📚 Year 11 SQA Statistics: Unit Test Mock Paper Walkthrough | 苏格兰SQA统计：单元测试模拟卷解析
Welcome to this detailed walkthrough of a mock unit test for the SQA Statistics award at Year 11 (National 5 level). This paper is designed to help you revise key topics such as descriptive statistics, probability, scatter graphs, normal distribution and sampling methods. Each question is presented with a full, step-by-step solution to clarify common exam techniques and avoid typical mistakes.
欢迎阅读本单元测试模拟卷的详细解析，对应SQA统计课程Year 11（National 5级别）。本试卷旨在帮助你复习描述性统计、概率、散点图、正态分布与抽样方法等重点内容。每一道题都配有完整的分步解答，帮助你掌握考试技巧并避开常见错误。
1. Question 1: Estimating the Mean and Median from a Grouped Frequency Table | 问题1：根据分组频数表估计平均数和中位数
Question: The marks of 30 students in a test are summarised in the grouped frequency table below. (a) Calculate an estimate for the mean mark. (b) Find the median class and estimate the median mark.
题目：30名学生的考试成绩汇总在下方的分组频数表中。(a) 计算平均分的估计值。(b) 找出中位数组并估计中位数分数。
| Marks | Frequency |
|---|---|
| 1 – 10 | 4 |
| 11 – 20 | 6 |
| 21 – 30 | 12 |
| 31 – 40 | 5 |
| 41 – 50 | 3 |
To estimate the mean we first find the midpoint of each class. For example, the midpoint of 1–10 is (1+10)/2 = 5.5. We then multiply each midpoint by the corresponding frequency and sum these products. Finally, we divide the total by the number of students, 30.
要估计平均数，我们先找出每组的组中点。例如，1–10的中点是(1+10)/2 = 5.5。然后将每个中点乘以相应的频数并求和。最后，将这个总和除以学生总数30。
Midpoints: 5.5, 15.5, 25.5, 35.5, 45.5. The fx totals are: 5.5×4 = 22; 15.5×6 = 93; 25.5×12 = 306; 35.5×5 = 177.5; 45.5×3 = 136.5. Sum = 22+93+306+177.5+136.5 = 735.
中点：5.5, 15.5, 25.5, 35.5, 45.5。fx乘积分别为：5.5×4=22；15.5×6=93；25.5×12=306；35.5×5=177.5；45.5×3=136.5。总和 = 22+93+306+177.5+136.5 = 735。
Estimated mean = 735 ÷ 30 = 24.5
估计平均数 = 735 ÷ 30 = 24.5
To find the median class we work out cumulative frequencies: 4, 10 (4+6), 22 (10+12), 27 (22+5), 30 (27+3). The median position is at (30+1)/2 = 15.5, i.e. between the 15th and 16th value. The cumulative frequency reaches 22 in the class 21–30, so this is the median class.
要找出中位数组，我们计算累积频数：4, 10, 22, 27, 30。中位数的位置在(30+1)/2 = 15.5，即介于第15和第16个数据之间。累积频数在21–30组达到22，因此该组即为中位数组。
Using the formula for an estimated median: L + [(n/2 – F) / f] × w, where L = lower class boundary (20.5), n = 30, F = cumulative frequency before the median class (10), f = frequency of median class (12), w = class width (10).
使用估计中位数的公式：L + [(n/2 – F) / f] × w，其中L = 组下限 (20.5)，n = 30，F = 中位数组之前的累积频数 (10)，f = 中位数组频数 (12)，w = 组距 (10)。
Median ≈ 20.5 + [(15 – 10) / 12] × 10 = 20.5 + (5/12)×10 ≈ 24.67
中位数估计值 ≈ 20.5 + [(15 – 10) / 12] × 10 = 20.5 + (5/12)×10 ≈ 24.67
2. Question 2: Calculating Standard Deviation and Range | 问题2：计算标准差与极差
Question: The following data set shows the number of hours six students spent on revision in one week: 5, 7, 8, 9, 11, 13. Find the range and the sample standard deviation.
题目：以下数据表示六名学生一周内用于复习的小时数：5, 7, 8, 9, 11, 13。求极差和样本标准差。
The range is the difference between the maximum and minimum values: 13 – 5 = 8 hours.
