Year 13 SQA Statistics: Unit Test Mock Paper Walkthrough | 高三 SQA 统计：单元模拟卷详解

📚 Year 13 SQA Statistics: Unit Test Mock Paper Walkthrough | 高三 SQA 统计：单元模拟卷详解

This article provides a complete walkthrough of a mock unit test designed for the Year 13 SQA Statistics course (Advanced Higher). Each question has been carefully crafted to reflect the style and depth required by SQA assessment standards, covering key topics such as probability distributions, hypothesis testing, confidence intervals, chi‑squared tests, correlation and regression, normal approximations, and combinations of random variables. Step‑by‑step solutions are presented in both English and Chinese, ensuring learners can follow the statistical reasoning in their preferred language.

本文为高三 SQA 统计（Advanced Higher）课程的单元模拟卷提供完整详解。试卷中的每一道题目都紧扣 SQA 评估要求的风格与深度，覆盖了概率分布、假设检验、置信区间、卡方检验、相关与回归、正态近似以及随机变量的组合等重要主题。每道题的解答都采用英文和中文逐步对照呈现，确保学习者可以无障碍地理解统计推理过程。

1. Question 1: Binomial Probability | 问题1：二项分布概率

Question: A biased coin with P(Head) = 0.35 is tossed 15 times. Find the probability of obtaining exactly 6 heads.

题目： 将一枚 P(正面) = 0.35 的有偏硬币抛掷 15 次。求恰好得到 6 次正面的概率。

Step 1: Recognise that X ~ B(15, 0.35). The required probability is P(X = 6).

步骤1：识别出 X ~ B(15, 0.35)。所求概率为 P(X = 6)。

Step 2: Apply the binomial probability formula: P(X = k) = nCk * p^k * (1-p)^(n-k). Here n=15, k=6, p=0.35.

步骤2：应用二项概率公式：P(X = k) = nCk * p^k * (1-p)^(n-k)。此处 n=15, k=6, p=0.35。

Step 3: Compute the combination 15C6 = 5005. Then (0.35)^6 * (0.65)^9 ≈ 0.001838 * 0.02071 ≈ 3.808 * 10^-5, multiplied by 5005 gives 0.1906.

步骤3：计算组合数 15C6 = 5005。然后 (0.35)^6 * (0.65)^9 ≈ 0.001838 * 0.02071 ≈ 3.808 × 10⁻⁵，再乘以 5005 得到 0.1906。

Step 4: Therefore, P(X = 6) = 0.191 (to 3 significant figures).

步骤4：因此，P(X = 6) = 0.191（保留三位有效数字）。

Key point: Always show the full substitution before rounding intermediate values.

关键点： 在舍入中间值之前，始终展示完整的代入过程。

2. Question 2: Poisson Distribution | 问题2：泊松分布

Question: The number of misprints per page in a book follows a Poisson distribution with mean 1.2. Find the probability that a randomly chosen page contains at most 2 misprints.

题目： 某书中每页的印刷错误数服从均值为 1.2 的泊松分布。求随机选取的一页包含至多 2 处错误的概率。

Step 1: Let X ~ Po(1.2). We need P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2).

步骤1：设 X ~ Po(1.2)。我们需要 P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)。

Step 2: Poisson probability formula: P(X=k) = (λ^k * e^(-λ)) / k! where λ = 1.2.

步骤2：泊松概率公式：P(X=k) = (λ^k * e^(-λ)) / k!，其中 λ = 1.2。

Step 3: P(X=0) = e^(-1.2) ≈ 0.3012; P(X=1) = 1.2 * e^(-1.2) ≈ 0.3614; P(X=2) = (1.2^2 / 2) * e^(-1.2) = 0.72 * 0.3012 ≈ 0.2169.

步骤3：P(X=0) = e⁻¹·² ≈ 0.3012；P(X=1) = 1.2 * e⁻¹·² ≈ 0.3614；P(X=2) = (1.2² / 2) * e⁻¹·² = 0.72 * 0.3012 ≈ 0.2169。

Step 4: Summing gives 0.3012 + 0.3614 + 0.2169 = 0.8795. Thus the probability is about 0.880 (3 s.f.).

步骤4：求和得 0.3012 + 0.3614 + 0.2169 = 0.8795。因此概率约为 0.880（保留三位有效数字）。

3. Question 3: Normal Distribution and Inverse Normal | 问题3：正态分布与逆向正态

Question: The volume of drink in a can is normally distributed with mean 330 ml and standard deviation 4 ml. (i) Find the proportion of cans containing less than 325 ml. (ii) Find the volume exceeded by 95% of cans.

