📚 Year 10 SQA Statistics: Unit Test Mock Exam Walkthrough | SQA 十年级统计：单元测试模拟卷解析
This mock exam walkthrough is designed to help Year 10 SQA Statistics students consolidate key concepts before the unit test. We tackle typical questions on data handling, averages, spread, correlation, probability and the normal distribution, with step-by-step solutions in both English and Chinese.
这份模拟卷解析专为 SQA 十年级统计单元测试复习设计，通过中英双语逐步讲解数据处理、平均数、离散程度、相关性、概率与正态分布等核心题型，帮助同学们查漏补缺。
1. Data Collection & Sampling Methods | 数据收集与抽样方法
A school council wants to find out students’ opinions on new cafeteria options. The student population is divided by year group (S1 to S6). Suggest a suitable sampling method and give one reason for your choice.
某学校学生会想了解学生对食堂新方案的意见，学生按年级分为 S1 至 S6。请建议一种合适的抽样方法并说明理由。
The most appropriate method is stratified random sampling. Students are already grouped into clear strata by year group, and opinions may differ across these strata. By taking a random sample from each year group proportional to its size, the survey results will be more representative of the whole school than a simple random sample.
最合适的方法是分层随机抽样。学生天然按年级分成不同层，各年级的看法可能不同。按照各层人数比例从每层随机抽取样本，可以使调查结果比简单随机抽样更具全校代表性。
Avoid a convenience sample (e.g. only asking friends) or a volunteer sample, as these introduce bias. The question asks for a suitable method and justification, which are commonly tested in SQA unit assessments.
避免使用便利抽样（例如只问朋友）或自愿抽样，这些方法会引入偏差。在 SQA 单元测试中，常要求提出并说明抽样方法，这正是考查的重点。
2. Mean, Median, Mode & Range | 平均数、中位数、众数与极差
The data set below shows the number of minutes 7 pupils spent on a statistics quiz: 12, 15, 14, 18, 21, 14, 16. Calculate the mean, median, mode and range.
下面一组数据记录了 7 名学生完成统计小测验所用的时间（分钟）：12, 15, 14, 18, 21, 14, 16。请计算平均数、中位数、众数和极差。
First, find the sum: 12 + 15 + 14 + 18 + 21 + 14 + 16 = 110. With 7 values, the mean = 110 ÷ 7 ≈ 15.7 minutes. To find the median, write the numbers in order: 12, 14, 14, 15, 16, 18, 21. The middle value (4th) is 15, so the median is 15 minutes. The mode is the most frequent value, which is 14 (it appears twice). The range = maximum − minimum = 21 − 12 = 9 minutes.
先求和：12 + 15 + 14 + 18 + 21 + 14 + 16 = 110。共 7 个数据，平均数 = 110 ÷ 7 ≈ 15.7 分钟。将数据从小到大排列：12, 14, 14, 15, 16, 18, 21，中间第 4 个是 15，因此中位数为 15 分钟。众数是出现次数最多的 14（出现 2 次）。极差 = 最大值 − 最小值 = 21 − 12 = 9 分钟。
In SQA questions you might be asked to state which average best describes the data. Here the mean and median are close, but with extreme values the median may be preferred. Remember to include units when the context involves measured quantities.
SQA 题目还可能问哪一个平均数最能代表数据。这组数据的平均数与中位数接近，但当数据含有极端值时中位数往往更合适。注意上下文带有计量单位时，答案一定要写出单位。
3. Quartiles & Box Plots | 四分位数与箱线图
Consider the data set: 3, 4, 5, 6, 8, 9, 11, 12, 15. Work out the median, lower quartile Q₁, upper quartile Q₃, the interquartile range (IQR) and describe how to draw a box plot.
考虑数据集：3, 4, 5, 6, 8, 9, 11, 12, 15。计算中位数、下四分位数 Q₁、上四分位数 Q₃ 和四分位距 IQR，并描述如何绘制箱线图。
There are 9 ordered values. Median position = (9 + 1) ÷ 2 = 5th value → median Q₂ = 8. For Q₁, consider the lower half: 3, 4, 5, 6. Its median is halfway between the 2nd and 3rd values: (4 + 5) ÷ 2 = 4.5, so Q₁ = 4.5. For Q₃, the upper half is 9, 11, 12, 15; (11 + 12) ÷ 2 = 11.5, so Q₃ = 11.5. IQR = Q₃ − Q₁ = 11.5 − 4.5 = 7. A box plot is drawn with a horizontal scale, a box from Q₁ to Q₃ and a vertical line inside at the median. Whiskers extend to the minimum (3) and maximum (15), provided no outliers.
