📚 SQA Statistics Unit Test Mock Exam Analysis | SQA统计单元测试模拟卷解析
This detailed analysis walks through a full SQA Statistics unit test mock exam designed for Year 12 learners. Each question is broken down with step‑by‑step reasoning, key formulas, and examiner tips to help you master the core topics: descriptive statistics, probability, distributions, confidence intervals, hypothesis testing, and regression. Use this as a revision guide to identify common pitfalls and strengthen your problem‑solving approach.
本文详细解析了一份为 Year 12 学生设计的 SQA 统计单元测试模拟卷。每道题目都配有分步推理、关键公式和阅卷技巧，帮助你掌握描述统计、概率、分布、置信区间、假设检验和回归等核心主题。将本篇解析作为复习指南，可以帮你发现常见错误，强化解题思路。
1. Descriptive Statistics & Box Plots | 描述统计与箱线图
8. The mean is Σx ÷ n = 139 ÷ 10 = 13.9 minutes. The median is the average of the 5th and 6th values: (14+14)/2 = 14. For quartiles, Q₁ = median of the lower half (10,12,12,13,14) = 12; Q₃ = median of the upper half (14,15,15,16,18) = 15. The interquartile range (IQR) = 15 – ith no outliers because 1.5×IQR = 4.5; upper fence = 15+4.5 = 19.5, so 18 is within range.
数据集为完成拼图所需时间（分钟）：12, 15, 14, 10, 18, 13, 16, 12, 14, 15。首先排序：10, 12, 12, 13, 14, 14, 15, 15, 16, 18。均值 = 总和 ÷ 个数 = 139 ÷ 10 = 13.9 分钟。中位数是第5和第6个数的平均值：(14+14)/2 = 14。四分位数：Q₁ = 下半部分中位数 (10,12,12,13,14) = 12；Q₃ = 上半部分中位数 (14,15,15,16,18) = 15。四分位距 IQR = 15 – 12 = 3。绘制箱线图：须线延伸至最小值10和最大值18，无异常值，因为1.5×IQR = 4.5，上须界限为15+4.5=19.5，18在界限内。
2. Probability with Playing Cards | 扑克牌概率
A card is drawn from a standard 52‑ g) = 4/52, P(Heart and King) = 1/52 (the King of Hearts). Using the addition rule: P(Heart ∪ King) = 13/52 + 4/52 – 1/52 = 16/52 = 4/13. Remember to subtract the intersection to avoid double‑counting.
从标准52张牌中抽一张，求抽到红心或国王的概率。P(红心)=13/52，P(国王)=4/52，P(红心且国王)=1/52（红心K）。使用加法法则：P(红心∪国王)=13/52+4/52–1/52=16/52=4/13。务必减去交集以避免重复计数。
3. Conditional Probability & Bayes' Theorem | 条件概率与贝叶斯定理
A disease affects 1% of the population. The test is 95% sensitive (true positive) and 90% ative). If a person tests positive, what is the probability they actually have the disease? Draw a tree: first branch – 5, negative with 0.05. For healthy, test positive with 0.10, negative with 0.90. P(Disease| 1085 ≈ 0.0876. The low positive predictive value highlights the effect of a low base rate.
一种疾病患病率为1%。检测敏感性95%（真阳性率），特异性90%（真阴性率）。若一人检测为阳性，他真正患病的概率是多少？画树状图：第一层分支——患病(0.01)或未患病(0.99)。患病者检测阳性概率0.95，阴性0.05；健康者检测阳性概率0.10，阴性0.90。P(患病|阳性) = (0.01×0.95) / (0.01×0.95 + 0.99×0.10) = 0.0095 / (0.0095+0.099) = 0.0095/0.1085 ≈ 0.0876。较低的阳性预测值体现了低基础比率的影响。
4. Discrete Random Variable: Expectation & Variance | 离散随机变量：期望与方差
ariance, first find E(X²) = 0²×0.2 + 1²×0.3 + 2²×0.4 + 3²×0.1 = 0 + 0.3 + 1.6 + 0.9 = 2.8. Then Var(X) = E(X²) – [E(X)]² = 2.8 – 1.4² = 2.8 – 1.96 = 0.84. Always present the final variance to an appropriate degree of accuracy.
概率分布：x = 0,1,2,3，对应p(x)=0.2,0.3,0.4,0.1。计算期望 E(X)=Σ x·p(x)=0×0.2+1×0.3+2×0.4+3×0.1=0+0.3+0.8+0.3=1.4。求方差先求E(X²)=0²×0.2+1²×0.3+2²×0.4+3²×0.1=0+0.3+1.6+0.9=2.8。然后 Var(X)=E(X²)–[E(X)]²=2.8–1.4²=2.8–1.96=0.84。最终方差保留恰当的精度。
5. Binomial Distribution Calculations | 二项分布计算
A machine produces 5% defective items. A sample of 20 is taken. Let X ~ 0.95¹⁸ ≈ 190 × 0.0025 × 0.3972 ≈ 0.1887. Probability of at least one defective: P(X≥1) = 1 – P(X=0) = 1 – 0.95²⁰ ≈ 1 – unctions (binompdf/binomcdf) in exams to save time, but show the formula for method marks.
