Edexcel数学 统计 概率 正态 假设检验 抽样
Edexcel A Level 数学的第三张卷子（9MA0/03）把统计与力学放在一起，2 小时 100 分，Section A 统计 50 分，Section B 力学 50 分，这张卷本身占最终成绩的 33.33%。Pearson 官方 spec（Issue 4，2017 年 9 月起施教，封面写明 first certification from 2018，同一册包含 AS 8MA0 与 A Level 9MA0）把统计部分切成五个 Topic：抽样、数据展示与解释、概率、统计分布、假设检验。计算器全程允许使用，spec 明确假设学生手上是带统计分布功能的机器，二项与正态的概率值都应该直接读出来。很多学生以为统计就是按计算器出数，题目却把分数压在假设怎么写、临界域怎么找、结论怎么落回情境上。阅卷时统计部分的失分很少出在算错，更多出在没有写出零假设与备择假设，或者结论只写拒绝而没有回到题目里的具体情境。
Statistics shares Paper 3 (9MA0/03) with mechanics: two hours, 100 marks, with Section A on statistics and Section B on mechanics at 50 marks each, and the paper carries 33.33% of the A Level. The Pearson specification (Issue 4, first teaching September 2017) splits the statistics content into sampling, data presentation and interpretation, probability, statistical distributions and hypothesis testing, and a calculator with distribution functions is assumed throughout. Marks are lost far more often on the written structure of a test, where the hypotheses, the critical region and a conclusion in context all earn credit, than on the arithmetic itself.
Paper 3 Section A 的 50 分怎么分布
Section A 通常是四到五道大题，每题内部再分成 (a) 到 (e) 的小问，括号里的分值直接告诉你该写多少行。(a) 往往是一两分的读数或者中间量，(d) 与 (e) 才是四五分的检验与评论，最后一个小问常写 using your answers to parts (b) and (c)，要求真的把前面两个结果接上来。Pearson 官方的 mark scheme 把方法分与答案分分开给，方法分只要步骤对就给，答案分必须建立在前面的方法分上，前一问算错时后面的答案分按 follow through 继续给，空着不写比写错更吃亏。官方样卷里有一道关于咖啡包装重量的题，14 分，先算样本均值与标准差，再用 5% 显著性水平检验卖家的说法，最后一分要求评价正态假设是否站得住，这种把计算、检验、评论串成一条线的题就是 Section A 的典型形状。50 分对应 60 分钟，平均一分钟一分左右，一道 14 分的题如果耗到 20 分钟，后面的检验题就来不及写。
Paper 3 (9MA0/03) is the shared statistics and mechanics paper: two hours, 100 marks, Section A on statistics and Section B on mechanics at 50 marks each, and the whole paper is worth 33.33% of the A Level. Section A usually holds four or five questions split into parts, with one-mark readings early and four or five-mark tests and comments at the end; the phrase using your answers to parts (b) and (c) means those earlier results have to be carried forward explicitly. Pearson's mark scheme separates method marks, which stand on their own, from accuracy marks, which follow on from the method marks, and wrong earlier answers are followed through rather than punished, so a blank line costs more than a wrong number.
抽样方法：随机 系统 分层 与配额
抽样这一章要求说清方法、操作过程和局限三个层面，spec 的 Topic 15 把它分成概率抽样与非概率抽样两类。简单随机抽样给每个个体同等的被选机会，用随机数表或者计算器随机数实现，800 人的名单抽 40 人，就是给 1 到 800 逐个编号之后取 40 个不重复的随机数。系统抽样先算间隔，总体除以样本量，1200 人抽 60 人时间隔是 20，从第一个间隔里随机取一个起点，之后每隔 20 个取一个，题目会要求你说明随机起点这一步，漏掉就变成有偏抽样。分层抽样按层在总体中的比例分配样本量，某个年级有 300 人、占总体的四分之一，60 人的样本里这一层分到 15 人，写的时候要把按比例分配的计算过程写出来。响应偏差来自问卷回收率低，抽样框偏差来自名单本身漏掉一部分人群，这两种偏差都不是靠加大样本量能修掉的。很多学生以为样本越大越准，遇到 500 人的有偏样本与 80 人的随机样本作比较时还按人数多少来选，这一分就丢了。
Sampling questions expect three things together: the method, the mechanism and the limitation, and Topic 15 of the specification separates probability sampling from the non-probability kind. Simple random sampling gives every individual an equal chance, systematic sampling uses an interval of population divided by sample size and starts from a random point inside the first interval, and stratified sampling allocates the sample in proportion to each stratum, so a stratum holding a quarter of the population takes a quarter of a 60-student sample. The quiet danger with a systematic sample is a periodic list, where every twentieth name from a list ordered by building may never reach the other building types, and the standard weaknesses of opportunity and quota sampling are coverage bias and the interviewer's own judgement; a larger sample cannot repair either.
