WJEC Pre-U Statistics Mock Paper Walkthrough | WJEC Pre-U 统计模拟卷解析

📚 WJEC Pre-U Statistics Mock Paper Walkthrough | WJEC Pre-U 统计模拟卷解析

This article provides a complete walkthrough of a WJEC Pre-U Statistics mock paper, covering representative exam-style questions. Each solution is structured with clear steps, allowing you to deepen your understanding of key statistical concepts and techniques. Read through the English and Chinese explanations side by side to master both theory and exam technique.

本文对一套WJEC Pre-U统计模拟卷进行逐题详解，涵盖典型考题。每道解析都分步说明，帮助你加深对核心统计概念和方法的理解。通过对照阅读中英文解释，掌握理论与应试技巧。

1. Probability and Binomial Distribution | 概率与二项分布

Question 1: A fair die is rolled 8 times. Find the probability of obtaining exactly 3 sixes.

题目 1：掷一枚均匀骰子8次，求恰好得到3次6点的概率。

This is a binomial experiment with n = 8 independent trials, each having success probability p = 1/6. Let X be the number of sixes observed. Then X ~ B(8, 1/6).

这是一个二项分布问题，n = 8 次独立试验，每次成功（得到6点）概率 p = 1/6。设 X 为出现6点的次数，则 X ~ B(8, 1/6)。

Use the binomial probability formula P(X = k) = ⁿCₖ pᵏ (1 − p)ⁿ⁻ᵏ. For k = 3 we have:

使用二项概率公式 P(X = k) = ⁿCₖ pᵏ (1 − p)ⁿ⁻ᵏ。当 k = 3 时：

P(X = 3) = ⁸C₃ (1/6)³ (5/6)⁵

Compute the combination: ⁸C₃ = 56. The powers: (1/6)³ = 1/216 and (5/6)⁵ = 3125/7776. Multiplying gives:

计算组合数：⁸C₃ = 56。幂次：(1/6)³ = 1/216，(5/6)⁵ = 3125/7776。相乘得：

P(X = 3) = 56 × (1/216) × (3125/7776) = 56 × 3125 / (216 × 7776) ≈ 175000 / 1679616 ≈ 0.1042

Thus, the probability of exactly three sixes in eight rolls is about 0.104.

因此，掷8次骰子恰好得到3次6点的概率约为0.104。

2. Poisson Approximation to the Binomial | 二项分布的泊松近似

Question 2: A large batch of components contains 2% defectives. A random sample of 200 is taken. Use a Poisson approximation to find the probability that there are at most 5 defective components.

题目 2：一批大量元件中有2%是次品。随机抽取200件，用泊松近似求次品数不超过5件的概率。

When n is large and p is small, the binomial distribution B(n, p) can be approximated by a Poisson distribution with mean λ = np. Here n = 200 and p = 0.02, so λ = 200 × 0.02 = 4.

当 n 很大而 p 很小时，二项分布 B(n, p) 可用均值为 λ = np 的泊松分布近似。本题 n = 200, p = 0.02，故 λ = 200 × 0.02 = 4。

Let Y ~ Poisson(4) represent the number of defectives. We need P(Y ≤ 5). This is the cumulative probability:

设 Y ~ Poisson(4) 表示次品数，要求 P(Y ≤ 5)。这是累积概率：

P(Y ≤ 5) = e⁻⁴ [1 + 4 + 4²/2! + 4³/3! + 4⁴/4! + 4⁵/5!]

Compute each term:

逐项计算：

1
4
4²/2 = 16/2 = 8
4³/6 = 64/6 ≈ 10.6667
4⁴/24 = 256/24 ≈ 10.6667
4⁵/120 = 1024/120 ≈ 8.5333

Sum of terms = 1 + 4 + 8 + 10.6667 + 10.6667 + 8.5333 = 42.8667. Multiplying by e⁻⁴ ≈ 0.01832 gives:

各项之和 = 1 + 4 + 8 + 10.6667 + 10.6667 + 8.5333 = 42.8667。乘以 e⁻⁴ ≈ 0.01832 得：

P(Y ≤ 5) ≈ 42.8667 × 0.01832 ≈ 0.785

The probability of at most 5 defectives is approximately 0.785.

次品数不超过5的概率约为0.785。

3. Inverse Normal Distribution | 正态分布的逆运算

Question 3: The weight of apples from an orchard is normally distributed with mean 150 g and standard deviation 20 g. Find the weight exceeded by only the heaviest 5% of apples.

