KS3 WJEC Statistics: Unit Test Mock Paper Analysis | KS3 WJEC 统计：单元测试模拟卷解析

📚 KS3 WJEC Statistics: Unit Test Mock Paper Analysis | KS3 WJEC 统计：单元测试模拟卷解析

This article provides a detailed walkthrough of a KS3 WJEC Statistics unit test mock paper. By working through each question, you can deepen your understanding of key topics such as bar charts, averages, pie charts, probability, scatter graphs, and data collection. Use this analysis to identify common pitfalls and master essential statistical skills.

本文详细解析了一份 KS3 WJEC 统计单元测试模拟卷。通过逐题演练，你可以加深对柱状图、平均数、饼图、概率、散点图以及数据收集等关键主题的理解。请利用这份解析发现常见错误，掌握核心统计技能。

1. Overview of the Mock Paper | 模拟卷概览

The mock paper consists of six questions designed to mirror the WJEC KS3 Statistics unit test. It covers data representation, measures of central tendency and spread, probability, and data handling. The total mark is 50, and the suggested time is 45 minutes. Each question targets different assessment objectives, helping you build confidence across the entire syllabus.

这份模拟卷包含六道题目，仿照 WJEC KS3 统计单元测试设计。内容涵盖数据表示、集中趋势和离散程度、概率以及数据处理。总分50分，建议用时45分钟。每道题针对不同的评估目标，帮助你在整个教学大纲中建立信心。

The questions are scaffolded: earlier ones test basic skills such as reading bar charts and calculating averages, while later ones assess interpretation, probability reasoning, and data collection design. Mark allocations are clearly shown alongside each sub-question.

题目编排由浅入深：前面测试读柱状图、算平均数等基础技能，后面评估解读、概率推理和收集数据的设计。每道小题旁都清晰标有分值。

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2. Question 1: Bar Charts & Frequency Tables | 题目1：柱状图与频数表

The first question presents a bar chart showing the number of pets owned by students in Ms. Lee’s class. The frequencies are: 0 pets – 5 students, 1 pet – 8 students, 2 pets – 6 students, 3 pets – 2 students, and 4 pets – 1 student. All parts are based on this data.

第一题展示了一张柱状图，显示李老师班上学生拥有的宠物数量。频数分布为：0只宠物—5人，1只宠物—8人，2只宠物—6人，3只宠物—2人，4只宠物—1人。所有小问都基于此数据。

a) To find the total number of students in the class, add all the frequencies: 5 + 8 + 6 + 2 + 1 = 22. There are 22 students.

a) 求班级总人数，将所有频数相加：5 + 8 + 6 + 2 + 1 = 22。共有22名学生。

b) The mode is the category with the highest frequency. The highest frequency is 8, which corresponds to 1 pet. So the mode is 1 pet.

b) 众数是出现频数最高的类别。最高频数为8，对应1只宠物。因此众数是1只宠物。

c) A frequency table must have two columns: “Number of pets” and “Frequency”. List the categories in order (0, 1, 2, 3, 4) with their corresponding frequencies (5, 8, 6, 2, 1). Always include a title and check that the frequencies sum to 22.

c) 频数表必须包含两列：“宠物数量”和“频数”。按顺序列出类别（0, 1, 2, 3, 4）及其频数（5, 8, 6, 2, 1）。务必加上标题，并检查频数总和是否为22。

d) How many students have more than 2 pets? Add the frequencies for 3 pets and 4 pets: 2 + 1 = 3 students.

d) 有多少名学生拥有的宠物数量超过2只？将3只和4只宠物的频数相加：2 + 1 = 3名学生。

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3. Question 2: Averages and Range | 题目2：平均数、中位数、众数和极差

Q2 provides a dataset: 12, 15, 12, 18, 20, 12, 10, 22. You need to calculate the mean, median, mode, and range. Working must be shown clearly.

第二题给出数据集：12, 15, 12, 18, 20, 12, 10, 22。需要计算平均数、中位数、众数和极差，并清晰展示计算过程。

a) For the mean, add all values: 12 + 15 + 12 + 18 + 20 + 12 + 10 + 22 = 121. There are 8 values, so mean = 121 ÷ 8 = 15.125.

a) 求平均数，先把所有数值相加：12 + 15 + 12 + 18 + 20 + 12 + 10 + 22 = 121。数据个数为8，所以平均数 = 121 ÷ 8 = 15.125。

b) To find the median, first put the numbers in order: 10, 12, 12, 12, 15, 18, 20, 22. Since there are 8 values (an even number), the median is the average of the 4th and 5th values: (12 + 15) ÷ 2 = 13.5.