极差是最大值与最小值的差：13 – 5 = 8小时。
Next calculate the sample mean: x̄ = (5+7+8+9+11+13) / 6 = 53/6 ≈ 8.833. For a sample standard deviation we use n–1 in the denominator.
接下来计算样本平均数：x̄ = (5+7+8+9+11+13) / 6 = 53/6 ≈ 8.833。样本标准差使用分母 n–1。
s = √[ Σ(x – x̄)² / (n – 1) ]
s = √[ Σ(x – x̄)² / (n – 1) ]
Compute each deviation and its square: (5–8.833)² = 14.694, (7–8.833)² = 3.361, (8–8.833)² = 0.694, (9–8.833)² = 0.028, (11–8.833)² = 4.694, (13–8.833)² = 17.361. Sum of squares = 40.832 (approx).
计算各偏差及其平方：(5–8.833)² = 14.694, (7–8.833)² = 3.361, (8–8.833)² = 0.694, (9–8.833)² = 0.028, (11–8.833)² = 4.694, (13–8.833)² = 17.361。平方和 ≈ 40.832。
s = √(40.832 / 5) = √8.1664 ≈ 2.86
s = √(40.832 / 5) = √8.1664 ≈ 2.86
The standard deviation is approximately 2.86 hours, indicating that typical revision times deviate from the mean by around 2.86 hours.
标准差约为2.86小时，表明典型的复习时间与平均值相差约2.86小时。
3. Question 3: Quartiles, Interquartile Range and Box Plots | 问题3：四分位数、四分位距与盒须图
Question: The sorted ages of 11 participants in a survey are: 12, 15, 17, 18, 19, 21, 23, 25, 28, 30, 35. Determine Q1, Q2, Q3 and the interquartile range (IQR). Identify any outliers and sketch a box plot.
题目：某项调查中11名参与者的年龄排序为：12, 15, 17, 18, 19, 21, 23, 25, 28, 30, 35。确定Q1, Q2, Q3以及四分位距(IQR)。识别异常值并画出盒须图。
The median (Q2) is the 6th value: 21. To find Q1 we take the lower half of data excluding the median: 12, 15, 17, 18, 19. The median of these five values is the 3rd, so Q1 = 17.
中位数(Q2)是第6个值：21。为求Q1，取中位数之前的下半部分数据：12, 15, 17, 18, 19。这五个值的中位数是第3个，因此Q1 = 17。
The upper half is 23, 25, 28, 30, 35; its median (Q3) is 28. IQR = Q3 – Q1 = 28 – 17 = 11.
上半部分数据为23, 25, 28, 30, 35；其中位数(Q3) = 28。IQR = Q3 – Q1 = 28 – 17 = 11。
Outlier boundaries: lower fence = Q1 – 1.5×IQR = 17 – 16.5 = 0.5; upper fence = Q3 + 1.5×IQR = 28 + 16.5 = 44.5. All data points lie between 0.5 and 44.5, so there are no outliers.
异常值边界：下限 = Q1 – 1.5×IQR = 17 – 16.5 = 0.5；上限 = Q3 + 1.5×IQR = 28 + 16.5 = 44.5。所有数据点均位于0.5与44.5之间，因此没有异常值。
The box plot would show a box from 17 to 28 with a median line at 21, and whiskers extending to the minimum (12) and maximum (35). This gives a visual summary of the spread and central tendency.
盒须图将显示方框从17到28，中位线在21，触须延伸至最小值12和最大值35。这为我们提供了数据分散程度与集中趋势的可视化概览。
4. Question 4: Probability Using a Two-Way Table | 问题4：利用双向表计算概率
Question: The two-way table below shows the preferences of 70 students about whether they like mathematics, broken down by gender. A student is chosen at random. Calculate: (a) P(likes maths), (b) P(female | likes maths), (c) P(male and does not like maths).
题目：以下双向表展示了70名学生对数学的喜好情况，并按性别分类。随机选择一名学生，计算：(a) P(喜欢数学), (b) P(女性 | 喜欢数学), (c) P(男性且不喜欢数学)。
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