题目： 某饮料罐的容量服从均值为 330 ml、标准差为 4 ml 的正态分布。(i) 求容量低于 325 ml 的罐子所占比例。(ii) 求被 95% 的罐子超过的那个容量值。

(i) Standardise: z = (325 – 330) / 4 = -1.25. P(Z < -1.25) = 1 - Φ(1.25) = 1 - 0.8944 = 0.1056.

(i) 标准化：z = (325 – 330) / 4 = -1.25。P(Z < -1.25) = 1 - Φ(1.25) = 1 - 0.8944 = 0.1056。

(ii) We need the 5th percentile because 95% exceed it. Find z such that P(Z < z) = 0.05. From tables, z ≈ -1.645 (or -1.64).

(ii) 我们需要第 5 百分位数，因为有 95% 超过它。找出使 P(Z < z) = 0.05 的 z 值。查表得 z ≈ -1.645（或 -1.64）。

Unstandardise: x = μ + z*σ = 330 + (-1.645)*4 = 330 – 6.58 = 323.42 ml. Rounded to 323 ml (or 323.4).

去标准化：x = μ + z*σ = 330 + (-1.645)*4 = 330 – 6.58 = 323.42 ml。四舍五入到 323 ml（或 323.4）。

Thus, about 10.6% of cans contain less than 325 ml, and the volume 323.4 ml is exceeded by 95% of cans.

因此，约 10.6% 的罐子容量低于 325 ml，且容量 323.4 ml 被 95% 的罐子超过。

4. Question 4: Hypothesis Testing (Binomial) | 问题4：假设检验（二项分布）

Question: A manufacturer claims that the proportion of defective items is at most 8%. In a random sample of 25 items, 4 are found to be defective. Test the claim at the 5% significance level, stating your hypotheses clearly.

题目： 某制造商声称其产品的不合格率至多为 8%。从随机抽取的 25 件产品中发现 4 件不合格。在 5% 显著性水平下检验该声明，需清晰地给出假设。

Step 1: H0: p = 0.08, H1: p > 0.08 (one‑tailed test).

步骤1：H0: p = 0.08，H1: p > 0.08（单尾检验）。

Step 2: Under H0, X ~ B(25, 0.08). Observed number of defectives is 4. We compute P(X ≥ 4).

步骤2：在 H0 下，X ~ B(25, 0.08)。观察到的不合格品数为 4。计算 P(X ≥ 4)。

Step 3: P(X ≥ 4) = 1 – P(X ≤ 3). Using tables or formula: P(X=0) = 0.92^25 ≈ 0.1244, P(X=1) = 25*0.08*0.92^24 ≈ 0.2703, P(X=2) = C(25,2)*0.08^2*0.92^23 ≈ 0.2821, P(X=3) = C(25,3)*0.08^3*0.92^22 ≈ 0.1880. Sum = 0.8648, so P(X ≥ 4) = 0.1352.

步骤3：P(X ≥ 4) = 1 – P(X ≤ 3)。查表或计算得：P(X=0) = 0.92²⁵ ≈ 0.1244，P(X=1) = 25×0.08×0.92²⁴ ≈ 0.2703，P(X=2) = C(25,2)×0.08²×0.92²³ ≈ 0.2821，P(X=3) = C(25,3)×0.08³×0.92²² ≈ 0.1880。和为 0.8648，故 P(X ≥ 4) = 0.1352。

Step 4: Since 0.1352 > 0.05, we do not reject H0. There is insufficient evidence to contradict the manufacturer’s claim.

步骤4：因为 0.1352 > 0.05，我们不拒绝 H0。没有足够证据否定制造商的声明。

Note: Always relate the conclusion to the context.

注意：结论必须结合问题的实际背景。

5. Question 5: Confidence Interval for the Mean | 问题5：均值的置信区间

Question: A random sample of 20 bags of sugar gives a mean weight of 502 g and a sample standard deviation of 5 g. Construct a 95% confidence interval for the population mean weight, stating any assumptions.

题目： 随机抽取 20 袋糖，样本均重为 502 g，样本标准差为 5 g。构造总体均重的 95% 置信区间，并说明所需要的假设。

Step 1: Since population variance is unknown, we use the t‑distribution with degrees of freedom ν = n‑1 = 19.

步骤1：由于总体方差未知，我们使用 t 分布，自由度 ν = n‑1 = 19。

Step 2: For a 95% CI, the critical value t(19) is 2.093 (two‑tailed).

步骤2：对于 95% 置信区间，临界值 t(19) = 2.093（双尾）。

Step 3: Standard error = s / √n = 5 / √20 ≈ 1.118. Margin of error = t * SE = 2.093 * 1.118 ≈ 2.34.

步骤3：标准误 = s / √n = 5 / √20 ≈ 1.118。误差界限 = t × SE = 2.093 × 1.118 ≈ 2.34。

Step 4: The CI is (502 – 2.34, 502 + 2.34) = (499.66, 504.34). Assumption: the weights are normally distributed (or sample size large enough, but we assume normal population for small n).