直方图 频率密度 与累积频率读数
直方图的纵轴是频率密度，频率密度等于频数除以组宽，所有矩形面积加起来等于 1，组宽不相等时这条规则才显出分量。体重落在 10 到 20 千克这一组的频数是 18，组宽 10，频率密度就是 1.8；同样 18 个数据落在 20 到 40 千克、组宽 20 的一组里，频率密度只有 0.9，两根柱子的高度完全不同。累积频率图的读法是纵向找 25%、50%、75%，横着画到曲线上再往下读横轴，总频数 200 的样本里中位数就落在这条线的交点位置，题面写 estimate 时允许一点读数误差，给出一个小区间也接受。箱线图要画须、箱体和中间那条中位线，两个箱线图并排比较时，中位线的位置差说明典型水平，箱体长度差说明中间一半数据的集中程度。离群值的判据是固定的两条，超过上四分位加 1.5 倍四分位间距，或者低于下四分位减 1.5 倍四分位间距，算完必须明确点名哪一个是离群值。学生最常把直方图纵轴标成频数，纵轴一错后面的分位估计全跟着错。
A histogram plots frequency density, which is frequency divided by class width, so each bar's area gives its frequency and the areas total one; 18 observations in a class of width 10 give a density of 1.8 while the same 18 in a class of width 20 give 0.9, and an interval is accepted rather than an exact value when a question says estimate. A cumulative frequency curve is read by tracing 25%, 50% and 75% across to the curve and down to the axis, and a box plot then compares a typical value through the median line with a spread through the box. Outliers are claimed only against the fixed rules of Q3 plus 1.5 times the interquartile range and Q1 minus 1.5 times the same range, and drawing the whiskers to the wrong value is the usual way a plotting mark disappears.
平均数 标准差 四分位与偏度
统计量分三个层次，位置看平均数与中位数，离散看极差、四分位间距与标准差，形状看偏度，题目问哪一个更合适时答案要跟着层次走。官方公式册给出 Sxx 等于所有 x 平方之和减去所有 x 之和的平方除以 n，标准差等于 Sxx 除以 n 减 1 再开根号，用 n 减 1 还是 n 取决于数据是样本还是总体，题面写 sample 时用 n 减 1。保留位数本身是分的一部分，题面已经用到 2 位有效数字时，最后的均值与标准差写 3 位也会被扣，前后保持一致才安全。偏度只看四分位之间的关系，上四分位到中位数的距离大于中位数到下四分位的距离就是正偏，此时平均数被拉向右侧，落在中位数右边。很多学生以为平均数总能代表一群数据，遇到房价与收入这类明显偏斜的分布还坚持用平均数比较，题目问哪个更合适时要答中位数。常见的错误是把方差与标准差混着写，答案要求标准差却给了方差，单位也对不上，阅卷时这一类写法拿不到最后那一分。
Summary statistics divide into location, spread and shape, and the formulae booklet gives Sxx as the sum of x squared minus the square of the sum of x divided by n, with the standard deviation as the square root of Sxx over n minus 1 when the data are a sample. Significant figures count as marks, so a final mean and standard deviation have to match the accuracy already used in the question rather than drifting to an extra figure. Quartiles follow the convention of halving the data and taking the median of each half, skewness is read off as a longer distance from the median to the upper quartile than to the lower one, and that longer tail pulls the mean above the median; quoting a variance where a standard deviation was asked for is the error examiners see most often.
概率 条件概率 与独立事件
概率部分公式册只给记号，加法定理要自己写，两个事件并集的概率等于各自概率之和减去交集的概率，条件概率把分母换成已知事件的概率，用交集概率除以那个已知事件的概率。独立事件的判据是两个事件同时发生的概率等于两个概率的乘积，互斥事件同时发生的机会是零，这两个概念最容易被写反，交集概率等于 0 说明互斥，等于两个概率的乘积才说明独立。不放回的抽取要用树状图把每一步的分数写清，5 个红球 3 个蓝球里连取两个，两次都是红球的概率是 5/8 乘 4/7，约为 0.357，第二次的分母从 8 变成 7，这一步漏掉整题就作废。韦恩图适合三个事件的情形，题面给的数常是并集或者补集的频数，先填中间的交集再倒推其余区域，填完把各区域加起来核对总频数。条件概率的题目读题时要认出给定条件，问的是在某个事件已经发生的前提下另一事件发生的概率，分母是那个已知事件的概率而不是全部结果的总数。学生最常把两个方向的条件概率写成同一个数，问法与语序就是线索，前后颠倒一般并不相等。
The formulae booklet offers only notation for probability, so the addition rule, that the probability of A or B is the sum of the individual probabilities minus the probability of both, has to be recalled along with the conditional form that divides the intersection by the probability of the given event. Independence means the probability of both events is the product of the two probabilities, while mutually exclusive events cannot happen together at all, and the two conditions are routinely swapped by candidates. Drawing without replacement needs a tree diagram with each fraction visible, so two red balls from five red and three blue come to 5/8 times 4/7, about 0.357, and a conditional probability question has to be read for its given event because the two directions are not equal.
