Statistiques 2
Pearson Edexcel · International A-Level · Mathematics Sujet 8 4:56 Narration en anglais · Sous-titres anglais + 中文 incrustés
Chapitres
Transcript
Paper WST02: 1 hour 30 minutes, 75 marks, with S1 assumed.
WST02 试卷：1 小时 30 分钟，75 分，默认已学 S1。
Memorise the binomial mean n p and variance n p times 1 minus p; the Poisson's mean and variance, both lambda; the density as the derivative of F; and the uniform's mean and variance.
记住二项分布的均值 np 和方差 np(1 − p)；泊松分布的均值和方差都是 λ；密度是 F 的导数；以及均匀分布的均值和方差。
The binomial needs n independent trials with a constant probability p.
二项分布需要 n 次独立试验，概率 p 不变。
The Poisson needs events occurring singly, independently, at a constant average rate, and the rate scales with time.
泊松分布需要事件单个、独立、以恒定平均速率发生，且速率随时间按比例变化。
Mean 3 and variance 2.55 give 1 minus p equals 0.85: p is 0.15 and n is 20.
均值 3、方差 2.55 得 1 − p = 0.85：p = 0.15，n = 20。
A bag of 40 at 20 percent is B of 40, 0.2, and P of X above 8 is 0.407.
40 颗、20% 的袋子服从 B(40, 0.2)，P(X > 8) = 0.407。
A bag of 80 at 10 percent: n is large and p small, so Po of 8 gives 0.112.
80 颗、10% 的袋子：n 大、p 小，用 Po(8) 得 0.112。
F of x is 6x squared minus x to the 4, over 5, on 0 to 1.
F(x) = (6x² − x⁴)/5，定义在 0 到 1 上。
It climbs from 0 to 1.
它从 0 升到 1。
Where F equals a half, read across and down: the median is 0.671.
在 F 等于一半处横向读再向下读：中位数是 0.671。
Differentiate F to get the density, 12x minus 4x cubed over 5.
对 F 求导得密度 (12x − 4x³)/5。
Its derivative is zero at x equals 1; minus 1 is outside the support.
它的导数在 x = 1 处为零；−1 不在定义域内。
The mean, 0.64, is below the median, 0.671, below the mode, 1: negative skew.
平均数 0.64 小于中位数 0.671，小于众数 1：负偏态。
The density rises as 2x over 21 up to k, then falls as 2 times 6 minus x over 15: two straight pieces.
密度在 k 之前按 2x/21 上升，之后按 2(6 − x)/15 下降：两段直线。
Setting the total area 1 gives a perfect square, so k is 3.5, with both pieces equalling a third there.
令总面积为 1 得到完全平方，所以 k = 3.5，两段在该点都等于 1/3。
Integrate x f over each piece and add: the mean is 19 over 6.
对每一段求 xf 的积分再相加：均值是 19/6。
With E of X squared given as 277 over 24, the variance is 277 over 24 minus 361 over 36: 109 over 72.
已知 E(X²) = 277/24，方差是 277/24 − 361/36：109/72。
P of X between 2 and 5 is 3 over 16, so the interval has length 16.
P(2 < X < 5) = 3/16，所以区间长度为 16。
The mean 6 gives a plus b equals 12, so X is uniform from minus 2 to 14.
均值 6 给出 a + b = 12，所以 X 在 −2 到 14 上均匀分布。
Then P of X above 8 is the length 6 over 16: three eighths.
于是 P(X > 8) 是长度 6 除以 16：3/8。
E of c X plus 1 equals 3 gives c is a third.
E(cX + 1) = 3 得 c = 1/3。
The variance is 16 squared over 12, 64 over 3, and adding the mean squared gives E of X squared, 172 over 3.
方差是 16²/12 = 64/3，加上均值的平方得 E(X²) = 172/3。
Rearranged, the last inequality is X above 8.
最后一个不等式整理后就是 X > 8。
A census measures everybody: complete, but costly, slow, and possibly destructive.
普查测量所有人：完整，但成本高、速度慢，还可能具有破坏性。
A sample measures part: cheaper and faster, at the price of sampling error.
抽样只测量一部分：更便宜、更快，代价是抽样误差。
A statistic uses the sample values only.
统计量只使用样本值。
Three beads give scores 3, 6, 9, 12, with binomial weights out of 343.
三颗珠子的得分是 3、6、9、12，按二项权重，分母为 343。
A 9-point set has probability 60 over 343, so the number of them in n plays is binomial, and at least one with 95 percent certainty needs n of 16.
得 9 分的概率是 60/343，所以 n 次中出现的次数服从二项分布，要以 95% 的把握至少出现一次需要 n = 16。
State the null hypothesis, the quoted parameter, and the alternative.
写出原假设（题中给出的参数）和备择假设。
Accumulate tail probabilities to find the critical region, report the actual significance level, and conclude in context.
累加尾部概率求临界域，报告实际显著性水平，并结合情境下结论。
Customers average 20 an hour, so a 30-minute interval is Po of 10.
每小时平均 20 位顾客，所以 30 分钟服从 Po(10)。
At 3 percent two-tailed, each tail holds at most 1.5 percent: the critical region is X at most 3 or at least 18, actual level 2.46 percent.
3% 双尾检验，每个尾部至多 1.5%：临界域是 X ≤ 3 或 X ≥ 18，实际水平 2.46%。
The observed 16 is outside it: do not reject.
观察值 16 在临界域外：不拒绝原假设。
For 40 customers, the largest r with P of fewer than r below 5 percent is 7.
40 位顾客时，使 P(X < r) 小于 5% 的最大 r 是 7。
For 200, the Normal approximation with a continuity correction gives t of 50.
200 位时，用带连续性修正的正态近似得 t = 50。
After the offer, 11 of 25 buy: P of X at least 11 is 0.0978, below 10 percent, so reject: the proportion has increased.
优惠之后 25 人中 11 人购买：P(X ≥ 11) = 0.0978，小于 10%，所以拒绝：比例上升了。
Check yourself.
自我检测。
A Poisson's mean and variance are both lambda.
泊松分布的均值和方差都是 λ。
The median solves F of m equals a half.
中位数满足 F(m) = 1/2。
The uniform variance is b minus a squared over 12.
均匀分布的方差是 (b − a)²/12。
X bar minus mu is not a statistic.
X̄ − μ 不是统计量。
And at least t becomes t minus a half.
“至少 t”变为 t − 0.5。