步骤4：置信区间为 (502 – 2.34, 502 + 2.34) = (499.66, 504.34)。假设：糖袋重量服从正态分布（对于小样本必须假设总体正态）。

6. Question 6: Chi‑squared Test for Association | 问题6：卡方关联性检验

Question: A survey investigates the relationship between exercise frequency and smoking status. Observed frequencies are given in the table. Test at the 1% significance level whether there is an association.

	Non‑smoker	Smoker	Total
Regular exercise	80	20	100
No regular exercise	50	50	100
Total	130	70	200

题目： 一项调查探究锻炼频率与吸烟状况之间的关系。观察频数如上表所示。在 1% 显著性水平下检验是否存在关联。

Step 1: H0: No association between exercise and smoking. H1: Association exists.

步骤1：H0：锻炼与吸烟之间无关联。H1：存在关联。

Step 2: Expected frequencies for each cell = (row total * column total) / grand total. E.g., cell (Regular, Non‑smoker) = (100*130)/200 = 65.

步骤2：每个单元格的期望频数 = (行合计 × 列合计) / 总计。例如，单元格 (规律锻炼，不吸烟) = (100×130)/200 = 65。

Step 3: Table of expected: Row1: 65, 35; Row2: 65, 35.

步骤3：期望频数表：第一行：65, 35；第二行：65, 35。

Step 4: Compute χ² = Σ (O-E)²/E. Row1: (80-65)²/65 = 3.4615, (20-35)²/35 ≈ 6.4286. Row2: (50-65)²/65 = 3.4615, (50-35)²/35 = 6.4286. Sum χ² ≈ 19.78.

步骤4：计算 χ² = Σ (O-E)²/E。第一行：(80-65)²/65 = 3.4615，(20-35)²/35 ≈ 6.4286。第二行：(50-65)²/65 = 3.4615，(50-35)²/35 = 6.4286。总和 χ² ≈ 19.78。

Step 5: Degrees of freedom = (2-1)*(2-1) = 1. Critical value at 1% from tables: χ²(1) = 6.635. Since 19.78 > 6.635, reject H0. Strong evidence of an association.

步骤5：自由度 = (2-1)×(2-1) = 1。1% 水平的临界值查表得 χ²(1) = 6.635。因 19.78 > 6.635，拒绝 H0。有很强的证据表明存在关联。

7. Question 7: Correlation and Regression | 问题7：相关与回归

Question: Data: x (hours studied): 2, 4, 5, 7, 9; y (test score): 50, 60, 65, 75, 90. (i) Calculate the Pearson correlation coefficient r. (ii) Find the equation of the regression line of y on x. (iii) Test, at 5% significance, whether the population correlation is zero.

题目： 数据：x (学习小时数)：2, 4, 5, 7, 9；y (测试分数)：50, 60, 65, 75, 90。(i) 计算皮尔逊相关系数 r。(ii) 求 y 对 x 的回归直线方程。(iii) 在 5% 显著性水平下检验总体相关系数是否为零。

Step 1: Compute sums: n=5, Σx=27, Σy=340, Σx²=175, Σy²=24150, Σxy=1960.

步骤1：计算累和：n=5, Σx=27, Σy=340, Σx²=175, Σy²=24150, Σxy=1960。

Step 2: Sxx = 175 – (27²)/5 = 29.2; Syy = 24150 – (340²)/5 = 1030; Sxy = 1960 – (27*340)/5 = 124.

步骤2：Sxx = 175 – (27²)/5 = 29.2；Syy = 24150 – (340²)/5 = 1030；Sxy = 1960 – (27×340)/5 = 124。

Step 3: r = Sxy / √(Sxx * Syy) = 124 / √(29.2 * 1030) ≈ 124 / √30076 ≈ 0.714. Very strong positive correlation.

步骤3：r = Sxy / √(Sxx × Syy) = 124 / √(29.2 × 1030) ≈ 124 / √30076 ≈ 0.714。具有很强的正相关。

Step 4: Regression equation y = a + bx. b = Sxy / Sxx = 124 / 29.2 ≈ 4.25; a = ȳ – b*x̄ = 68 – 4.25*5.4 = 68 – 22.95 = 45.05. So y ≈ 45.0 + 4.25x.

步骤4：回归方程 y = a + bx。b = Sxy / Sxx = 124 / 29.2 ≈ 4.25；a = ȳ – b×x̄ = 68 – 4.25×5.4 = 68 – 22.95 = 45.05。因此 y ≈ 45.0 + 4.25x。

Step 5: Test H0: ρ = 0. Test statistic t = r * √(n-2) / √(1-r²) = 0.714 * √3 / √(1-0.5098) ≈ 0.714 * 1.732 / 0.700 ≈ 1.77. Critical t(3) two‑tailed = 3.182. Since 1.77 < 3.182, do not reject H0. Not significant at 5%.

