. (Option (a) $0.0176$ is the complement $P(X = 0)$ itself.)用补集：$P(X \ge 1) = 1 - P(X = 0) = 1 - \binom{200}{0}(0.02)^0(0.98)^{200} = 1 - (0.98)^{200} \approx 1 - 0.0176 = 0.9824$。（选项 (a) $0.0176$ 是补集 $P(X = 0)$ 本身。）
"At least one" is the complement of "none". Compute $P(X = 0) = (1-p)^n$ first, then subtract from $1$. Don't confuse $1 - p$ (single-trial failure) with $(1 - p)^n$ (all $n$ trials fail)."至少一个"是"零个"的补集。先算 $P(X = 0) = (1-p)^n$，再用 $1$ 减。注意区分 $1 - p$（单次失败）与 $(1 - p)^n$（$n$ 次全失败）。
Section 6 · The normal distribution (intro)第 6 节 · 正态分布（入门）

The Normal Distribution: $\mu$, $\sigma$, and $z$-Scores正态分布：$\mu$、$\sigma$ 与 $z$-分数

Three things to know cold.三件必须脱稿掌握的事。
  • Shape and parameters.形状与参数。 A normal distribution $N(\mu, \sigma^2)$ is a symmetric bell-shaped curve centred at $\mu$ (mean = median = mode) with spread controlled by $\sigma$. Total area under the curve is $1$.正态分布 $N(\mu, \sigma^2)$ 以 $\mu$ 为对称中心呈钟形（均值 = 中位数 = 众数），离散度由 $\sigma$ 控制。曲线下总面积 $= 1$。
  • $68$-$95$-$99.7$ rule (empirical rule).$68$-$95$-$99.7$ 法则（经验法则）。 $P(|X - \mu| \le \sigma) \approx 0.68$, $P(|X - \mu| \le 2\sigma) \approx 0.95$, $P(|X - \mu| \le 3\sigma) \approx 0.997$. About $2/3$ within $1\sigma$, $95\%$ within $2\sigma$, virtually all within $3\sigma$.$P(|X - \mu| \le \sigma) \approx 0.68$；$P(|X - \mu| \le 2\sigma) \approx 0.95$；$P(|X - \mu| \le 3\sigma) \approx 0.997$。约 $2/3$ 在 $\pm 1\sigma$、$95\%$ 在 $\pm 2\sigma$、几乎全部在 $\pm 3\sigma$。
  • $z$-score.$z$-分数。
$$ z \;=\; \frac{x - \mu}{\sigma}. $$

$z$ is the number of standard deviations $x$ is above ($z > 0$) or below ($z < 0$) the mean. Standardising lets one $z$-table serve every normal distribution.$z$ 表示 $x$ 高于（$z > 0$）或低于（$z < 0$）均值多少个标准差。标准化后一张 $z$-表通用于所有正态分布。

  • Standard normal.标准正态。 $Z \sim N(0, 1)$. $P(Z \le z)$ is read from a $z$-table or calculator (e.g. normalcdf).$Z \sim N(0, 1)$。$P(Z \le z)$ 由 $z$-表或计算器（如 normalcdf）读取。
  • Symmetry.对称性。 $P(Z \le -z) = 1 - P(Z \le z)$ , lets you handle negatives with a positive table.$P(Z \le -z) = 1 - P(Z \le z)$ , 用正值表即可处理负值。
AB Math 30-2 Statistics General Outcome 1 ("normal distribution, including standard deviation, $z$-scores") names indicators 1.1-1.9: explain $\sigma$ via examples; properties of a normal curve (mean, median, mode, symmetry, area); $z$-scores from given values; contextual problems. The curriculum's parenthetical: "the focus of this outcome [is] on interpretation of data rather than on statistical calculations." Match that emphasis.AB Math 30-2 统计总目标 1（"正态分布，含标准差、$z$-分数"）的指标 1.1-1.9：用例子解释 $\sigma$；正态曲线性质（均值、中位数、众数、对称、面积）；由值算 $z$-分数；应用题。课纲附注："本目标重点在数据解读，而非统计计算"。要匹配此侧重。
Worked Example 6 · SAT score above 1400例题 6 · SAT 分数高于 1400

SAT scores are approximately normal with $\mu = 1060$ and $\sigma = 200$. What proportion of students score above $1400$? Use the empirical rule first, then a $z$-table for refinement.SAT 分数近似正态，$\mu = 1060$、$\sigma = 200$。多少比例学生分数高于 $1400$？先用经验法则，再用 $z$-表细化。

Standardise.标准化。

$$ z = \frac{1400 - 1060}{200} = \frac{340}{200} = 1.70. $$

Empirical-rule rough estimate.经验法则粗估。 $z = 1.70$ is between $1\sigma$ and $2\sigma$. $P(Z \ge 1) \approx (1 - 0.68)/2 = 0.16$; $P(Z \ge 2) \approx (1 - 0.95)/2 = 0.025$. So the answer is somewhere between $0.025$ and $0.16$, closer to $0.025$.$z = 1.70$ 介于 $1\sigma$ 与 $2\sigma$ 之间。$P(Z \ge 1) \approx (1 - 0.68)/2 = 0.16$；$P(Z \ge 2) \approx (1 - 0.95)/2 = 0.025$。答案应在 $0.025$ 到 $0.16$ 之间，靠近 $0.025$。

$z$-table lookup.$z$-表查询。 $P(Z \le 1.70) \approx 0.9554$, so $P(Z \ge 1.70) = 1 - 0.9554 = 0.0446$.$P(Z \le 1.70) \approx 0.9554$，故 $P(Z \ge 1.70) = 1 - 0.9554 = 0.0446$。

Interpret.解读。 About $4.5\%$ of students score above $1400$ , consistent with the empirical-rule range.约 $4.5\%$ 学生分数高于 $1400$ , 与经验法则区间吻合。