题目 3：某果园苹果重量服从均值为150克、标准差为20克的正态分布。求只有最重的5%苹果能超过的重量。

We need the 95th percentile of the N(150, 20²) distribution. Let X ~ N(150, 20²) and find w such that P(X > w) = 0.05, i.e. P(X ≤ w) = 0.95.

需要求 N(150, 20²) 的第95百分位数。设 X ~ N(150, 20²)，求 w 使得 P(X > w) = 0.05，即 P(X ≤ w) = 0.95。

Standardise: Z = (X − 150)/20 ~ N(0,1). From tables, the z-value for a cumulative probability of 0.95 is approximately 1.645. Hence:

标准化：Z = (X − 150)/20 ~ N(0,1)。查表得累积概率0.95对应的z值约为1.645。因此：

(w − 150) / 20 = 1.645 ⟹ w = 150 + 1.645 × 20 = 150 + 32.9 = 182.9 g

The weight exceeded by only the heaviest 5% is about 182.9 grams.

只有最重的5%苹果能超过的重量约为182.9克。

4. Confidence Interval for the Mean (t-distribution) | 均值的置信区间（t分布）

Question 4: A random sample of 12 students has a mean test score of 78 with a standard deviation of 10. Assuming the scores are normally distributed, construct a 95% confidence interval for the population mean.

题目 4：随机抽取12名学生，测验平均分为78，标准差为10。假设成绩服从正态分布，构造总体均值的95%置信区间。

Since the population standard deviation is unknown and the sample size is small, use the t-distribution with n − 1 = 11 degrees of freedom. The 95% confidence interval is given by:

由于总体标准差未知且样本量较小，使用 t 分布，自由度为 n − 1 = 11。95%置信区间公式为：

x̄ ± t(0.025, 11) × (s / √n)

From the t-table, the critical value for 11 df at the 2.5% two-tail value (or 0.975 cumulative) is t = 2.201.

查 t 分布表，自由度为11时双侧2.5%临界值（或累积概率0.975）为 t = 2.201。

Compute the margin of error: s/√n = 10/√12 ≈ 10/3.464 = 2.887. Multiply: 2.201 × 2.887 ≈ 6.354.

计算误差边际：s/√n = 10/√12 ≈ 10/3.464 = 2.887。相乘：2.201 × 2.887 ≈ 6.354。

Thus the confidence interval is (78 − 6.354, 78 + 6.354) = (71.646, 84.354).

因此置信区间为 (78 − 6.354, 78 + 6.354) = (71.65, 84.35) 近似。

We are 95% confident that the true population mean test score lies between 71.65 and 84.35.

我们有95%的把握认为总体平均测验成绩介于71.65至84.35之间。

5. Hypothesis Test for a Population Proportion | 总体比例的假设检验

Question 5: A company claims that 40% of customers prefer its new product. In a survey of 500 customers, 220 say they prefer it. Test at the 5% significance level whether the true proportion is greater than 40%.

题目 5：某公司声称40%的顾客偏爱其新产品。在一项500名顾客的调查中，220人表示偏爱该产品。在5%显著性水平下检验真实比例是否大于40%。

Set up hypotheses: H₀: p = 0.40, H₁: p > 0.40 (one-tailed test). Sample proportion p̂ = 220/500 = 0.44.

设立假设：H₀: p = 0.40，H₁: p > 0.40（单侧检验）。样本比例 p̂ = 220/500 = 0.44。

Under H₀ the standard error is SE = √[p₀(1 − p₀)/n] = √(0.40 × 0.60 / 500) = √(0.24/500) = √(0.00048) ≈ 0.02191.

在H₀下标准误为 SE = √[p₀(1 − p₀)/n] = √(0.40 × 0.60 / 500) = √(0.24/500) = √(0.00048) ≈ 0.02191。

Test statistic: z = (p̂ − p₀) / SE = (0.44 − 0.40) / 0.02191 ≈ 0.04 / 0.02191 ≈ 1.826.

检验统计量：z = (p̂ − p₀) / SE = (0.44 − 0.40) / 0.02191 ≈ 0.04 / 0.02191 ≈ 1.826。

The critical z-value for a one-tailed 5% test is 1.645. Since 1.826 > 1.645, we reject H₀.

单侧5%检验的临界z值为1.645。因为1.826 > 1.645，拒绝H₀。

There is sufficient evidence at the 5% level to conclude that the true proportion of customers preferring the new product is greater than 40%.

