IGCSE AQA Statistics Unit Test Mock Paper Analysis | IGCSE AQA 统计单元测试模拟卷解析

📚 IGCSE AQA Statistics Unit Test Mock Paper Analysis | IGCSE AQA 统计单元测试模拟卷解析

This article walks you through a complete IGCSE AQA Statistics unit test mock paper, with fully worked solutions and examiner-style commentary. Every question targets a core specification topic – from descriptive statistics and data representation to probability, correlation and discrete distributions – so you can assess your readiness and sharpen your exam technique.

本文逐题解析一份 IGCSE AQA 统计单元测试模拟卷，配以完整的推导过程和阅卷人风格的点评。每道题都紧扣考纲核心主题——从描述统计、数据表示到概率、相关及离散分布，帮助你检验备考水平并打磨应试技巧。

1. Mean, Median, Mode and Range | 第1题：平均数、中位数、众数与极差

Question: The ages of eight students in a math club are: 12, 15, 14, 10, 18, 14, 16, 13. Calculate the mean, median, mode and range of this set of ages.
Answer: Mean = (12+15+14+10+18+14+16+13) ÷ 8 = 112 ÷ 8 = 14. Ordered list: 10, 12, 13, 14, 14, 15, 16, 18. Median = (14+14) ÷ 2 = 14. Mode = 14 (appears twice). Range = 18 – 10 = 8.

题目：数学俱乐部中 8 名学生的年龄为：12, 15, 14, 10, 18, 14, 16, 13。计算这组年龄的平均数、中位数、众数和极差。
解答：平均数 = (12+15+14+10+18+14+16+13) ÷ 8 = 112 ÷ 8 = 14。排序后为 10, 12, 13, 14, 14, 15, 16, 18，中位数 = (14+14) ÷ 2 = 14。众数 = 14（出现两次）。极差 = 18 – 10 = 8。

2. Estimated Mean from a Frequency Table | 第2题：根据频数表估计平均数

Question: The table shows the time spent on homework by 28 students. Calculate an estimate of the mean time.

Answer: Midpoints: 5, 15, 25, 35. Σf = 28. Σfx = 5×5 + 15×12 + 25×8 + 35×3 = 25 + 180 + 200 + 105 = 510. Estimated mean = 510 ÷ 28 ≈ 18.2 minutes (1 d.p.).

题目：下表显示了 28 名学生做作业所花的时间。请估计平均时间。

解答：组中值：5, 15, 25, 35。Σf = 28。Σfx = 5×5 + 15×12 + 25×8 + 35×3 = 25 + 180 + 200 + 105 = 510。估计平均数 = 510 ÷ 28 ≈ 18.2 分钟（保留一位小数）。

3. Box Plot and Outliers | 第3题：箱线图与异常值

Question: A box plot is drawn using the five-number summary: minimum = 2, Q₁ = 5, median = 8, Q₃ = 12, maximum = 20. Calculate the interquartile range (IQR) and determine whether any outliers are present using the 1.5 × IQR rule.
Answer: IQR = Q₃ – Q₁ = 12 – 5 = 7. Lower fence = Q₁ – 1.5 × IQR = 5 – 10.5 = –5.5. Upper fence = Q₃ + 1.5 × IQR = 12 + 10.5 = 22.5. All data values lie between 2 and 20, which are inside the fences, so there are no outliers.