Adult heights are normal with $\mu = 170$ cm, $\sigma = 8$ cm. Roughly what proportion are between $162$ and $178$ cm? Use the empirical rule.成年人身高近似正态，$\mu = 170$ cm、$\sigma = 8$ cm。大致多少比例身高在 $162$ 至 $178$ cm 之间？用经验法则。
§6 · Q1
$\approx 0.34$
$\approx 0.95$
$\approx 0.68$
$\approx 0.50$
$162$ and $178$ are exactly $\mu \pm \sigma$. By the $68$-$95$-$99.7$ rule, $\approx 68\%$ of the data lies within $\pm 1\sigma$.$162$ 与 $178$ 恰为 $\mu \pm \sigma$。由 $68$-$95$-$99.7$ 法则，约 $68\%$ 数据落在 $\pm 1\sigma$ 内。
Compute how many $\sigma$ each bound is from the mean. Both are at $\pm 1\sigma$, so the empirical-rule $68\%$ applies.先算每个端点离均值多少个 $\sigma$。两端均为 $\pm 1\sigma$，故套用 $68\%$。
Test scores are $N(\mu = 75, \sigma = 10)$. A score of $88$ has $z$-score equal to:考试分数 $N(\mu = 75, \sigma = 10)$。$88$ 分的 $z$-分数为：
§6 · Q2
$0.88$
$1.30$
$13$
$-1.30$
$z = (x - \mu)/\sigma = (88 - 75)/10 = 13/10 = 1.30$. The score sits $1.3\sigma$ above the mean.$z = (x - \mu)/\sigma = (88 - 75)/10 = 13/10 = 1.30$。该分数高于均值 $1.3\sigma$。
Subtract the mean, then divide by $\sigma$. Sign tells direction (above mean = positive).先减均值，再除以 $\sigma$。正负表方向（高于均值为正）。
Section 7 · Descriptive statistics第 7 节 · 描述统计

Describing a Data Set: Centre, Spread, and Shape描述数据集：中心、离散与形状

Five summaries you should be able to compute by hand on small data.小数据集上应能手算的五个统计量。
  • Mean均值 $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$. Sensitive to outliers.对异常值敏感。
  • Median.中位数。 Sort, take the middle ($n$ odd) or the average of the two middles ($n$ even). Resistant to outliers , use median when the data is skewed.排序后取中间（$n$ 奇）或两中间值的平均（$n$ 偶）。对异常值稳健 , 数据偏态时用中位数。
  • Mode.众数。 Most-frequent value. Can be none, one, or multiple. Most useful for categorical or discrete data.最高频值。可无、可一、可多。最适合分类或离散数据。
  • Range极差 $= x_{\max} - x_{\min}$. Simplest spread; uses only two values, so extremely sensitive to outliers.最简单的离散度量；只用两个值，对异常值极度敏感。
  • IQRIQR（四分位距） $= Q_3 - Q_1$. Resistant spread , the middle $50\%$ of the data. CCSSM HSS-ID.A.3: "interpret differences in shape, center, and spread in the context of the data sets, accounting for possible effects of extreme data points (outliers)."稳健离散度 , 中间 $50\%$ 数据。CCSSM HSS-ID.A.3："在数据集背景下解释形状、中心、离散度的差异，并考虑极端数据点（异常值）的可能影响"。

Standard deviation (sample):标准差（样本）：

$$ s \;=\; \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}, \qquad \text{variance} \;=\; s^2. $$

Use $n$ in the denominator for a population, $n - 1$ for a sample (Bessel's correction , the curriculum usually says "sample standard deviation" or assumes technology).总体用分母 $n$；样本用 $n - 1$（贝塞尔修正 , 课纲通常写"样本标准差"或假设用技术工具）。

  • Boxplot (five-number summary):箱线图（五数概括）： min, $Q_1$, median, $Q_3$, max. Box from $Q_1$ to $Q_3$ (length $=$ IQR); whiskers to min and max (or to inside $1.5 \cdot $ IQR, with outliers shown as dots).最小值、$Q_1$、中位数、$Q_3$、最大值。箱体从 $Q_1$ 到 $Q_3$（长度 $=$ IQR）；须延伸至最大 / 最小（或 $1.5 \cdot$ IQR 内，异常值另点出）。
  • Histogram.直方图。 Bars over equal-width bins; height shows frequency or density. Shape labels: symmetric, right-skewed (tail to the right, mean $>$ median), left-skewed.等宽箱内的柱形；高度为频数或密度。形状标签：对称、右偏（尾在右，均值 $>$ 中位数）、左偏。
  • Comparison rule.比较规则。 CCSSM HSS-ID.A.2: "use statistics appropriate to the shape of the data distribution to compare center and spread of two or more different data sets." If data is skewed, compare median + IQR; if symmetric, mean + standard deviation.CCSSM HSS-ID.A.2："使用与数据分布形状相匹配的统计量，比较两个或多个数据集的中心与离散度"。偏态时比较中位数 + IQR；对称时比较均值 + 标准差。
Worked Example 7 · Full five-number summary on a small data set例题 7 · 小数据集上的完整五数概括

Quiz scores out of $10$: $4, 6, 7, 7, 8, 8, 8, 9, 10$ ($n = 9$). Find mean, median, mode, range, IQR, and sample standard deviation. State the boxplot summary.十分制小测成绩：$4, 6, 7, 7, 8, 8, 8, 9, 10$（$n = 9$）。求均值、中位数、众数、极差、IQR、样本标准差。给出箱线图概括。

Sort first.先排序。 Already sorted: $4, 6, 7, 7, 8, 8, 8, 9, 10$.已排序：$4, 6, 7, 7, 8, 8, 8, 9, 10$。

Mean.均值。 $(4 + 6 + 7 + 7 + 8 + 8 + 8 + 9 + 10)/9 = 67/9 \approx 7.44$.$(4 + 6 + 7 + 7 + 8 + 8 + 8 + 9 + 10)/9 = 67/9 \approx 7.44$。

Median.中位数。 $n = 9$ is odd; the middle is position $5$: median $= 8$.$n = 9$ 为奇，中位在第 $5$ 位：中位数 $= 8$。

Mode.众数。 $8$ appears three times , mode $= 8$.$8$ 出现三次 , 众数 $= 8$。

Range.极差。 $10 - 4 = 6$.$10 - 4 = 6$。

Quartiles.四分位数。 $Q_1$ = median of lower half $\{4, 6, 7, 7\}$ = $(6 + 7)/2 = 6.5$. $Q_3$ = median of upper half $\{8, 8, 9, 10\}$ = $(8 + 9)/2 = 8.5$. IQR $= 8.5 - 6.5 = 2$.$Q_1$ = 下半 $\{4, 6, 7, 7\}$ 的中位 $= (6 + 7)/2 = 6.5$。$Q_3$ = 上半 $\{8, 8, 9, 10\}$ 的中位 $= (8 + 9)/2 = 8.5$。IQR $= 8.5 - 6.5 = 2$。

