IGCSE WJEC Statistics: Unit Test Mock Paper Walkthrough | IGCSE WJEC 统计：单元测试模拟卷解析

📚 IGCSE WJEC Statistics: Unit Test Mock Paper Walkthrough | IGCSE WJEC 统计：单元测试模拟卷解析

This article provides a complete walkthrough of a mock unit test for the IGCSE WJEC Statistics course, covering core topics such as data representation, measures of central tendency, probability, and interpretation of statistical diagrams. Each question is broken down step‑by‑step to reinforce exam technique and conceptual understanding.

本文提供一份针对 IGCSE WJEC 统计课程的单元测试模拟卷的完整解析，涵盖数据表示、集中趋势度量、概率以及统计图表的解读等核心主题。每道题都逐步分解，以强化应试技巧和概念理解。

1. Question 1: Mean from a Frequency Table | 第1题：根据频数表求平均数

The table shows the number of books read by 30 students in a month. Calculate the mean number of books read.

下表显示了30名学生在一个月内阅读的书籍数量。计算所读书籍数量的平均数。

Number of books (x)	Frequency (f)
0	4
1	8
2	10
3	6
4	2

To find the mean, multiply each number of books (x) by its frequency (f), sum these products, then divide by the total frequency (Σf = 30).

要求平均数，先将每类书籍数量 (x) 乘以其频数 (f)，求出这些乘积之和，再除以总频数 (Σf = 30)。

Σfx = 0×4 + 1×8 + 2×10 + 3×6 + 4×2 = 0+8+20+18+8 = 54

Mean = Σfx / Σf = 54 / 30 = 1.8 books

The mean number of books read is 1.8. Since the data are discrete, you can state the answer as 1.8 without rounding to a whole number.

平均阅读量为 1.8 本书。由于数据是离散的，答案可以保留为 1.8，不必四舍五入到整数。

2. Question 2: Median and Interquartile Range | 第2题：中位数与四分位距

Using the same frequency table, find the median and the interquartile range (IQR) of the books read.

使用同一频数表，求所读书籍数量的中位数和四分位距 (IQR)。

The median position is (n+1)/2 = 31/2 = 15.5th value. Cumulatively, the 15th and 16th values both fall in the x=2 category, so the median is 2. The lower quartile Q₁ is at position (n+1)/4 = 7.75th value, which lies in x=1; Q₁ = 1. The upper quartile Q₃ is at position 3(n+1)/4 = 23.25th value, which lies in x=3; Q₃ = 3.

中位数的位置为 (n+1)/2 = 31/2 = 第15.5个值。从累积频数看，第15和第16个值都在 x=2 这一组，因此中位数为 2。下四分位数 Q₁ 的位置为 (n+1)/4 = 第7.75个值，落在 x=1 组，故 Q₁ = 1。上四分位数 Q₃ 的位置为 3(n+1)/4 = 第23.25个值，落在 x=3 组，故 Q₃ = 3。

Thus the interquartile range (IQR) = Q₃ − Q₁ = 3 − 1 = 2 books.

因此四分位距 IQR = Q₃ − Q₁ = 3 − 1 = 2 本书。

The median tells us that half the students read at most 2 books; the IQR of 2 indicates the spread of the middle 50% of the students.

中位数告诉我们，一半学生最多读了 2 本书；IQR 为 2 表明中间 50% 学生的阅读量分布范围。

3. Question 3: Bar Chart Construction | 第3题：绘制条形图

Draw a bar chart to represent the data from the frequency table. Make sure to label axes clearly and use an appropriate scale.

绘制条形图来表示频数表中的数据。确保清晰地标注坐标轴并使用合适的比例。

The horizontal axis represents the number of books (0 to 4), and the vertical axis represents frequency (0 to 10). Each bar is drawn with equal widths, and the height corresponds to the frequency. Unlike a histogram, there are gaps between the bars because the data are discrete.

横轴表示书籍数量（0 至 4），纵轴表示频数（0 至 10）。每个条形的宽度相等，高度对应频数。与直方图不同，条形之间留有空隙，因为这些数据是离散的。

In the exam, use a ruler and a sharp pencil. Label the axes as ‘Number of books’ and ‘Frequency’. The title could be ‘Books read by students in a month’.