共 9 个已排序数据。中位数位置 = (9 + 1) ÷ 2 = 第 5 个数 → 中位数 Q₂ = 8。下四分位数取下半部分 3, 4, 5, 6 的中位数，即第 2 和第 3 个数的均值 (4 + 5) ÷ 2 = 4.5，故 Q₁ = 4.5。上四分位数取上半部分 9, 11, 12, 15 的中位数 (11 + 12) ÷ 2 = 11.5，故 Q₃ = 11.5。IQR = Q₃ − Q₁ = 11.5 − 4.5 = 7。绘制箱线图时，先画一条数值轴，以 Q₁ 和 Q₃ 为边界画箱体，在箱内中位数位置画竖线，须线（晶须）延伸至最小值 3 和最大值 15（假设无离群值）。
SQA often defines a quartile using the (n+1) method. If asked to comment on spread, the IQR is a measure of variation that is not affected by outliers, unlike the range.
SQA 通常采用 (n+1) 方法计算四分位数。若要求描述数据的分散程度，可用 IQR，它与极差不同，不受离群值影响，更适合描述中间 50% 的分布。
4. Standard Deviation | 标准差
A small data set shows the number of hours five students revised: 5, 7, 8, 9, 11. Calculate the sample mean and the sample standard deviation. Use the formula s = √[ Σ(x − x̄)² / (n−1) ].
一个小型数据集记录了 5 名学生的复习小时数：5, 7, 8, 9, 11。计算样本平均数和样本标准差，使用公式 s = √[ Σ(x − x̄)² / (n−1) ]。
Mean x̄ = (5 + 7 + 8 + 9 + 11) ÷ 5 = 40 ÷ 5 = 8 hours. Subtract the mean from each value and square: (5−8)² = 9, (7−8)² = 1, (8−8)² = 0, (9−8)² = 1, (11−8)² = 9. Sum of squared deviations = 9 + 1 + 0 + 1 + 9 = 20. The sample size n = 5, so divide by (5−1) = 4, giving 20 ÷ 4 = 5. Finally, take the square root: s = √5 ≈ 2.236 hours (to 3 decimal places).
平均数 x̄ = (5 + 7 + 8 + 9 + 11) ÷ 5 = 40 ÷ 5 = 8 小时。将每个值减去平均数再平方：(5−8)² = 9，(7−8)² = 1，(8−8)² = 0，(9−8)² = 1，(11−8)² = 9。离差平方和 = 20。样本大小 n = 5，除以 n−1 = 4，得 20 ÷ 4 = 5。最后开平方：s = √5 ≈ 2.236 小时（保留三位小数）。
Always use the sample standard deviation (dividing by n−1) unless the data represents the whole population. The standard deviation has the same units as the data, so here it is hours.
除非数据代表整个总体，否则一律使用样本标准差（除以 n−1）。标准差的单位与原始数据相同，此处为小时。在比较两组数据时，较大的标准差意味着数据更分散。
5. Scatter Graphs & Correlation | 散点图与相关性
The table shows outdoor temperature (°C) and the number of ice creams sold at a kiosk:
| Temperature (°C) | 10 | 12 | 14 | 16 | 18 | 20 |
| Ice creams sold | 22 | 28 | 34 | 40 | 44 | 52 |
(a) Plot a scatter graph. (b) Describe the correlation. (c) A day has a temperature of 17 °C; use any trend to predict sales.
表格列出了室外温度（°C）和售出冰激凌数量：(a) 绘制散点图；(b) 描述相关性；(c) 某天温度为 17 °C，利用趋势预测销量。
Plotting the points (10,22), (12,28), … shows a clear upward pattern. As temperature increases, the number of ice creams sold also increases. This indicates a strong positive correlation. We can draw a line of best fit by eye that balances the points. At 17 °C, the line roughly suggests sales of about 46 ice creams (answers may vary slightly depending on the line).
在坐标图上描出点 (10,22)、(12,28)…… 可以看到明显的上升趋势。温度越高，冰激凌销量越大，呈现强正相关。可以用目测法画出一条最佳拟合线，使点均匀分布在直线两侧。在 17 °C 处读取直线，预计销量约为 46 个冰激凌（根据所画直线可能略有不同）。
You do not need to calculate the equation of the regression line at this stage unless the question specifically asks. The key is to recognise that correlation does not imply causation, but in this case there is a plausible causal link.
除非题目明确要求，否则现阶段无需计算回归直线方程。关键是理解相关性并不意味因果关系，但在本例中温度影响销量符合常识。
6. Basic Probability | 基础概率
A bag contains 4 red, 3 blue and 2 green marbles. A marble is taken at random. (a) Find P(red or blue). (b) The marble is NOT replaced and a second marble is drawn. Find the probability that both marbles are blue.