Sample standard deviation.样本标准差。 Deviations from $\bar{x} \approx 7.444$: $-3.444, -1.444, -0.444, -0.444, 0.556, 0.556, 0.556, 1.556, 2.556$. Squared sum $\approx 11.86 + 2.09 + 0.20 + 0.20 + 0.31 + 0.31 + 0.31 + 2.42 + 6.53 \approx 24.22$. $s^2 = 24.22 / 8 \approx 3.03$. $s \approx 1.74$.与 $\bar{x} \approx 7.444$ 的偏差：$-3.444, -1.444, -0.444, -0.444, 0.556, 0.556, 0.556, 1.556, 2.556$。平方和 $\approx 11.86 + 2.09 + 0.20 + 0.20 + 0.31 + 0.31 + 0.31 + 2.42 + 6.53 \approx 24.22$。$s^2 = 24.22 / 8 \approx 3.03$，$s \approx 1.74$。

Boxplot five-number summary.箱线图五数概括。 min $= 4$, $Q_1 = 6.5$, median $= 8$, $Q_3 = 8.5$, max $= 10$. The box ($Q_1$ to $Q_3$) is narrow; the lower whisker (from $Q_1$ down to $4$) is much longer than the upper whisker , suggests a slight left skew, consistent with mean ($7.44$) $<$ median ($8$).最小 $= 4$、$Q_1 = 6.5$、中位 $= 8$、$Q_3 = 8.5$、最大 $= 10$。箱体（$Q_1$ 至 $Q_3$）较窄；下须（$Q_1$ 至 $4$）远长于上须 , 提示轻微左偏，与均值（$7.44$）$<$ 中位（$8$）一致。

A data set is strongly right-skewed (e.g. household incomes). Which pair best summarises centre + spread?数据集强烈右偏（如家庭收入）。哪一对最适合概括中心 + 离散？
§7 · Q1
Mean + standard deviation均值 + 标准差
Median + IQR中位数 + IQR
Mode + range众数 + 极差
Mean + range均值 + 极差
Mean and standard deviation are pulled by the long right tail; median and IQR are resistant. CCSSM HSS-ID.A.2 names this matching rule explicitly: "use statistics appropriate to the shape of the data distribution."均值与标准差被长右尾拉偏；中位数与 IQR 稳健。CCSSM HSS-ID.A.2 明确点名此匹配规则："使用与数据分布形状相匹配的统计量"。
Skewed $\Rightarrow$ use the resistant pair (median, IQR). Symmetric $\Rightarrow$ use mean and standard deviation.偏态 $\Rightarrow$ 用稳健对（中位数、IQR）。对称 $\Rightarrow$ 用均值与标准差。
Data: $2, 4, 4, 5, 7, 9, 11$. Find the median and IQR.数据：$2, 4, 4, 5, 7, 9, 11$。求中位数与 IQR。
§7 · Q2
median $= 4$, IQR $= 7$中位 $= 4$、IQR $= 7$
median $= 7$, IQR $= 5$中位 $= 7$、IQR $= 5$
median $= 5$, IQR $= 5$中位 $= 5$、IQR $= 5$
median $= 5$, IQR $= 9$中位 $= 5$、IQR $= 9$
$n = 7$, sorted, middle (position $4$) is $5$, so median $= 5$. Lower half (positions 1-3): $\{2, 4, 4\}$, $Q_1 = 4$. Upper half (positions 5-7): $\{7, 9, 11\}$, $Q_3 = 9$. IQR $= 9 - 4 = 5$.$n = 7$，已排序，中间（第 $4$ 位）为 $5$，故中位 $= 5$。下半（1-3 位）$\{2, 4, 4\}$，$Q_1 = 4$；上半（5-7 位）$\{7, 9, 11\}$，$Q_3 = 9$。IQR $= 9 - 4 = 5$。
$n$ odd: median is the single middle value, then $Q_1$ and $Q_3$ are medians of the halves excluding the overall median.$n$ 奇：中位为单中间值；$Q_1$、$Q_3$ 是不含整体中位时上下半的中位。
Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Setup discipline解题前的纪律
  • Always picture the structure.先画结构图。 Tree diagram for sequential events. Two-way table for conditional probability. Histogram or boxplot for descriptive stats. The picture forces you to declare independence (or not) and to identify the conditioning event.序列事件画树图。条件概率画双向表。描述统计画直方图或箱线图。图能强迫你声明独立性（或非独立性）并明确条件事件。
  • Name the random variable.命名随机变量。 Before any binomial / normal calculation, write a one-line statement: "$X$ = number of successes in $n$ trials, $X \sim B(n, p)$" or "$X$ = scoring distribution, $X \sim N(\mu, \sigma^2)$". This is the markscheme's first communication point in AP Stats and IB Math AI HL.做任何二项 / 正态计算前，先写一句话："$X$ 等于 $n$ 次试验的成功次数，$X \sim B(n, p)$" 或 "$X$ 等于分数分布，$X \sim N(\mu, \sigma^2)$"。AP Stats 与 IB Math AI HL 评分表的第一沟通分就给这一步。
  • State assumptions explicitly.明确写出假设。 "Assume independence" before $P(A \cap B) = P(A)P(B)$. "Assume approximately normal" before using $z$-scores. The grader rewards the disclosure even if the assumption is questionable.用 $P(A \cap B) = P(A)P(B)$ 前先写"假设独立"。用 $z$-分数前先写"假设近似正态"。即便假设可疑，写明本身就有分。
Probability rules (§1-§3)概率规则（§1-§3）
  • Complement first.先想补集。 If the question says "at least one", compute $1 - P($ none $)$. Direct summation almost always wastes time.题目说"至少一个"时，算 $1 - P($ 一个也没有 $)$。直接求和几乎都浪费时间。
  • "And" vs "or"."与" 与 "或"。 "And" $\to$ intersection $\to$ multiply (if independent) or multiplication rule with conditional. "Or" $\to$ union $\to$ add and subtract overlap."与" $\to$ 交集 $\to$ 独立时相乘，相依时用条件乘法。"或" $\to$ 并集 $\to$ 相加再减去重叠。
  • Independent ≠ mutually exclusive.独立 ≠ 互斥。 Different rules. Mutually exclusive events with positive probability are not independent , knowing one occurred forces the other to have probability $0$.两条不同的规则。具有正概率的互斥事件不独立 , 知道一者发生即强制另一者概率为 $0$。
Distributions (§4-§6) Honors — US A2 / AB 30-2分布（§4-§6）荣誉 — US A2 / AB 30-2
  • Binomial flag.二项识别。 If the question has "fixed $n$ trials", "success / failure", "same $p$ each time", and "independent" $\Rightarrow$ binomial. Reach for $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$.题目含"固定 $n$ 次"、"成功 / 失败"、"每次 $p$ 相同"、"独立" $\Rightarrow$ 二项。立刻套 $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$。
  • Empirical rule before $z$-table.先用经验法则再查 $z$-表。 For "between $\mu - \sigma$ and $\mu + \sigma$", you don't need a table , $68\%$ by inspection. Reserve the $z$-table for non-round multiples of $\sigma$."$\mu - \sigma$ 与 $\mu + \sigma$ 之间"无需查表 , 一眼 $68\%$。$\sigma$ 的非整数倍才查 $z$-表。
  • Variance vs standard deviation.方差与标准差。 When asked for "standard deviation", take the square root. When asked for "variance", do not. Half the points lost in distribution problems come from confusing these.问"标准差"开根号；问"方差"不开。分布题失分一半来自混淆二者。
Descriptive statistics (§7)描述统计（§7）
  • Sort, then compute.先排序，再计算。 Median, $Q_1$, $Q_3$, and IQR are only correct on sorted data. Forgetting to sort first is the single most common error in this block.中位数、$Q_1$、$Q_3$、IQR 必须基于排序数据。不排序就算是本块最常见错误。
  • Match the summary to the shape.概括量要匹配形状。 Symmetric $\Rightarrow$ mean and standard deviation. Skewed or outlier-heavy $\Rightarrow$ median and IQR. CCSSM HSS-ID.A.2 grades this choice explicitly.对称 $\Rightarrow$ 均值与标准差；偏态或异常值多 $\Rightarrow$ 中位数与 IQR。CCSSM HSS-ID.A.2 显式评分此选择。
  • Read the boxplot picture.读箱线图。 A long whisker on one side indicates skew toward that side. The box width is IQR. Outliers (if marked separately) sit beyond $1.5 \cdot $ IQR from $Q_1$ or $Q_3$.一侧须长 $\Rightarrow$ 偏向该侧。箱宽即 IQR。异常值（若另点出）位于 $Q_1$ 或 $Q_3$ 外 $1.5 \cdot$ IQR 之外。
  • Sanity-check standard deviation.合理性核验标准差。 For most data sets, $s$ is roughly $\frac{1}{4}$ of the range. If your computed $s$ is wildly outside that, recheck for a missed squaring or a forgotten $n - 1$.对大多数数据集，$s$ 约为极差的 $1/4$。若结果偏离过大，回查漏的平方或漏的 $n - 1$。
Active Recall主动复习