在考试中，使用直尺和尖铅笔。将坐标轴分别标为“书籍数量”和“频数”。标题可以写“学生一个月内所读书籍”。

For checking, make sure bar for 2 books reaches exactly 10 on the frequency axis, and bar for 4 books reaches 2.

检查时，确保 2 本书的条形正好达到纵轴 10 的高度，4 本书的条形达到 2。

4. Question 4: Pie Chart Calculation | 第4题：饼图计算

The data below show how a group of 60 students travel to school. Represent the data using a pie chart by calculating the angle for each sector.

以下数据展示了一组 60 名学生的上学方式。请通过计算每个扇形的角度，用饼图表示数据。

Method of travel	Frequency
Walk	18
Bus	24
Car	12
Bicycle	6

Each sector angle = (frequency / total) × 360°. Total students = 60.

每个扇形的角度 = (频数 / 总数) × 360°。学生总数为 60。

Walk: (18/60) × 360° = 108°

步行： (18/60) × 360° = 108°
Bus: (24/60) × 360° = 144°

公交车： (24/60) × 360° = 144°
Car: (12/60) × 360° = 72°

小汽车： (12/60) × 360° = 72°
Bicycle: (6/60) × 360° = 36°

自行车： (6/60) × 360° = 36°

Check: sum of angles = 108° + 144° + 72° + 36° = 360°, which confirms the calculation. Draw the circle, measure angles carefully with a protractor, label each sector or provide a key.

验证：角度总和 = 108° + 144° + 72° + 36° = 360°，确认计算无误。画出圆形，用量角器小心测量角度，标注每个扇形或提供图例。

5. Question 5: Probability Basics | 第5题：基础概率

A bag contains 5 red, 3 blue and 2 green marbles. One marble is chosen at random. Find the probability that it is (a) red, (b) not blue, (c) either red or green.

一个袋子里有 5 个红球、3 个蓝球和 2 个绿球。随机取出一个球。求该球是 (a) 红色，(b) 不是蓝色，(c) 红色或绿色的概率。

Total marbles = 5 + 3 + 2 = 10.

球的总数 = 5 + 3 + 2 = 10。

(a) P(red) = 5/10 = 1/2 or 0.5

(a) P(红色) = 5/10 = 1/2 或 0.5
(b) ‘Not blue’ means red or green: 5+2 = 7 favourable outcomes. P(not blue) = 7/10 = 0.7

(b) “不是蓝色”意味着红色或绿色：5+2 = 7 个有利结果。P(不是蓝色) = 7/10 = 0.7
(c) P(red or green) is also 7/10, because red and green are mutually exclusive (you cannot pick a marble that is both red and green). Simply add their probabilities: 5/10 + 2/10 = 7/10.

(c) P(红色或绿色) 也等于 7/10，因为红色和绿色互斥（不可能同时是红色又是绿色）。只需将它们的概率相加：5/10 + 2/10 = 7/10。

Always express probabilities as fractions, decimals or percentages; simplify fractions where possible.

概率总是以分数、小数或百分数表示；可能的话对分数进行约分。

6. Question 6: Cumulative Frequency Graph | 第6题：累积频数图

The table below shows the heights of 40 plants. Construct a cumulative frequency table and draw a cumulative frequency graph (ogive). Use your graph to estimate the median height.

下表显示了 40 株植物的高度。制作累积频数表并绘制累积频数图（肩形图）。利用图估计高度中位数。

Height (cm)	Frequency
0 ≤ h < 10	2
10 ≤ h < 20	8
20 ≤ h < 30	14
30 ≤ h < 40	10
40 ≤ h < 50	6

First, add a cumulative frequency column. Cumulative frequency after each class: 2, 2+8=10, 10+14=24, 24+10=34, 34+6=40. Plot points at the upper class boundaries (10,2), (20,10), (30,24), (40,34), (50,40). Join the points with a smooth curve. The median is the height corresponding to half the total frequency: 40/2 = 20th value. From the graph, draw a horizontal line from cumulative frequency 20 to the curve, then drop vertically to the axis – approximate median is 26 cm.

