Year 13 AQA Statistics: Unit Test Mock Paper Solutions | 13年级AQA统计学：单元测试模拟卷解析

📚 Year 13 AQA Statistics: Unit Test Mock Paper Solutions | 13年级AQA统计学：单元测试模拟卷解析

This article provides detailed solutions to a mock unit test designed for the Year 13 AQA Statistics syllabus. Each question targets a core topic, including binomial hypothesis testing, Poisson approximation, normal distribution calculations, correlation and regression, conditional probability, chi-squared tests for independence, continuous random variables, sampling distributions, normal hypothesis tests for the mean, and the geometric distribution. Step-by-step reasoning is presented in both English and Chinese to reinforce understanding and exam technique.

本文为13年级AQA统计学设计的单元测试模拟卷提供详细解析。每道题针对一个核心主题，涵盖二项分布的假设检验、泊松近似、正态分布计算、相关与回归、条件概率、独立性卡方检验、连续随机变量、抽样分布、均值的正态假设检验以及几何分布。讲解以中英双语配对呈现，逐步拆解思路，帮助巩固理解并打磨应考技巧。

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1. Binomial Distribution Hypothesis Test | 二项分布假设检验

A manufacturer claims that 80% of light bulbs are defect‑free. In a random sample of 20 bulbs, 14 are found to have no defects. Test, at the 5% significance level, whether there is evidence that the proportion of defect‑free bulbs is less than 80%.

某制造商声称灯泡无缺陷的比例为80%。从一批产品中随机抽取20个灯泡，发现14个无缺陷。在5%显著性水平下，检验是否有证据表明无缺陷灯泡的比例低于80%。

Let X be the number of defect‑free bulbs in a sample of 20. Under the null hypothesis, X ~ B(20, 0.8).

设X为20个灯泡中无缺陷的个数。在原假设下，X ~ B(20, 0.8)。

H₀: p = 0.8 H₁: p < 0.8 (one‑tailed test). We need P(X ≤ 14) for a binomial distribution with n = 20, p = 0.8.

H₀: p = 0.8 H₁: p < 0.8（单侧检验）。需计算二项分布(20, 0.8)下P(X ≤ 14)的概率。

mula or cumulative tables: P(X ≤ 14) = 0.0210 (3 d.p.). Since 0.0210 < 0.05, we reject H₀.

利用二项分布公式或累积表得：P(X ≤ 14) = 0.0210（保留3位小数）。由于0.0210 < 0.05，拒绝原假设。

There is sufficient evidence at the 5% level to suggest the proportion of defect‑free bulbs is less than 80%.

在5%显著性水平下有足够证据表明无缺陷灯泡的比例低于80%。

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

A warehouse inspector notes that 2% of a large batch of items are damaged. A sample of 150 items is taken. Using a suitable approximation, find the probability that at most 4 items are damaged, and comment on the validity of the approximation.

仓库检验员发现一批大量货物中有2%的物品损坏。抽取150件样品。用适当的近似方法求最多有4件物品损坏的概率，并评价该近似的合理性。

X ~ use n is large and p is small, we can approximate with Poisson(λ = np = 150 × 0.02 = 3.0).

X ~ B(150, 0.02)。由于n很大且p很小，可用泊松分布近似，参数λ = np = 150 × 0.02 = 3.0。

Then X ~ Po(3) approximately. P(X ≤ 4) = 0.8153 (4 d.p.) from cumulative Poisson tables.

于是近似有X ~ Po(3)。由泊松累积表得P(X ≤ 4) = 0.8153（4位小数）。

The approximation is valid because n is large (150) and p is small (0.02). The condition np < 10 is satisfied (3 < 10).

该近似合理，因为n大而p小，且满足np < 10的条件（3 < 10）。

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3. Normal Distribution: Finding Probabilities | 正态分布：求概率

The mass of cereal in a packet is normally distributed with mean 505 g and standard deviation 5 g. A packet is chosen at random. Find the probability that its mass is (a) less than 498 g, (b) between 503 g and 512 g.

包装谷物质量服从正态分布，均值为505 g，标准差为5 g。随机选取一包，求其质量 (a) 小于498 g 的概率， (b) 介于503 g与512 g之间的概率。

Let M ~ N(505, 5²). Standardise: Z = (M − 505) / 5.

设M ~ N(505, 5²)。标准化：Z = (M − 505) / 5。

(a) P(M < 498) = P(Z < (498−505)/5 = −1.4) = 1 − Φ(1.4) = 1 − 0.9192 = 0.0808.