Flashcards闪卡

0 / 16 flipped0 / 16 已翻
Classical probability?经典概率？
$$P(A) = \frac{|A|}{|S|}$$ (equally likely outcomes)（等可能结果）
Complement rule?补集规则？
$$P(A^c) = 1 - P(A)$$
Addition rule?加法规则？
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ CCSSM HSS-CP.B.7CCSSM HSS-CP.B.7
Conditional probability?条件概率？
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$
Multiplication rule?乘法规则？
$$P(A \cap B) = P(A \mid B) P(B) = P(B \mid A) P(A)$$
Bayes' rule?贝叶斯法则？
$$P(A \mid B) = \frac{P(B \mid A) P(A)}{P(B)}$$
Independence test?独立性检验？
$$P(A \cap B) = P(A) P(B)$$ CCSSM HSS-CP.A.2CCSSM HSS-CP.A.2
Expected value $E(X)$?期望 $E(X)$？
$$E(X) = \mu = \sum_i x_i \, P(X = x_i)$$
Variance shortcut?方差捷径？
$$\mathrm{Var}(X) = E(X^2) - \mu^2$$ $\sigma = \sqrt{\mathrm{Var}(X)}$$\sigma = \sqrt{\mathrm{Var}(X)}$
Binomial PMF, $X \sim B(n, p)$?二项分布 PMF？
$$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$
Binomial mean and variance?二项分布均值与方差？
$$E(X) = np, \;\; \mathrm{Var}(X) = np(1-p)$$
$z$-score?$z$-分数？
$$z = \frac{x - \mu}{\sigma}$$
$68$-$95$-$99.7$ rule?$68$-$95$-$99.7$ 法则？
$\pm 1\sigma$: $68\%$. $\pm 2\sigma$: $95\%$. $\pm 3\sigma$: $99.7\%$.$\pm 1\sigma$：$68\%$；$\pm 2\sigma$：$95\%$；$\pm 3\sigma$：$99.7\%$。
Sample standard deviation?样本标准差？
$$s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2}$$
IQR?IQR？
$$\text{IQR} = Q_3 - Q_1$$ middle $50\%$ of the data中间 $50\%$ 数据
When to use median + IQR?何时用中位数 + IQR？
When data is skewed or has outliers. CCSSM HSS-ID.A.2: match summary to shape.数据偏态或多异常值时。CCSSM HSS-ID.A.2：概括量要匹配形状。
Mixed Practice综合练习