首先，添加累积频数列。各组的累积频数：2, 2+8=10, 10+14=24, 24+10=34, 34+6=40。在组上界处描点：(10,2), (20,10), (30,24), (40,34), (50,40)。用光滑曲线连接各点。中位数是总频数一半所对应的高度：40/2 = 第20个值。在图上，从累积频数 20 画水平线与曲线相交，再垂直下落到横轴 – 估计中位数约为 26 cm。

In the exam, be precise with plotting and drawing lines. Always label axes: ‘Height (cm)’ and ‘Cumulative frequency’.

在考试中，描点和画线要精确。始终标注坐标轴：“高度 (cm)”和“累积频数”。

7. Question 7: Box‑and‑Whisker Plot Interpretation | 第7题：箱线图解读

Given the five‑number summary from the height data: minimum = 3 cm, Q₁ = 18 cm, median = 26 cm, Q₃ = 35 cm, maximum = 48 cm, draw a box‑and‑whisker plot and comment on the skewness.

给定高度数据集的五数汇总：最小值 = 3 cm，Q₁ = 18 cm，中位数 = 26 cm，Q₃ = 35 cm，最大值 = 48 cm，绘制箱线图并评论其偏态。

The box is drawn from Q₁ to Q₃, with a vertical line at the median. The whiskers extend to the minimum and maximum unless there are outliers. The distance Q₃ − median = 9 cm, median − Q₁ = 8 cm, which are roughly equal, but the right whisker (48 − 35 = 13 cm) is longer than the left whisker (18 − 3 = 15 cm) – wait: left whisker is 15, right is 13, so they are fairly symmetric. Actually the box itself is slightly skewed to the right because the median is closer to Q₁? Let’s recalc: box length = 17, left part of box = 8, right part = 9. Very slight positive skew. Overall the distribution is quite symmetric. A comment: ‘The box plot shows a slight positive skew, as the median is closer to the lower quartile than to the upper quartile, and the right whisker is slightly longer.’

箱体从 Q₁ 到 Q₃ 绘制，中位数处画一条竖线。须线延伸到最小值和最大值，除非有异常值。距离 Q₃ − 中位数 = 9 cm，中位数 − Q₁ = 8 cm，两者大致相等，但右须 (48 − 35 = 13 cm) 相比左须 (18 − 3 = 15 cm) 稍短。实际上箱体本身略呈右偏，因为中位数更靠近 Q₁。整体分布相当对称。评论：“箱线图显示微弱的正偏态，因为中位数离下四分位数比离上四分位数更近，且右须稍长。”

Always use a scale and label the axis with the variable name. Box plots are excellent for comparing distributions.

始终使用比例尺并标上变量名。箱线图非常适合比较分布。

8. Question 8: Data Collection Methods | 第8题：数据收集方法

A school wants to investigate students’ opinions on a new canteen menu. Describe a suitable data collection method, identify the type of data collected, and discuss one advantage and one disadvantage of the method.

某所学校想调查学生对新食堂菜单的意见。描述一种合适的数据收集方法，指出所收集数据的类型，并讨论该方法的一个优点和一个缺点。

A suitable method is a questionnaire with closed questions (e.g., rating scale 1‑5). The data collected are primary and quantitative (can also include qualitative if open‑ended questions are used). Advantage: quick to administer to many students; responses can be easily summarised. Disadvantage: response bias may occur if students don’t answer truthfully, or low response rate if not compulsory.

一种合适的方法是使用封闭式问题的问卷（例如 1‑5 评分量表）。收集的数据是原始数据，且为定量数据（如果使用开放式问题也可包含定性数据）。优点：能快速对大量学生实施；回答易于汇总。缺点：如果学生不真实回答，可能出现回答偏差；若非强制性，可能回收率低。

Alternative methods could be an interview (time‑consuming but detailed) or observation. For this scenario, a questionnaire is practical and efficient.

替代方法可以是访谈（耗时但详尽）或观察。在此情境下，问卷既实用又高效。

Published by TutorHao | Statistics Revision Series | aleveler.com

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IGCSE WJEC Statistics: Unit Test Mock Paper Walkthrough | IGCSE WJEC 统计：单元测试模拟卷解析