(a) P(M < 498) = P(Z < (498−505)/5 = −1.4) = 1 − Φ(1.4) = 1 − 0.9192 = 0.0808。

512−505)/5 ) = P(−0.4 < Z < 1.4) = Φ(1.4) − (1 − Φ(0.4)) = 0.9192 − (1 − 0.6554) = 0.5746.

b) P(503 < M < 512) = P(−0.4 < Z < 1.4) = Φ(1.4) − (1 − Φ(0.4)) = 0.9192 − 0.3446 = 0.5746。

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4. Correlation and Linear Regression | 相关与线性回归

culate the product moment correlation coefficient. (b) Obtain the equation of the regression line of y on x. (c) Predict the mark for a student who revises for 10 hours.

六名学生的复习小时数 (x) 与测验分数 (y) 的数据汇总如下：Σx = 48，Σy = 360，Σx² = 480，Σy² = 22 800，Σxy = 3 180。(a) 计算积矩相关系数。(b) 求y对x的回归直线方程。(c) 预测复习10小时的学生的分数。

n = 6. S_xx = Σx² − (Σx)²/n = 480 − 48²/6 = 96. S_yy = 22 800 − 360²/6 = 1 200. S_xy = 3 180 − (48×360)/6 = 300.

n = 6。S_xx = 480 − (48²)/6 = 96。S_yy = 22 800 − (360²)/6 = 1 200。S_xy = 3 180 − (48×360)/6 = 300。

(a) r = S_xy / √S_xxS_yy = 300 / √96×1200 = 0.8839 (4 d.p.). Strong positive correlation.

(a) r = 300 / √96×1200 = 0.8839（4位小数），呈强正相关。

(b) b = S_xy / S_xx = 300/96 = 3.125. a = ȳ − b x̄ = (360/6) − 3.125×(48/6) = 60 − 25 = 35. Equation: y = 35 + 3.125x.

(b) b = 300/96 = 3.125。a = 60 − 3.125×8 = 35。方程为 y = 35 + 3.125x。

(c) When x = 10, predicted y = 35 + 3.125×10 = 66.25. Marks are likely to be around 66.

(c) 当x = 10时，预测y = 35 + 31.25 = 66.25，分数预计约为66分。

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5. Conditional Probability Using Venn Diagrams | 条件概率与文氏图

For events A and B, P(A) = 0.3, P(B) = 0.5 and P(A ∩ B) = 0.2. Evaluate (a) P(A|B), (b) P(B'|A), (c) P(A ∪ B). State whether A and B are independent.

对于事件A与B，已知P(A) = 0.3，P(B) = 0.5，P(A ∩ B) = 0.2。计算 (a) P(A|B)，(b) P(B'|A)，(c) P(A ∪ B)，并判断A与B是否独立。

(a) P(A|B) = P(A∩B) / P(B) = 0.2/0.5 = 0.4.

(a) P(A|B) = 0.2/0.5 = 0.4。

(b) P(B'|A) = P(A∩B') / P(A) = (0.3 − 0.2)/0.3 = 0.1/0.3 = 1/3 ≈ 0.333.

(b) P(B'|A) = (P(A) − P(A∩B)) / P(A) = 0.1/0.3 = 1/3 ≈ 0.333。

(c) P(A ∪ B) = 0.3 + 0.5 − 0.2 = 0.6.

(c) P(A ∪ B) = 0.3 + 0.5 − 0.2 = 0.6。

Check independence: P(A) × P(B) = 0.3 × 0.5 = 0.15 ≠ 0.2, so A and B are not independent.

检验独立性：P(A)×P(B) = 0.15 ≠ 0.2，因此A与B不独立。

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6. Chi-Squared Test for Independence | 独立性卡方检验

A survey of 200 people collected data on gender and preference for a new product:

一项对200人的调查收集了性别与新产品偏好的数据：

	Like	Dislike	Total
Male	60	40	100
Female	80	20	100
Total	140	60	200

Test at the 1% level whether gender and preference are independent.

在1%显著性水平下检验性别与偏好是否独立。

Expected frequencies are calculated by (row total × column total)/grand total. For Male‑Like: (100×140)/200 = 70. The expected table:

期望频数由 (行合计×列合计)/总计 计算。如男性-喜欢：(100×140)/200 = 70。期望频数表：

	Like	Dislike
Male	70	30
Female	70	30

70 + (40−30)²/30 + (80−70)²/70 + (20−30)²/30 = 1.4286 + 3.3333 + 1.4286 + 3.3333 = 9.5238.