Practice Quiz综合测验

A bag has $7$ white and $3$ black marbles. Two are drawn without replacement. Find $P($ both white $)$.袋中 $7$ 白 $3$ 黑弹珠。无放回抽两个。求 $P($ 两白 $)$。
Q1
$49/100$
$42/90 = 7/15$
$7/10$
$1/2$
Dependent: $P($ 1st W $) = 7/10$; $P($ 2nd W $|$ 1st W $) = 6/9$. $P($ both W $) = (7/10)(6/9) = 42/90 = 7/15 \approx 0.467$. (Option (a) $49/100$ would be correct with replacement.)相依：$P($ 1st 白 $) = 7/10$；$P($ 2nd 白 $|$ 1st 白 $) = 6/9$。$P($ 两白 $) = (7/10)(6/9) = 42/90 = 7/15 \approx 0.467$。（选项 (a) $49/100$ 是有放回的答案。）
Multiplication rule with conditional. After removing one white, $6$ whites in $9$ remain.带条件的乘法规则。取走一白后剩 $6$ 白共 $9$ 个。
$P(A) = 0.6$, $P(B) = 0.5$, $P(A \cup B) = 0.8$. Find $P(A \cap B)$ and $P(A | B)$.$P(A) = 0.6$、$P(B) = 0.5$、$P(A \cup B) = 0.8$。求 $P(A \cap B)$ 与 $P(A | B)$。
Q2
$P(A \cap B) = 0.10$, $P(A | B) = 0.50$
$P(A \cap B) = 0.30$, $P(A | B) = 0.50$
$P(A \cap B) = 0.30$, $P(A | B) = 0.60$
$P(A \cap B) = 0.30$, $P(A | B) = 0.30$
Addition rule reversed: $P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.8 = 0.30$. Then $P(A | B) = 0.30 / 0.50 = 0.60$.逆用加法规则：$P(A \cap B) = P(A) + P(B) - P(A \cup B) = 0.6 + 0.5 - 0.8 = 0.30$。再由 $P(A | B) = 0.30 / 0.50 = 0.60$。
Solve $P(A \cap B)$ from the addition rule, then divide by $P(B)$ for $P(A | B)$.用加法规则解 $P(A \cap B)$，再除以 $P(B)$ 得 $P(A | B)$。
A factory's components have defective rate $0.03$. A sample of $50$ items is taken. Find $P($ exactly $2$ defective $)$ to four decimals.工厂次品率 $0.03$。抽 $50$ 件。求 $P($ 恰好 $2$ 件次品 $)$，保留四位小数。
Q3
$0.5000$
$0.2555$
$0.0030$
$0.0900$
$X \sim B(50, 0.03)$. $P(X = 2) = \binom{50}{2} (0.03)^2 (0.97)^{48} = 1225 \cdot 0.0009 \cdot (0.97)^{48}$. $(0.97)^{48} \approx 0.2317$. So $P \approx 1225 \cdot 0.0009 \cdot 0.2317 \approx 0.2555$.$X \sim B(50, 0.03)$。$P(X = 2) = \binom{50}{2} (0.03)^2 (0.97)^{48} = 1225 \cdot 0.0009 \cdot (0.97)^{48}$。$(0.97)^{48} \approx 0.2317$，故 $P \approx 1225 \cdot 0.0009 \cdot 0.2317 \approx 0.2555$。
Binomial PMF: $\binom{n}{k} p^k (1-p)^{n-k}$. The $(0.97)^{48}$ tail is what controls the magnitude here.二项 PMF：$\binom{n}{k} p^k (1-p)^{n-k}$。本题数量级由 $(0.97)^{48}$ 控制。
$X \sim B(100, 0.4)$. Find $E(X)$ and $\sigma$. 🇨🇦 ON MDM4U / AP Stats$X \sim B(100, 0.4)$。求 $E(X)$ 与 $\sigma$。🇨🇦 ON MDM4U / AP Stats
Q4
$E(X) = 40$, $\sigma = 24$
$E(X) = 40$, $\sigma = 6$
$E(X) = 40$, $\sigma \approx 4.90$
$E(X) = 0.4$, $\sigma = 4.9$
$E(X) = np = 100 \cdot 0.4 = 40$. $\mathrm{Var}(X) = np(1-p) = 100 \cdot 0.4 \cdot 0.6 = 24$. $\sigma = \sqrt{24} \approx 4.90$.$E(X) = np = 100 \cdot 0.4 = 40$。$\mathrm{Var}(X) = np(1-p) = 100 \cdot 0.4 \cdot 0.6 = 24$。$\sigma = \sqrt{24} \approx 4.90$。
Variance is $np(1-p) = 24$, but $\sigma$ takes one more square root. Don't stop at variance.方差 $np(1-p) = 24$，但 $\sigma$ 还需开根。别停在方差。
Heights are $N(\mu = 170, \sigma = 10)$ cm. Find $P(160 \le X \le 185)$ to four decimals.身高 $N(\mu = 170, \sigma = 10)$ cm。求 $P(160 \le X \le 185)$，保留四位小数。
Q5
$\approx 0.7745$
$\approx 0.6800$
$\approx 0.5000$
$\approx 0.9500$
$z_1 = (160 - 170)/10 = -1$, $z_2 = (185 - 170)/10 = 1.5$. $P(-1 \le Z \le 1.5) = \Phi(1.5) - \Phi(-1) \approx 0.9332 - 0.1587 = 0.7745$.$z_1 = (160 - 170)/10 = -1$，$z_2 = (185 - 170)/10 = 1.5$。$P(-1 \le Z \le 1.5) = \Phi(1.5) - \Phi(-1) \approx 0.9332 - 0.1587 = 0.7745$。
Standardise each end to a $z$, then subtract cumulative probabilities $\Phi(z_2) - \Phi(z_1)$.两端各标准化为 $z$，再用累积概率差 $\Phi(z_2) - \Phi(z_1)$。
Data set: $3, 7, 7, 8, 9, 12, 15$. Find the mean.数据集 $3, 7, 7, 8, 9, 12, 15$。求均值。
Q6
$8$
$\approx 8.71$
$7$
$12$
$\bar{x} = (3 + 7 + 7 + 8 + 9 + 12 + 15)/7 = 61/7 \approx 8.71$. (Median $= 8$ , do not confuse with mean.)$\bar{x} = (3 + 7 + 7 + 8 + 9 + 12 + 15)/7 = 61/7 \approx 8.71$。（中位 $= 8$ , 勿混淆。）
Sum and divide by $n = 7$. Mean is not the middle value , that's the median.求和除以 $n = 7$。均值不是中间值 , 那是中位数。
In a class, $60\%$ of students play sports, $30\%$ play music, and $20\%$ do both. A student is chosen at random. Find $P($ music $|$ sports $)$.某班 $60\%$ 学生参加体育，$30\%$ 参加音乐，$20\%$ 两者都参加。随机抽一名学生。求 $P($ 音乐 $|$ 体育 $)$。
Q7
$1/3$
$2/3$
$0.20$
$0.30$
$P($ music $|$ sports $) = P($ both $) / P($ sports $) = 0.20 / 0.60 = 1/3$. (Independence would require $P($ music $|$ sports $) = P($ music $) = 0.30$; since $1/3 \approx 0.333 \ne 0.30$, the events are slightly dependent.)$P($ 音乐 $|$ 体育 $) = P($ 两者 $) / P($ 体育 $) = 0.20 / 0.60 = 1/3$。（若独立则应等于 $P($ 音乐 $) = 0.30$；$1/3 \approx 0.333 \ne 0.30$，故微相依。）
Conditional probability divides the intersection by the conditioning event: $P($ both $)/P($ sports $)$.条件概率用交集除以条件事件：$P($ 两者 $)/P($ 体育 $)$。
A game pays $\$10$ if a die shows $6$, $\$2$ if it shows $4$ or $5$, and $\$0$ otherwise. Find the expected payout per roll.某游戏掷骰子：出 $6$ 赔 $\$10$；出 $4$ 或 $5$ 赔 $\$2$；否则 $\$0$。求每掷期望支出。
Q8
$\$5.00$
$\$2.00$
$\$2.33$
$\$12.00$
$E($ payout $) = 10 \cdot 1/6 + 2 \cdot 2/6 + 0 \cdot 3/6 = 10/6 + 4/6 = 14/6 \approx 2.33$.$E($ 支出 $) = 10 \cdot 1/6 + 2 \cdot 2/6 + 0 \cdot 3/6 = 10/6 + 4/6 = 14/6 \approx 2.33$。
Weighted sum: each payout times its probability ($1/6$ or $2/6$ or $3/6$), then add.加权求和：每个支出乘其概率（$1/6$、$2/6$、$3/6$），再相加。
In a test, $X \sim N(70, 64)$ (so $\sigma = 8$). The top $10\%$ of students earn an A. Approximately what is the cutoff score? Use $z_{0.90} \approx 1.28$.考试 $X \sim N(70, 64)$（即 $\sigma = 8$）。前 $10\%$ 获 A。A 的分数线大致为多少？用 $z_{0.90} \approx 1.28$。
Q9
$78$
$80$
$86$
$90$
Invert $z = (x - \mu)/\sigma$: $x = \mu + z \sigma = 70 + 1.28 \cdot 8 = 70 + 10.24 \approx 80$. Cutoff $\approx 80$.逆解 $z = (x - \mu)/\sigma$：$x = \mu + z \sigma = 70 + 1.28 \cdot 8 = 70 + 10.24 \approx 80$。分数线 $\approx 80$。
Find the $z$ matching the desired tail, then convert back via $x = \mu + z\sigma$.先找匹配尾概率的 $z$，再用 $x = \mu + z\sigma$ 反求。
A disease has prevalence $2\%$. A test has sensitivity $95\%$ and false-positive rate $3\%$. What is $P($ disease $|$ positive $)$? 🇨🇦 ON MDM4U / AP Stats疾病患病率 $2\%$。检测灵敏度 $95\%$，假阳性率 $3\%$。求 $P($ 患病 $|$ 阳性 $)$。🇨🇦 ON MDM4U / AP Stats
Q10
$0.95$
$0.50$
$\approx 0.39$
$\approx 0.02$
$P(+) = 0.95 \cdot 0.02 + 0.03 \cdot 0.98 = 0.019 + 0.0294 = 0.0484$. Bayes: $P(D | +) = (0.95 \cdot 0.02) / 0.0484 = 0.019 / 0.0484 \approx 0.393$. Despite a $95\%$-sensitive test, only about $39\%$ chance of disease , base-rate effect.$P(+) = 0.95 \cdot 0.02 + 0.03 \cdot 0.98 = 0.019 + 0.0294 = 0.0484$。贝叶斯：$P(D | +) = (0.95 \cdot 0.02) / 0.0484 = 0.019 / 0.0484 \approx 0.393$。尽管灵敏度 $95\%$，患病概率仍仅约 $39\%$ , 基础概率效应。
Bayes' rule: $P(D | +) = P(+ | D) P(D) / P(+)$, with $P(+)$ from total probability. Rare disease + small false-positive rate still gives low posterior probability.贝叶斯：$P(D | +) = P(+ | D) P(D) / P(+)$，$P(+)$ 由全概率公式。稀有疾病 + 即使假阳性率小，后验概率仍偏低。
Self-check自查