离散随机变量与二项分布
离散随机变量给出一张概率分布表，表里所有概率加起来必须等于 1，缺的那个格子往往就是第一小问。二项分布要求试验次数固定、每次只有成功与失败两种结果、每次成功的概率相同、各次试验相互独立，题面出现 independent 或者 constant probability 就是在提示你写这些假设。分布写成 B(n, p) 之后，单个取值的概率是组合数乘 p 的 r 次方再乘 1 减 p 的 n 减 r 次方，20 次试验成功率 0.3 的模型里恰好 5 次成功的概率约为 0.1789，期望 6，方差 4.2，这些数用计算器的分布功能几秒就能出。写出模型那一步不能省，只给一个数不给 B(20, 0.3) 的记号会被扣分，后面几问的分也跟着没地方落。学生最常把不放回抽样也当成二项分布，不放回时每次概率都在变，这时候的模型不是二项，题面写 without replacement 时尤其要注意。常见的错误是算累积概率时用错尾端，at least 与 at most 对应两个相反的求和方向，答案差得很多。
A discrete random variable comes with a probability table whose entries must total one, which is usually the first mark on the question, and the expectation is the sum of each value multiplied by its probability while the variance is quickest as the mean of the squares minus the square of the mean. A binomial model needs a fixed number of trials, two outcomes, a constant probability and independent trials, and the notation B(n, p) itself carries a mark; for B(20, 0.3) exactly five successes come to about 0.1789 with an expectation of 6 and a variance of 4.2. Sampling without replacement breaks the constant-probability requirement, so the binomial is the wrong model there, and confusing at least with at most reverses the tail being summed and costs the following parts as well.
正态分布 标准化 与逆向查表
正态分布的记号是 N(μ, σ²)，括号里第二个数是方差不是标准差，题面给标准差时写成方差要先平方，这一步写错后面全偏。标准化把任意正态变量变成标准正态，z 等于取值减去均值再除以标准差，X 服从 N(30, 25) 时求 X 大于 38 的概率，就是求 z 大于 1.6 的概率，查表或者计算器得到 0.0548。反过来的问题是给出概率求取值，用计算器的逆正态功能，要找 X 小于某个数的概率等于 0.9 的分位点，z 是 1.2816，代回去这个数约为 36.41。正态近似的条件是 n 大并且 p 接近 0.5，同时 np 与 n 乘 1 减 p 都大于 5，写成连续性更正时边界要移动半个单位，求至少 20 的概率近似成大于 19.5 才对。很多学生以为查表得到的概率就是最终答案，题目要求四位小数或者要求写出不等式时还要看指令词，写反了照样丢分。常见的错误是标准化时把分母写成方差，或者忘记把答案从 z 值换回原来的单位，答案 36.41 与 1.2816 之间差着题目的量纲。
The normal model is written N(mu, sigma squared), so the second parameter is a variance and a question quoting a standard deviation has to be squared before use. Standardising divides the difference between the value and the mean by the standard deviation, which turns X greater than 38 in an N(30, 25) model into z greater than 1.6 and a probability of 0.0548, while the inverse problem uses the calculator's inverse normal function, where a probability of 0.9 below a value gives z = 1.2816 and x about 36.41. Approximating a binomial by a normal requires a large n with p near 0.5 and both np and n(1 minus p) above 5, and the continuity correction shifts the boundary by half a unit, so at least 20 becomes greater than 19.5.