某机器生产5%的次品，抽取20件。设 X ~ B(20, 0.05)。恰好2件次品的概率：P(X=2)=C(20,2)×0.05²×0.95¹⁸≈190×0.0025×0.3972≈0.1887。至少一件次品的概率：P(X≥1)=1–P(X=0)=1–0.95²⁰≈1–0.3585=0.6415。次品期望数：E(X)=np=20×0.05=1。考试时可用计算器函数（binompdf/binomcdf）节省时间，但仍需写出公式以获得方法分。
6. Poisson Distribution as an Approximation | 泊松分布及其近似
re receives an average of 3 calls per hour. Find the probability of exactly 5 calls in a 2‑hour period. The mean for 2 hours is λ = 3×2 = 6. X ~ Po(6), so P(X=5) = e⁻⁶ × 6⁵ / 5! nd p is small; here λ = np. Remember that the variance equals the mean in a Poisson model.
某呼叫中心平均每小时接3个电话。求2小时内恰好5个电话的概率。2小时均值 λ = 3×2 = 6。X ~ Po(6)，P(X=5)=e⁻⁶×6⁵/5!≈0.1606。当二项分布中n很大而p很小时，泊松分布可作为近似，此时λ=np。切记泊松模型中方差等于均值。
7. Normal Distribution & Z‑scores | 正态分布与Z分数
Let X ~ N(100, 15²). Find P(85 < X < 115) and P(X > 130). Convert to Z: for 85, z = (85–100)/15 = –1; for 115, z = (115–100)/15 = 1. P(–1 < Z < 1) ≈ 0.6827 (68% within 1σ). For 130: z = (130–100)/15 = 2. P(Z > 2) = 1 – Φ(2) ≈ 1 – rmal tables or the inverse normal function on your calculator, but always write down the z‑score equation for method marks.
设 X ~ N(100, 15²)。求 P(85 < X < 115) 和 P(X > 130)。转换为Z值：85对应 z=(85–100)/15=–1；115对应 z=(115–100)/15=1。P(–1
8. Confidence Interval for the Mean (σ Known) | 均值的置信区间（σ已知）
mean mass of 105 g. The population standard deviation is known to be 15 g. Construct a 95% confidence interval for μ. The formula is x̄ ± z* × σ/√n. For 95%, z* = 1.96. Standard error = 15/√25 = 3. Margin of error = 1.96 × 3 = 5.88. Thus the 95% CI = 105 ± 5.88, giving (99.12 g, 110.88 g). We are 95% es within this interval. Interpretation must refer to the population mean, not the sample.
抽检25袋面粉，样本均重105 g，已知总体标准差15 g。构建μ的95%置信区间。公式：x̄ ± z* × σ/√n。95%对应 z*=1.96。标准误 = 15/√25 = 3。误差边际 = 1.96×3=5.88。因此95% CI = 105 ± 5.88，即 (99.12 g, 110.88 g)。我们可以有95%的把握认为总体均值落在此区间内。解释时需指向总体均值，而非样本。
9. Hypothesis Test for a Single Mean (σ Known) | 单样本均值假设检验（σ已知）
A filling machine is set to 500 g. A sample of 25 packets gives x̄ = 498 g. The population σ is 5 g. Test at the 5% significance level whether the mean is below 500 g. H₀: μ = 500; H₁: μ < 500 (one‑tailed). Test statistic z = (498 – 500) / (5/√25) = –2. Critical value for α = 0.05 left‑tailed is –1.645. Since –2 < – eject H₀. There is sufficient evidence that the mean fill weight is less than 500 g. The p‑value is P(Z < –2) = 0.0228, also less than 0.05. Always state the conclusion in context.
装填机标称500 g。抽检25包得到样本均值498 g，总体标准差5 g。在5%显著性水平下检验均值是否低于500 g。H₀: μ = 500；H₁: μ < 500（单尾检验）。检验统计量 z = (498–500)/(5/√25) = –2。α=0.05左尾临界值为 –1.645。因 –2 < –1.645，拒绝H₀。有充分证据表明平均装填重量低于500 g。p值为 P(Z<–2)=0.0228，也小于0.05。务必结合背景陈述结论。
10. Correlation & Linear Regression | 相关与线性回归
y): (1,2), (2,4), (3,5), (4,4), (5,6). Compute the correlation coefficient r: first find x̄=3, ȳ=4.2. Σ(x–x̄)(y–ȳ) = (1–3)(2–4.2)+(2–3)(4–4.2)+... = 4, Σ(x–x̄)² = 10, Σ(y–ȳ)² = 8.8. r = 4/√10×8.8 ≈ 0.426. For the regression line y = a + bx: b = Sxy/Sxx = 4/10 = 0.4, a = ȳ – b x̄ = 4.2 – ncreases by 0.4 on average. Always check for a linear pattern before relying on the model.
给定五对数据 (x,y)：(1,2), (2,4), (3,5), (4,4), (5,6)。计算相关系数r：先求 x̄=3，ȳ=4.2。Σ(x–x̄)(y–ȳ)= (1–3)(2–4.2)+(2–3)(4–4.2)+... = 4，Σ(x–x̄)²=10，Σ(y–ȳ)²=8.8。r = 4/√10×8.8 ≈ 0.426。回归直线 y = a + bx：b = Sxy/Sxx = 4/10 = 0.4，a = ȳ – b x̄ = 4.2 – 0.4×3 = 3.0。方程为 y = 3.0 + 0.4x。解释：x每增加1个单位，y平均增加0.4。依赖模型前务必检查是否存在线性趋势。
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