b) 求中位数，先将数字排序：10, 12, 12, 12, 15, 18, 20, 22。因为有8个数值（偶数个），中位数是第4和第5个值的平均数：(12 + 15) ÷ 2 = 13.5。

c) The mode is the value that appears most often. Here, 12 appears three times, more than any other number. So the mode is 12.

c) 众数是出现次数最多的值。此处12出现三次，多于任何其他数值，所以众数是12。

d) The range is the difference between the largest and smallest values: maximum = 22, minimum = 10, range = 22 − 10 = 12.

d) 极差是最大值与最小值之间的差值：最大值 = 22，最小值 = 10，极差 = 22 − 10 = 12。

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4. Question 3: Interpreting Pie Charts | 题目3：饼图解读

Question 3 gives a pie chart showing how students travel to school: 45% walk, 30% take the bus, 15% travel by car, and 10% cycle. The total number of students is 300.

第三题给出一个饼图，显示学生上学交通方式的比例：45%步行，30%坐公交车，15%乘小汽车，10%骑自行车。学生总数为300人。

a) Calculate the number of students for each mode: Walk = 45% of 300 = 0.45 × 300 = 135; Bus = 0.30 × 300 = 90; Car = 0.15 × 300 = 45; Cycle = 0.10 × 300 = 30. Always check that these numbers sum to 300.

a) 计算每种方式的学生人数：步行 = 300的45% = 0.45 × 300 = 135；公交车 = 0.30 × 300 = 90；小汽车 = 0.15 × 300 = 45；自行车 = 0.10 × 300 = 30。务必检查人数之和为300。

b) Identify one mode of transport that has exactly twice as many students as another. Bus (90) is twice Car (45). You could also compare Walk and Car, but Walk is three times Car, not twice.

b) 指出一种交通方式的人数恰好是另一种的两倍。公交车（90人）是小汽车（45人）的两倍。也可以比较步行和小汽车，但步行是小汽车的三倍。

c) If you were to draw the pie chart, the angle for each sector is: Walk = 0.45 × 360° = 162°; Bus = 108°; Car = 54°; Cycle = 36°. Use a protractor to measure these angles accurately.

c) 如果要绘制饼图，每个扇区的角度为：步行 = 0.45 × 360° = 162°；公交车 = 108°；小汽车 = 54°；自行车 = 36°。使用量角器精确测量这些角度。

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5. Question 4: Probability Scale & Simple Probability | 题目4：概率尺度与简单概率

A bag contains 3 red balls, 5 blue balls and 2 green balls. A ball is taken at random. All outcomes are equally likely. This question tests probability as a fraction and placing it on a probability scale.

一个袋子里有3个红球、5个蓝球和2个绿球。随机抽取一个球，所有结果等可能。此题考查用分数表示概率并将其标注在概率尺度上。

a) Find the probability of picking a red ball. Total balls = 3 + 5 + 2 = 10. P(red) = number of red balls / total number of balls = 3/10. On a probability scale from 0 to 1, 3/10 is between ‘unlikely’ and ‘even chance’.

a) 求抽到红球的概率。总球数 = 3 + 5 + 2 = 10。P(红) = 红球数量 / 总球数 = 3/10。在0到1的概率尺度上，3/10介于“不太可能”和“等可能性”之间。

b) What is the probability that the ball is not blue? Not blue means red or green. So favourable outcomes = 3 + 2 = 5. P(not blue) = 5/10 = 1/2. This sits exactly at ‘even chance’ on the scale.

b) 抽到的球不是蓝色的概率是多少？不是蓝色意味着是红色或绿色。所以有利结果数 = 3 + 2 = 5。P(不是蓝) = 5/10 = 1/2。这在概率尺度上正好落在“等可能性”位置。

c) If the bag had 6 blue balls and 4 red balls, the probability of picking blue would be 6/10 or 3/5. This is more likely than not. Being able to compare probabilities and relate them to everyday language is a key skill.

c) 如果袋中有6个蓝球和4个红球，抽到蓝球的概率为6/10即3/5。这比“不抽到”的可能性更大。能够比较概率并将其与日常语言联系起来是一项关键技能。

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6. Question 5: Scatter Graphs & Correlation | 题目5：散点图与相关性

The scatter graph plots hours of revision against test scores. Six points are plotted: (2,30), (3,45), (4,55), (5,65), (6,70), (8,80). You must describe the correlation, sketch a line of best fit, and use it to estimate.

散点图以复习小时数为横轴、考试成绩为纵轴绘制。六个数据点为：(2,30), (3,45), (4,55), (5,65), (6,70), (8,80)。你需要描述相关性，画出最佳拟合线，并用它进行估计。

a) Describe the relationship. As the number of revision hours increases, test scores tend to increase. The points rise from bottom left to top right, showing positive correlation. It is fairly strong because the points lie close to a straight line.

a) 描述变量关系。随着复习时间增加，考试成绩呈上升趋势。数据

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