/70 + (40−30)²/30 + (80−70)²/70 + (20−30)²/30 = 1.4286 + 3.3333 + 1.4286 + 3.3333 = 9.5238。

Degrees of freedom = (2−1)(2−1) = 1. Critical value at 1%: χ²₁,₀.₀₁ = 6.635. Since 9.52 > 6.635, reject H₀. There is evidence of association.

自由度 = (2−1)(2−1) = 1。1%临界值χ²₁,₀.₀₁ = 6.635。9.52 > 6.635，拒绝原假设，偏好与性别存在关联。

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7. Continuous Random Variables – PDF and CDF | 连续随机变量 – 概率密度函数与累积分布函数

nd k. (b) Determine the cumulative distribution function F(x). (c) Obtain the median of X.

连续随机变量X的概率密度函数为 f(x) = kx(2 − x)，0 ≤ x ≤ 2，其余为0。(a) 求k。(b) 求累积分布函数F(x)。(c) 求X的中位数。

kx(2−x) dx = 1. Compute ∫₀² (2x − x²) dx = [x² − x³/3]₀² = (4 − 8/3) = 4/3. So k × 4/3 = 1 ⇒ k = 3/4.

(a) 总面积 ∫₀² kx(2−x) dx = 1。计算 ∫₀² (2x − x²) dx = [x² − x³/3]₀² = 4 − 8/3 = 4/3，因此k × 4/3 = 1 ⇒ k = 3/4。

(b) F(x) = ∫₀ˣ (3/4)(2t − t²) dt = (3/4)[t² − t³/3]₀ˣ = (3/4)(x² − x³/3) for 0 ≤ x ≤ 2.

(b) F(x) = ∫₀ˣ (3/4)(2t − t²) dt = (3/4)(x² − x³/3)，0 ≤ x ≤ 2。

m−1)(m² − 2m − 2) = 0 gives m = 1. Check root: m = 1 works, and 1 is in [0,2]. Median = 1.

(c) 中位数m满足F(m)=0.5，即(3/4)(m² − m³/3)=0.5 → m² − m³/3 = 2/3 → 3m² − m³ = 2 → m³ − 3m² + 2 = 0。在[0,2]内解方程，因式分解(m−1)(m²−2m−2)=0，得m=1，中位数为1。

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8. Sampling Distribution and Central Limit Theorem | 抽样分布与中心极限定理

The diameter of a component is normally distributed with mean 12.00 mm and standard deviation 0.15 mm. A random sample of size 36 is taken. Find the probability that the sample mean is less than 11.95 mm. Explain how the Central Limit Theorem applies if the population were not normal.

某种零件的直径服从均值为12.00 mm、标准差为0.15 mm的正态分布。随机抽取36个零件，求样本均值小于11.95 mm的概率。如果总体不是正态分布，解释中心极限定理如何应用。

Sample mean X̄ ~ N(μ, σ²/n) = N(12.00, 0.15²/36) = N(12.00, 0.000625). Standard error = 0.15/6 = 0.025.

样本均值X̄ ~ N(μ, σ²/n) = N(12.00，0.15²/36) = N(12.00, 0.000625)。标准误 = 0.025。

P(X̄ < 11.95) = P( Z < (11.95−12.00)/0.025 = −2.00 ) = 1 − Φ(2.00) = 0.0228.

P(X̄ < 11.95) = P(Z < (11.95−12.00)/0.025 = −2.00) = 1 − 0.9772 = 0.0228。

rmal distribution as n increases, provided n is large (n ≥ 30 is typical). Here n=36, so X̄ is approximately normal even if the population is skewed.

若总体非正态，中心极限定理指出当样本量n增大时，样本均值的分布趋于正态。此处n=36，满足大样本条件，即使总体偏态，X̄也近似服从正态分布。

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9. Normal Hypothesis Test for the Mean | 均值的正态假设检验

A factory claims that the mean filling mass is 250 g. A consumer group suspects the mean is less than 250 g. A random sample of 25 packets gives a mean of 248 g. The population standard deviation is known to be 6 g. Test at the 2.5% significance level.

某工厂声称填充质量的均值为250 g。消费者团体怀疑均值低于250 g。随机抽取25包，样本均值为248 g。已知总体标准差为6 g。在2.5%显著性水平下检验。

H₀: μ = 250 H₁: μ < 250 (one‑tailed). Test statistic Z = (248 − 250) / (6/√25) = −2 / 1.2 = −1.667.

H₀: μ = 250 H₁: μ < 250（单侧）。检验统计量 Z = (248 − 250) / (6/√25) = −2 / 1.2 = −1.667。

Critical value at 2.5% left‑tail: −1.96. Since −1.667 > −1.96, we do not reject H₀. There is insufficient evidence

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