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。
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。
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。
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。
Sample Spaces, Events, and the Classical Probability Formula样本空间、事件与经典概率公式
- 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.1names this verbatim.$S$ 的子集。"掷出偶数"即事件 $A = \{2, 4, 6\} \subseteq S$。CCSSMHSS-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):
- 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。CCSSMHSS-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)。
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。
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
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。
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.6writes 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 $A$。CCSSMHSS-CP.B.6原文表述此法。
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"。
Independence, Dependence, and Tree Diagrams独立性、相依性与树图
- 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"。CCSSMHSS-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。
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。
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
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。
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。
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。
The Binomial Distribution: $X \sim B(n, p)$二项分布:$X \sim B(n, p)$
$X \sim B(n, p)$ 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:
- $n$ trials,$n$ 次试验, fixed in advance.事先固定。
- Each trial is "success" or "failure"每次试验只有"成功"或"失败" (Bernoulli).(伯努利试验)。
- $P(\text{success}) = p$ is the same on every trial.每次 $P($ 成功 $) = p$ 不变。
- 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。
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。
The Normal Distribution: $\mu$, $\sigma$, and $z$-Scores正态分布:$\mu$、$\sigma$ 与 $z$-分数
- 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。
$z$ 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。
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。
Describing a Data Set: Centre, Spread, and Shape描述数据集:中心、离散与形状
- 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。CCSSMHSS-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.CCSSMHSS-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。
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。
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"。Exam Strategy and Common Pitfalls考试策略与常见陷阱
- 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。
- 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 $0$。
- 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。
- 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.2grades 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。CCSSMHSS-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 $n - 1$。
Flashcards闪卡
HSS-CP.B.7CCSSM HSS-CP.B.7HSS-CP.A.2CCSSM HSS-CP.A.2HSS-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。Practice Quiz综合测验
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。
- List the sample space of a small experiment and compute a probability via $P(A) = |A|/|S|$ without notes.无笔记下列出小型实验的样本空间,并用 $P(A) = |A|/|S|$ 求概率。
- Apply the complement rule $P(A^c) = 1 - P(A)$ to "at least one" problems instead of summing. 🇺🇸 HSS-CP.A.1将"至少一个"问题用 $P(A^c) = 1 - P(A)$ 算,而非直接相加。🇺🇸 HSS-CP.A.1
- Apply the addition rule $P(A \cup B) = P(A) + P(B) - P(A \cap B)$, including the case where the events are mutually exclusive.运用加法规则 $P(A \cup B) = P(A) + P(B) - P(A \cap B)$(含互斥情形)。
- Compute $P(A | B) = P(A \cap B) / P(B)$ from a two-way frequency table. 🇺🇸 HSS-CP.B.6从双向频数表算 $P(A | B) = P(A \cap B) / P(B)$。🇺🇸 HSS-CP.B.6
- Honors Apply Bayes' rule and the law of total probability to invert a conditional probability (e.g. the medical-test base-rate problem).Honors 用贝叶斯法则与全概率公式反演条件概率(如医学检测基础概率题)。
- Test independence via $P(A \cap B) = P(A) P(B)$; distinguish independence from mutual exclusivity. 🇺🇸 HSS-CP.A.2 / 🇨🇦 AB 30-2 indicator 3.1用 $P(A \cap B) = P(A) P(B)$ 检验独立性;区分独立与互斥。🇺🇸 HSS-CP.A.2 / 🇨🇦 AB 30-2 指标 3.1
- Draw a tree diagram for a multi-stage experiment and compute path probabilities (multiply along, add across).为多阶段实验画树图,并求路径概率(沿枝相乘,跨枝相加)。
- Honors Compute $E(X)$ and $\mathrm{Var}(X)$ for a small discrete random variable using a PMF table. 🇺🇸 HSS-MD.A (+) / 🇨🇦 ON MDM4U Probability DistributionsHonors 用 PMF 表算小型离散随机变量的 $E(X)$ 与 $\mathrm{Var}(X)$。🇺🇸 HSS-MD.A (+) / 🇨🇦 ON MDM4U 概率分布
- Recognise when a problem is binomial ($n$ fixed, success / failure, fixed $p$, independent), write $X \sim B(n, p)$, and compute $P(X = k)$, $E(X) = np$, $\mathrm{Var}(X) = np(1-p)$.识别二项问题($n$ 固定、成功 / 失败、$p$ 固定、独立),写 $X \sim B(n, p)$,并算 $P(X = k)$、$E(X) = np$、$\mathrm{Var}(X) = np(1-p)$。
- Apply the $68$-$95$-$99.7$ rule by inspection at $\pm 1\sigma, \pm 2\sigma, \pm 3\sigma$, and convert a value to a $z$-score $z = (x - \mu)/\sigma$. 🇨🇦 AB 30-2 Statistics GO 1, indicators 1.3, 1.8$\pm 1\sigma, \pm 2\sigma, \pm 3\sigma$ 处一眼应用 $68$-$95$-$99.7$ 法则,并由值转 $z$-分数 $z = (x - \mu)/\sigma$。🇨🇦 AB 30-2 统计 GO 1,指标 1.3、1.8
- Compute mean, median, mode, range, IQR, and sample standard deviation by hand on a data set of $n \le 12$.在 $n \le 12$ 数据集上手算均值、中位数、众数、极差、IQR、样本标准差。
- Choose between (mean, standard deviation) and (median, IQR) based on whether the distribution is symmetric or skewed; read shape from a histogram or boxplot. 🇺🇸 HSS-ID.A.2, .A.3根据分布是对称还是偏态,在(均值、标准差)与(中位数、IQR)之间选择;从直方图或箱线图读形状。🇺🇸 HSS-ID.A.2、.A.3
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。
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。