题目：根据以下五数概括绘制了箱线图：最小值 = 2，下四分位数 Q₁ = 5，中位数 = 8，上四分位数 Q₃ = 12，最大值 = 20。计算四分位距 (IQR)，并用 1.5 × IQR 法则判断是否存在异常值。
解答：IQR = Q₃ – Q₁ = 12 – 5 = 7。下内限 = Q₁ – 1.5 × IQR = 5 – 10.5 = –5.5。上内限 = Q₃ + 1.5 × IQR = 12 + 10.5 = 22.5。所有数据均处于 2 到 20 之间，落在内限范围内，因此不存在异常值。

4. Probability Without Replacement | 第4题：不放回概率

Question: A bag contains 3 red balls and 5 blue balls. Two balls are drawn at random without replacement. Calculate the probability that both balls are red. Illustrate your calculation with a tree diagram.
Answer: P(Red first) = 3/8. After one red is taken, 2 red and 5 blue remain, so P(Red second | Red first) = 2/7. P(Both red) = (3/8) × (2/7) = 6/56 = 3/28. Tree diagram: first branch R (3/8) and B (5/8); from R, second branch R (2/7) and B (5/7), giving the combined probability for RR as shown.

题目：一个袋子里装有 3 个红球和 5 个蓝球。从中不放回地随机抽取两个球，求两个球都是红色的概率，并用树状图说明计算过程。
解答：第一次抽到红球的概率为 3/8。取走一个红球后，剩余 2 红 5 蓝，因此在第一次为红的条件下第二次仍为红的概率为 2/7。两球皆红的概率 = (3/8) × (2/7) = 6/56 = 3/28。树状图：第一层分支 R (3/8) 和 B (5/8)；从 R 延伸出第二层分支 R (2/7) 和 B (5/7)，RR 路径的乘积即为上述结果。

5. Pearson’s Correlation Coefficient | 第5题：皮尔逊相关系数

Question: For six pairs of observations, the following summary statistics are obtained: n = 6, Σx = 30, Σy = 36, Σx² = 190, Σy² = 254, Σxy = 210. Calculate Pearson’s product-moment correlation coefficient, r. Interpret the result.
Answer: r = [nΣxy – ΣxΣy] / √[(nΣx² – (Σx)²)(nΣy² – (Σy)²)]. Numerator = 6×210 – 30×36 = 1260 – 1080 = 180. nΣx² – (Σx)² = 6×190 – 30² = 1140 – 900 = 240. nΣy² – (Σy)² = 6×254 – 36² = 1524 – 1296 = 228. Denominator = √(240 × 228) = √54720 ≈ 233.92. Hence r ≈ 180 ÷ 233.92 ≈ 0.770 (3 s.f.). This indicates a strong positive linear correlation.

题目：对于 6 组观测值，得到以下汇总统计量：n = 6, Σx = 30, Σy = 36, Σx² = 190, Σy² = 254, Σxy = 210。计算皮尔逊积矩相关系数 r，并解释结果。
解答：r = [nΣxy – ΣxΣy] / √[(nΣx² – (Σx)²)(nΣy² – (Σy)²)]。分子 = 6×210 – 30×36 = 1260 – 1080 = 180。nΣx² – (Σx)² = 6×190 – 30² = 1140 – 900 = 240。nΣy² – (Σy)² = 6×254 – 36² = 1524 – 1296 = 228。分母 = √(240 × 228) = √54720 ≈ 233.92。因此 r ≈ 180 ÷ 233.92 ≈ 0.770（保留三位有效数字）。这表明存在强正线性相关。

6. Regression Line | 第6题：回归直线

Question: Using the statistics from Question 5, find the equation of the regression line of y on x in the form y = a + bx. Use your equation to predict y when x = 7.
Answer: b = [nΣxy – ΣxΣy] / [nΣx² – (Σx)²] = 180 / 240 = 0.75. Mean of x = 30/6 = 5, mean of y = 36/6 = 6. a = ȳ – b x̄ = 6 – 0.75×5 = 6 – 3.75 = 2.25. Equation: y = 2.25 + 0.75x. When x = 7, predicted y = 2.25 + 0.75×7 = 2.25 + 5.25 = 7.5.

题目：利用第5题中的统计量，求 y 对 x 的回归直线方程，形式为 y = a + bx。并用该方程预测当 x = 7 时 y 的值。
解答：b = [nΣxy – ΣxΣy] / [nΣx² – (Σx)²] = 180 / 240 = 0.75。x

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