步骤5：检验 H0: ρ = 0。检验统计量 t = r × √(n-2) / √(1-r²) = 0.714 × √3 / √(1-0.5098) ≈ 0.714 × 1.732 / 0.700 ≈ 1.77。双尾 t(3) 临界值 = 3.182。因为 1.77 < 3.182，不拒绝 H0。在 5% 水平下不显著。

8. Question 8: Normal Approximation to Binomial | 问题8：二项分布的正态近似

Question: X ~ B(200, 0.35). Use a normal approximation with continuity correction to estimate P(60 < X <= 75).

题目： X ~ B(200, 0.35)。使用带连续性修正的正态近似估计 P(60 < X ≤ 75)。

Step 1: Check conditions: np = 70 > 5, nq = 130 > 5, suitable.

步骤1：检查条件：np = 70 > 5, nq = 130 > 5，适合使用近似。

Step 2: Approximate by Y ~ N(μ = 70, σ² = npq = 200*0.35*0.65 = 45.5, σ ≈ 6.745).

步骤2：用 Y ~ N(μ = 70, σ² = 200×0.35×0.65 = 45.5, σ ≈ 6.745) 近似。

Step 3: With continuity correction: P(60 < X ≤ 75) becomes P(60.5 < Y ≤ 75.5).

步骤3：使用连续性修正：P(60 < X ≤ 75) 变为 P(60.5 < Y ≤ 75.5)。

Step 4: Standardise: z1 = (60.5 – 70) / 6.745 ≈ -1.408; z2 = (75.5 – 70) / 6.745 ≈ 0.815.

步骤4：标准化：z₁ = (60.5 – 70) / 6.745 ≈ -1.408；z₂ = (75.5 – 70) / 6.745 ≈ 0.815。

Step 5: P = Φ(0.815) – Φ(-1.408) = 0.7929 – 0.0797 = 0.7132.

步骤5：概率 = Φ(0.815) – Φ(-1.408) = 0.7929 – 0.0797 = 0.7132。

So the approximate probability is 0.713.

因此近似概率为 0.713。

9. Question 9: Combining Random Variables | 问题9：随机变量的组合

Question: The length of a pipe, L, is N(30, 2²) and the length of a connector, C, is N(5, 0.8²). They are joined end‑to‑end, and a gap of exactly 1 cm is left. Find the probability that the total assembly length exceeds 36 cm. Pipe and connector are independent.

题目： 管子的长度 L ~ N(30, 2²)，连接件的长度 C ~ N(5, 0.8²)。两者首尾相接，并留出恰好 1 cm 的间隙。求总组装长度超过 36 cm 的概率。管子和连接件相互独立。

Step 1: Define total length T = L + C + 1. Then E(T) = 30 + 5 + 1 = 36 cm. Var(T) = Var(L) + Var(C) = 4 + 0.64 = 4.64.

步骤1：定义总长度 T = L + C + 1。那么 E(T) = 30 + 5 + 1 = 36 cm，Var(T) = Var(L) + Var(C) = 4 + 0.64 = 4.64。

Step 2: Since L and C are normal, T ~ N(36, 4.64). Standard deviation σ = √4.64 ≈ 2.154.

步骤2：因为 L 和 C 服从正态分布，T ~ N(36, 4.64)。标准差 σ = √4.64 ≈ 2.154。

Step 3: P(T > 36) = P(Z > (36 – 36) / 2.154) = P(Z > 0) = 0.5.

步骤3：P(T > 36) = P(Z > (36 – 36)/2.154) = P(Z > 0) = 0.5。

Because the mean of the assembly length is exactly 36, the probability of exceeding 36 is exactly 0.5 by symmetry.

由于总组装长度的均值恰好为 36，根据对称性，超过 36 的概率恰好为 0.5。

10. Key Formula Reminders | 关键公式提醒

Ensure you are confident with the binomial probability formula P(X=k) = nCk * p^k * (1-p)^(n-k) and its cumulative use.

务必熟练掌握二项概率公式 P(X=k) = nCk * p^k * (1-p)^(n-k) 及其累积用法。

Poisson probabilities P(X=k) = (λ^k * e^(-λ)) / k! are essential for rare event modeling.

泊松概率 P(X=k) = (λ^k * e^(-λ)) / k! 对于稀有事件建模至关重要。

The standardising formula z =

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Year 13 SQA Statistics: Unit Test Mock Paper Walkthrough | 高三 SQA 统计：单元模拟卷详解