Readiness Checklist准备就绪清单

Tick each item when you can do it cold, without notes, on a first attempt.能在无笔记、首次尝试下完成，再勾选每一项。

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Next Steps后续单元

What This Feeds Into本单元的去向

Probability and descriptive statistics are the foundation for every later course that does inference, modelling, or empirical reasoning. The two natural in-repo continuations are IB Math (AA / AI) statistics topics and AP Statistics , both of which assume fluency on §1-§7 from day one. The binomial in §5 and the normal in §6 are also the workhorse distributions in AP Physics measurement-error problems and AP CSA Monte-Carlo-style projects. The cross-references below point at units in this repo that build directly on this material.概率与描述统计是所有后续涉及推断、建模或实证推理课程的基础。仓库内的两个自然延续是 IB Math（AA / AI）的统计主题与 AP Statistics , 二者均自第一天起默认你熟练 §1-§7。§5 二项与 §6 正态也是 AP Physics 测量误差与 AP CSA 蒙特卡洛风格项目中的主力分布。下面链接指向本仓库直接建基于此的相关单元。

Within High School Math.在 HS Math 内部。

Unit 11 (Combinatorics and the Binomial Theorem) is the direct prerequisite for §5: it builds $\binom{n}{k}$ as the count of $k$-element subsets and proves the Binomial Theorem $(a + b)^n = \sum_k \binom{n}{k} a^{n-k} b^k$. The probability-mass function in §5 is exactly the $k$-th term of $(p + q)^n$ where $q = 1 - p$ , same algebra, new interpretation. Unit 14 (Statistical Inference, if your row sends you there) extends §7 to sampling distributions, confidence intervals, and hypothesis tests; this guide is the prerequisite reading.Unit 11（组合学与二项式定理）是 §5 的直接前置：它把 $\binom{n}{k}$ 建为 $k$ 元子集的计数，并证明二项式定理 $(a + b)^n = \sum_k \binom{n}{k} a^{n-k} b^k$。§5 的概率质量函数恰是 $(p + q)^n$ 的第 $k$ 项（$q = 1 - p$）, 代数相同，解释不同。Unit 14（统计推断，如果你的行指向那里）把 §7 推广到抽样分布、置信区间与假设检验；本指南是其前置阅读。

Across the AP and IB feeders in this repo.本仓库中的 AP 与 IB 衔接单元。

HS Math Unit 11 · Combinatorics and the Binomial Theorem (the $\binom{n}{k}$ infrastructure that §5 rests on)HS Math Unit 11 · 组合学与二项式定理（$\binom{n}{k}$ 基础，§5 的依赖） IB Math HL D1 · Counting, Probability, Statistics (HL-level probability rules and one-variable stats at HL depth)IB Math HL D1 · 计数、概率、统计（HL 级概率规则与单变量统计深度） IB Math HL D3 · Probability Distributions (binomial and normal at HL depth: standardisation, inverse normal, distribution of $\bar{X}$)IB Math HL D3 · 概率分布（HL 级二项与正态：标准化、逆正态、$\bar{X}$ 的分布） AP Physics Unit 1 · Kinematics (measurement-uncertainty paragraph uses $\sigma$ and the $68$-$95$-$99.7$ rule from §6)AP Physics Unit 1 · 运动学（测量不确定性段使用 §6 的 $\sigma$ 与 $68$-$95$-$99.7$ 法则） AP CSA Unit 2 · Using Objects (Math.random( ) underpins Monte-Carlo simulation; binomial trials are the natural exercise)AP CSA Unit 2 · 使用对象（Math.random( ) 支撑蒙特卡洛模拟；二项试验是天然练习）