假设检验：临界域 p 值与结论写法
假设检验的答案有固定骨架，先写清零假设与备择假设，把参数与数值都写进去，再标明显著性水平与单尾还是双尾，然后算检验统计量或者找出临界域，最后把结论落回题面情境。举一个二项检验的例子，20 次试验里观察到 14 次成功，零假设是成功概率 0.5，单尾 5% 水平下至少 14 次成功的概率约为 0.0577，比 0.05 大，结论是不能拒绝零假设；临界域是至少 15 次，因为至少 15 次的概率约 0.0207 才小于 0.05。结论的写法要用题目里的情境词，写 reject H0 而不写这对咖啡包装重量意味着什么，就拿不到最后那一两分。学生最常把 p 值与显著性水平的关系说反，p 值小于显著性水平才拒绝零假设，大于是不拒绝，判据只有这一条。常见的错误是把备择假设写成小于等于，或者漏掉参数定义，state your hypotheses clearly 这句要求写清总体参数，不是写样本结果。
A hypothesis test follows a fixed script: state the null and alternative hypotheses with the parameter named, fix the significance level and the number of tails, calculate a test statistic or a critical region, then interpret the outcome in the context of the question. A two-tailed 5% test splits into 2.5% in each tail with a critical value of 1.96 in the normal case, while a one-tailed 5% test uses 1.645, and both numbers recur throughout Section A. For a binomial test with 20 trials, a null probability of 0.5 and 14 observed successes, the upper-tail probability is about 0.0577, which is above 0.05 and so does not reject the null; the critical region is 15 or more, where the tail probability falls to 0.0207, and a conclusion written without the context words earns no final mark.
相关系数 回归线 与编码变换
相关系数 r 的算式是协方差和除以两个平方和的平方根的乘积，取值永远在负 1 与正 1 之间，接近 0 说明线性关系很弱。回归线的斜率等于协方差和除以自变量的平方和，截距等于因变量均值减去斜率乘自变量均值，这两步都用 Sxx 与 Sxy 这对量，公式册里能找到记号但不会给出代换结果。编码是统计部分的高频考点，把 x 换成 x 减 100 再除以 10 这样的新变量，均值与标准差按同样的比例缩小，相关系数一分不变，回归线的斜率放大 10 倍，截距也跟着变。回归线只适合在数据范围内插值，超出数据范围外推要谨慎，涉及 2020 年以后数据的题会要求说明为什么预测不可靠。学生最常把编码后的标准差当成没变，或者把相关系数也按比例改写，这两处要按变换规则分别处理。常见的错误是求回归线时把自变量与因变量弄反，题目写 y 关于 x 的回归线时斜率要拿 Sxy 除以 Sxx。
The product moment correlation coefficient divides the covariance term by the square root of the two sums of squares, always lands between minus 1 and 1, and a value near zero means almost no linear relationship. The regression line takes its gradient from the covariance term over Sxx and its intercept from the mean of y minus that gradient times the mean of x, and the formula booklet gives only the notation rather than the substitution. Coding changes the mean and standard deviation by the same scale, leaves r untouched and multiplies the gradient by the reciprocal of that scale, regression is safe only as interpolation inside the data range, and correlation without causation is the sentence a comment question is looking for.
大样本表与四周复习节奏
统计部分的题常常引用官方 large data set，这份表里是天气数据，英国有 Camborne、Heathrow、Hurn、Leeming、Leuchars 五个站点，海外有 Beijing、Jacksonville、Perth 三个，样本期是 1987 年与 2015 年这两年的 5 月到 10 月，变量包括每日最高温与最低温、日照小时数、降雨量、风速与风向、相对湿度、云量与气压类别。考场上数据表会提供，需要记住的不是数值而是结构：哪个站点靠北、缺失值怎么标、风向记录的是风来的方向而不是去的方向。四周的复习节奏可以这样排，第一周把抽样与数据展示做透，每周写两次抽样方法的优缺点；第二周专攻概率与离散分布，把二项模型的四个假设默写出来；第三周做正态与假设检验，专门练结论的情境写法；第四周整套限时训练，每周一次 60 分钟做完整的 50 分统计部分，再用 40 分钟对着官方 mark scheme 归因。失分最多的四类位置是假设没写清、临界域找反、结论没有回到情境、保留位数不合题面要求，这四处的分数加起来通常在 6 到 10 分，落在 A 与 B 之间。
Statistics questions frequently lean on the official large data set of weather records, which holds five UK stations (Camborne, Heathrow, Hurn, Leeming and Leuchars) and three overseas ones (Beijing, Jacksonville and Perth) for May to October in 1987 and 2015, covering daily maximum and minimum temperature, sunshine hours, rainfall, wind speed and direction, humidity, cloud cover and pressure. The data are supplied in the exam, so what has to be remembered is the structure: which stations sit further north, how missing values appear, and that wind direction records where the wind comes from rather than where it goes. A four-week plan works well, moving from sampling and data presentation through probability and discrete distributions to the normal model and hypothesis testing, and finishing with one timed 60-minute run through a full 50-mark statistics section each week followed by 40 minutes of mark-scheme analysis.
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