在5%显著性水平下有足够证据表明偏爱新产品的真实顾客比例大于40%。

6. Chi-squared Test for Independence | 卡方独立性检验

Question 6: A researcher surveys 200 people, recording their favourite music genre (Pop, Rock, Classical) and age group (Under 30, 30–50, Over 50). The data are summarised in a 3×3 contingency table. Perform a chi-squared test for independence at the 1% significance level: state the hypotheses, the degrees of freedom, the critical value, and outline the steps without computing all expected frequencies.

题目 6：研究人员调查了200人，记录他们最喜爱的音乐类型（流行、摇滚、古典）和年龄组（30岁以下、30-50岁、50岁以上）。数据汇总在一个3×3列联表中。在1%显著性水平下进行卡方独立性检验：陈述假设、自由度、临界值，并概述步骤（无需计算所有期望频数）。

Hypotheses – H₀: Favourite music genre and age group are independent. H₁: They are not independent.

假设 – 原假设 H₀：最喜爱音乐类型与年龄组独立；备择假设 H₁：不独立。

Degrees of freedom: For an r × c table, df = (r − 1)(c − 1). Here r = 3, c = 3, so df = 2 × 2 = 4.

自由度：对于 r × c 表格，df = (r − 1)(c − 1)。此处 r = 3, c = 3，因此 df = 2 × 2 = 4。

At the 1% significance level, the critical value from the χ² distribution with 4 df is 13.277.

在1%显著性水平下，自由度为4的卡方分布临界值为13.277。

Procedure: Compute expected frequencies for each cell using (row total × column total) / grand total. Then calculate the test statistic χ² = Σ (O − E)² / E. Compare the calculated χ² with 13.277; if it exceeds 13.277, reject H₀, concluding that there is a significant association between music preference and age group.

步骤：利用（行合计 × 列合计）/ 总计计算每个单元格的期望频数。然后计算检验统计量 χ² = Σ (O − E)² / E。将所得 χ² 与13.277比较；若大于13.277，则拒绝 H₀，认为音乐偏好与年龄组之间存在显著关联。

7. Hypothesis Test for Correlation Coefficient | 相关系数的假设检验

Question 7: For a set of n = 10 data points, the Pearson product-moment correlation coefficient is found to be r = 0.68. Test at the 5% significance level whether there is evidence of positive correlation.

题目 7：对于一组 n = 10 的数据点，皮尔逊积矩相关系数为 r = 0.68。在5%显著性水平下检验是否存在正相关。

Hypotheses: H₀: ρ = 0 (no correlation), H₁: ρ > 0 (positive correlation). This is a one-tailed test.

假设：H₀: ρ = 0（无相关），H₁: ρ > 0（正相关）。此为单侧检验。

Use the Pearson correlation coefficient table. For n = 10, the 5% one-tailed critical value is 0.5494 (this can be found in standard statistical tables).

使用皮尔逊相关系数临界值表。对于 n = 10，单侧5%临界值为0.5494（可在标准统计表中查得）。

Compare r = 0.68 with the critical value: 0.68 > 0.5494. Therefore, we reject H₀ at the 5% level.

比较 r = 0.68 与临界值：0.68 > 0.5494。因此在5%水平下拒绝 H₀。

There is sufficient evidence to conclude that there is a significant positive linear correlation between the two variables.

有足够证据表明两变量之间存在显著的正线性相关。

8. Discrete Random Variable and Expectation | 离散随机变量与期望

Question 8: A discrete random variable X has probability distribution given by P(X = x) = k(x + 1) for x = 0, 1, 2, 3. Determine the value of k and compute E(X).

题目 8：离散随机变量 X 的概率分布为 P(X = x) = k(x + 1)，其中 x = 0, 1, 2, 3。求 k 的值并计算 E(X)。

The sum of all probabilities must equal 1. Set up the equation:

所有概率之和必须等于1。列出方程：

Σ P(X = x) = k(0+1) + k(1+1) + k(2+1) + k(3+1) = k(1 + 2 + 3 + 4) = 10k = 1

Thus k = 1/10 = 0.1.

因此 k = 1/10 = 0.1。

The probability mass function is P(X = x) = (x + 1)/10. Expected value:

概率分布函数为 P(X = x) = (x + 1)/10。期望值：

E(X) = Σ x · P(X = x) = 0×(1/10) + 1×(2/10) + 2×(3/10) + 3×(4/10) = (0 + 2 + 6 + 12)/10 = 20/10 = 2

The expected value of X is 2.

X 的期望值为2。

Published by TutorHao | Statistics Revision Series | aleveler.com

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WJEC Pre-U Statistics Mock Paper Walkthrough | WJEC Pre-U 统计模拟卷解析