If you are aiming for AP Statistics, this guide covers about three weeks of the AP Stats syllabus condensed; budget more time on the binomial-cumulative and normal-table mechanics. If you are aiming for IB Math AA HL, Topic 4 (Statistics and Probability) picks up exactly where this leaves off , especially the discrete-random-variable framework from §4 and the binomial from §5. If you are aiming for IB Math AI HL, the statistical content is heavier still: Topic 4 plus the additional inference work at HL. If you are aiming for the SAT, expect basic probability and a handful of "mean / median / range" stat-reading items in the calculator section , §1, §3, and §7 cover all of that fluently.备考 AP Statistics：本指南覆盖 AP Stats 课程约三周内容（压缩版）；二项累积形式与正态查表机械操作上多投入时间。备考 IB Math AA HL：Topic 4（统计与概率）从此接续 , 尤其 §4 的离散随机变量框架与 §5 的二项分布。备考 IB Math AI HL：统计内容更重：Topic 4 加 HL 级额外推断。备考 SAT：计算器节会出现基础概率与少量"均值 / 中位 / 极差"统计读图题 , §1、§3、§7 已熟练覆盖。

This is the ladder we coach students up: from these high school foundations into AP and IB, and on to first-year university. If you would rather have a guide for the next step than take it on trust, Dingrui Scholars tutors teach exactly this path in one-on-one sessions and small-group classes. A free assessment is the place to start.这正是我们陪伴学生攀登的阶梯：从这些高中基础，迈入 AP 与 IB，再走向大学一年级。如果你希望在下一步有人引路，而不是凭感觉前行，Dingrui Scholars 的导师正是沿着这条路径，以一对一辅导与小班课授课。欢迎从免费评估开始。

Common Questions常见问题

What is conditional probability?什么是条件概率？

Conditional probability is the chance of one event given that another has happened, found by dividing the probability of both events by the probability of the given event.条件概率是在另一事件已发生的前提下某事件发生的概率，等于两事件同时发生的概率除以已知事件的概率。

How do I calculate expected value?如何计算期望值？

Expected value is the long-run average: multiply each outcome by its probability and add the products together across all possible outcomes.期望值是长期平均值：把每个结果乘以它的概率，再把所有可能结果的乘积相加。

When do I use the binomial versus the normal distribution?什么时候用二项分布，什么时候用正态分布？

Use the binomial distribution for a fixed number of independent yes-or-no trials, and use the normal distribution for continuous data that clusters symmetrically around a mean.对固定次数、相互独立的是非试验用二项分布，对围绕均值对称聚集的连续数据用正态分布。

Syllabus alignment & how to use this guide大纲对照与使用说明
Read me first先读这里

How to use this guide如何使用本指南

Probability and statistics is the unit whose home course varies the most across our four curricula. Ontario has a whole Grade 12 course for it (MDM4U). Alberta splits it: Math 30-1 carries the binomial-coefficient infrastructure under Permutations, Combinations and Binomial Theorem, while Math 30-2 carries the probability and statistics strands explicitly. BC's Pre-Calc track skips it entirely; BC students meet it in the parallel Foundations of Mathematics (FPM) stream. US Common Core distributes it across HSS-CP (conditional probability and the rules), HSS-MD (distributions, the (+) cluster), HSS-IC (inference), and HSS-ID (interpreting data). The table tells you which sections match your row.概率与统计的"对口课程"在我们对照的四套大纲中差异最大。安大略有专门的 12 年级课程 MDM4U；阿尔伯塔分两门：Math 30-1 在"排列、组合与二项式定理"单元承担二项系数基础，Math 30-2 显式承担概率与统计；BC 的 Pre-Calc 主线完全跳过，BC 学生在并行的 Foundations of Mathematics（FPM）课程中学习；美国共同核心则把它分散在 HSS-CP（条件概率与规则）、HSS-MD（分布，(+) 簇）、HSS-IC（推断）、HSS-ID（解释数据）。下表告诉你当前大纲下应重点学习哪些节。