袋中有 4 个红球、3 个蓝球和 2 个绿球。随机取出一个球。(a) 求取出红色或蓝色的概率。(b) 取出的球不放回，再取一个，求两个都是蓝球的概率。
Total marbles = 4 + 3 + 2 = 9. (a) Favorable outcomes for red or blue = 4 + 3 = 7. So P(red or blue) = 7/9. (b) For both blue without replacement: P(first blue) = 3/9 = 1/3. After taking one blue, 2 blue remain out of 8, so P(second blue | first blue) = 2/8 = 1/4. Multiply: (1/3) × (1/4) = 1/12.
总球数 = 4 + 3 + 2 = 9。(a) 红色或蓝色的有利结果数 = 4 + 3 = 7，因此 P(红色或蓝色) = 7/9。(b) 不放回两个蓝球：P(第一个蓝) = 3/9 = 1/3。取走一个蓝球后，袋中剩 2 个蓝球、共 8 个球，故 P(第二个蓝 | 第一个蓝) = 2/8 = 1/4。相乘得 (1/3) × (1/4) = 1/12。
Always check whether events are ‘with replacement’ or ‘without replacement’. The probabilities change because the sample space shrinks. SQA often expects you to express probabilities as fractions in their simplest form.
务必区分“有放回”和“不放回”。不放回时样本空间缩小，概率随之改变。SQA 通常要求用最简分数表示概率结果。
7. Normal Distribution | 正态分布
Assessment scores in a large year group follow a normal distribution with mean μ = 60 and standard deviation σ = 12. Using the empirical rule (68–95–99.7%), estimate the percentage of students who scored more than 72.
某年级一次测验成绩服从正态分布，平均数 μ = 60，标准差 σ = 12。用经验法则（68–95–99.7%）估计成绩高于 72 分的学生约占百分之多少。
The value 72 is exactly one standard deviation above the mean: μ + σ = 60 + 12 = 72. In a normal distribution, about 68% of data lies within ±1σ of the mean. This leaves 100% − 68% = 32% in the two tails together. Since the normal curve is symmetric, the upper tail (above μ + σ) contains half of that 32%, i.e. 16%. Therefore, approximately 16% of students scored more than 72.
数值 72 恰好比平均数高一个标准差：μ + σ = 60 + 12 = 72。在正态分布中，约 68% 的数据落在平均数 ±1σ 范围内，两侧尾部总共占 32%。由于正态曲线对称，右侧尾部（高于 μ + σ）占 16%。因此大约有 16% 的学生成绩高于 72 分。
For other cut-offs, e.g. between 48 and 72, you would use the within-1σ band (68%). The empirical rule is essential for SQA; you may also be given a standard normal table, but the rule of thumb is often sufficient in unit tests.
若问 48 到 72 之间（μ ± σ），直接用 68% 即可。经验法则在 SQA 非常关键；有时会提供标准正态表，但单元测试中凭经验法则多半能直接作答。
8. Survey Question Design & Critique | 调查问题设计与评价
A student writes the following questionnaire item: ‘Don’t you agree that the new school lunches are tastier and healthier? Yes / No.’ Identify two problems with this question and rewrite it in an unbiased form.
一名学生设计了以下问卷题目：“难道你不认为新的学校午餐更美味、更健康吗？是 / 否。”请指出该题目的两个问题，并以无偏的形式重写。
Problem 1: The question is a leading question — it suggests a preferred answer by starting ‘Don’t you agree…’ and using positive adjectives like ‘tastier and healthier’. Problem 2: It is a double-barrelled question, asking about taste and healthiness in a single item, so a respondent who finds the food tasty but unhealthy cannot answer properly. A better version: ‘How would you rate the new school lunches?’ with a 5-point scale from ‘Very good’ to ‘Very poor’.
问题一：该题是诱导性问题，以“难道你不认为……”开头并使用“更美味、更健康”等褒义词语，暗示了期望的答案。问题二：它是一道双重问题，同时询问口感和健康两个维度，若受访者觉得好吃但不健康，就无法准确作答。改进版可写为：“您如何评价新的学校午餐？”并配以五级量表，从“非常好”到“非常差”。
In SQA, you may be asked to spot bias in data collection. Always look out for leading language, double questions, overlapping response options, and whether the sample is representative. Providing a rewritten question shows understanding of fair survey design.
SQA 常要求识别数据收集中的偏见。关注诱导性语言、多重问题、选项重叠以及样本代表性是关键。给出重写题目能体现对无偏调查设计的理解。
Published by TutorHao | Statistics Revision Series | aleveler.com
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