If you are in…如果你在… Focus on these sections重点学习 Defer / skip可推迟 Source依据
🇺🇸 US Algebra II / Stats , HSS-CP, HSS-ID美国 代数 II / 统计 , HSS-CP、HSS-ID §1, §2, §3, §7§1、§2、§3、§7 §4, §5, §6 (HSS-MD distributions are (+) honors content; defer to a Stats or AP Stats course)§4、§5、§6（HSS-MD 分布属 (+) 荣誉级；推迟至统计选修或 AP Stats） ccssm_hs_math.pdf , HSS-CP.A.1-5 (sample space, conditional, independence), HSS-CP.B.6-7 (rules), HSS-ID.A.2-3 (centre, spread, outliers), HSS-CP.A.1-5（样本空间、条件、独立）、HSS-CP.B.6-7（规则）、HSS-ID.A.2-3（中心、离散、异常值）
🇺🇸 US AP-feeder / AP Stats prep美国 AP 衔接 / AP Stats 备考 All seven sections at full depth. §5 binomial and §6 normal are the AP Stats core inference triggers全部 7 节完整深度。§5 二项与 §6 正态是 AP Stats 推断的核心触发点 Nothing , AP Stats expects fluency on all seven sections from week 1无 , AP Stats 自第一周即要求熟练全部 7 节 ccssm_hs_math.pdf , the (+) cluster HSS-MD.A is explicitly AP-feeder; HSS-IC.A inference is the AP Stats prerequisite, (+) 簇 HSS-MD.A 正是 AP 衔接；HSS-IC.A 推断是 AP Stats 前置
🇨🇦 ON Grade 12 , MDM4U安大略 12 年级 , MDM4U All seven sections. MDM4U is the dedicated home course , treat this guide as a single-resource pre-read for the full course全部 7 节。MDM4U 是对口课程 , 本指南可作为整门课程的单文档预读 Nothing , MDM4U's three strands (Counting & Probability, Probability Distributions, Statistical Analysis) all appear here无 , MDM4U 三大单元（计数与概率、概率分布、统计分析）此处均覆盖 math_grades_11-12.pdf , MDM4U strands explicit; full expectation codes in the curriculum PDF (cited at strand level here), MDM4U 单元清晰；完整期望代码在课纲 PDF 内（此处按单元级引用）
🇨🇦 ON Grade 11 , MCR3U / MBF3C安大略 11 年级 , MCR3U / MBF3C §1, §3, §7 (light intro to probability and descriptive stats)§1、§3、§7（概率与描述统计的轻量入门） §2 Bayes, §4 distributions, §5 binomial, §6 normal , these are MDM4U content, deferred from Grade 11§2 贝叶斯、§4 分布、§5 二项、§6 正态 , 均为 MDM4U 内容，11 年级推迟 math_grades_11-12.pdf , MCR3U is functions-focused; probability + stats is owned by MDM4U in Grade 12, MCR3U 以函数为主；概率 + 统计归 12 年级 MDM4U
🇨🇦 BC Pre-Calc 11 / 12 (PC 11, PC 12)BC Pre-Calc 11 / 12 None of these sections are PC 11 / PC 12 content. If a PC student wants this material, treat it as enrichment本节内容均不在 PC 11 / PC 12 范围内。PC 学生如需学习，可作为拓展 All seven (within PC) , verified by zero matches on "probability" or "statistics" in pc11_elab.txt / pc12_elab.txt全部 7 节（PC 范畴内）, pc11_elab.txt / pc12_elab.txt 关键词检索为零 pc11_elab.txt, pc12_elab.txt , grep on "probability", "statistics", "distribution" returns zero matches in PC content, "probability"、"statistics"、"distribution" 在 PC 内容中 grep 返回零
🇨🇦 BC Foundations , FPM 11 / 12BC 基础数学 , FPM 11 / 12 All seven sections, with FPM 12 emphasis on §6 normal distribution. FPM is the BC home for probability and statistics全部 7 节，FPM 12 重点是 §6 正态分布。FPM 是 BC 概率与统计对口课 Nothing in scope , the FPM stream is where this content lives in BC无 , FPM 主线是 BC 承载本内容的课程 BC FPM curriculum (not in our local extract) , cited at course-name level onlyBC FPM 课纲（不在本地节选中） , 仅按课程名引用
🇨🇦 AB Grade 12 , Math 30-1阿尔伯塔 12 年级 , Math 30-1 §5 only , Math 30-1 carries the binomial-coefficient infrastructure under Permutations, Combinations and Binomial Theorem; the binomial distribution itself is a natural application仅 §5 , Math 30-1 在"排列、组合与二项式定理"单元承担二项系数基础；二项分布是其自然应用 §1-§4, §6, §7 (Math 30-1 doesn't carry general probability or statistics , those live in Math 30-2)§1-§4、§6、§7（Math 30-1 不承担一般概率或统计 , 归 Math 30-2） pos_10-12_indicators.pdf , Math 30-1 Permutations, Combinations and Binomial Theorem General Outcome (indicators 1.1-6.4 in the curriculum), Math 30-1 "排列、组合与二项式定理"总目标（课纲指标 1.1-6.4）
🇨🇦 AB Grade 12 , Math 30-2阿尔伯塔 12 年级 , Math 30-2 §1, §2, §3, §6, §7. Math 30-2 is the AB home for probability and the normal distribution§1、§2、§3、§6、§7。Math 30-2 是 AB 概率与正态分布对口课 §4, §5 (general discrete distributions and the binomial are Math 30-1 content, not Math 30-2)§4、§5（一般离散分布与二项分布属 Math 30-1，不属 Math 30-2） pos_10-12_indicators.pdf , Math 30-2 Probability General Outcomes 1-3 plus Statistics General Outcomes 1-2 (normal distribution, $z$-scores, confidence intervals), Math 30-2 概率总目标 1-3 + 统计总目标 1-2（正态分布、$z$-分数、置信区间）

Once you have located your row, use the two cards below for the speed at which you should work through the recommended sections.找到所在行后，用下面两张卡片决定推进速度。

!
If you are cramming the night before如果你在临阵磨枪

Memorise five things: the classical-probability formula $P(A) = |A|/|S|$ and complement rule $P(A^c) = 1 - P(A)$; the conditional definition $P(A | B) = P(A \cap B)/P(B)$; the independence test $P(A \cap B) = P(A) P(B)$; the binomial PMF $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ with $E(X) = np$, $\mathrm{Var}(X) = np(1-p)$; and the $68$-$95$-$99.7$ rule with the $z$-score formula $z = (x - \mu)/\sigma$. Read every cram-cheat box. Skip the proofs and the going-deeper integrals.背熟五件事：经典概率公式 $P(A) = |A|/|S|$ 与补集规则 $P(A^c) = 1 - P(A)$；条件概率定义 $P(A | B) = P(A \cap B)/P(B)$；独立性检验 $P(A \cap B) = P(A) P(B)$；二项分布概率质量函数 $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$ 及 $E(X) = np$、$\mathrm{Var}(X) = np(1-p)$；以及 $68$-$95$-$99.7$ 法则与 $z$-分数公式 $z = (x - \mu)/\sigma$。读每个速记框，跳过证明与积分细节。

*
If you are going for the top mark如果你目标顶分

Always draw the picture: tree diagram for sequential events, two-way table for conditional probability, histogram or boxplot for descriptive stats. Always state independence as an assumption before applying $P(A \cap B) = P(A)P(B)$ , it's the single most over-applied rule. Practise the binomial cumulative ("at least", "at most", "exactly") and the normal $z$-table lookup until reflexive. Read both derivations: $E(X) = np$ for the binomial via expectation of indicator variables, and the empirical-rule percentages from integrating the standard normal density. These re-appear verbatim in AP Stats and IB Math AI / AA SL.永远先画图：序列事件画树图（tree diagram），条件概率画双向表（two-way table），描述统计画直方图或箱线图。应用 $P(A \cap B) = P(A)P(B)$ 之前一定先把独立性作为假设说出来 , 这是被滥用最严重的规则。反复练二项的累积形式（"至少"、"至多"、"恰好"）以及正态的 $z$-表查询，直到形成条件反射。两个推导都要会：用指示变量的期望和推出二项的 $E(X) = np$；用标准正态密度的积分推出经验法则百分比。这两件事在 AP Stats、IB Math AI / AA SL 中原样再现。

Honors flag.荣誉级标记。 Sections 13.2 (Bayes' rule) and 13.4 (variance and full discrete distributions) carry the Honors chip for US Algebra II / Stats and AB Math 30-2 , in those courses Bayes is named but not derived, and variance is named but rarely computed by hand. They are core, not honors, in ON MDM4U and on the AP Stats track. If your row above sends you to §2 and §4, treat them as required content, not enrichment.§13.2（贝叶斯法则）与 §13.4（方差与完整离散分布）在美国 代数 II / 统计、AB Math 30-2 中标为 Honors , 这些课程提到贝叶斯但不推导，提到方差但少手算。但在 ON MDM4U 与 AP Stats 主线中，它们是核心而非荣誉内容。如果你的行指向 §2 与 §4，就把它们视为必学，不是拓展。
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High School Math · Probability and Statistics Foundations · v1高中数学 · 概率与统计基础 · v1