Year 7 WJEC Statistics: Unit Test Mock Paper Analysis | Year 7 WJEC 统计：单元测试模拟卷解析

📚 Year 7 WJEC Statistics: Unit Test Mock Paper Analysis | Year 7 WJEC 统计：单元测试模拟卷解析

Welcome to this detailed walkthrough of a Year 7 WJEC Statistics mock paper. In this analysis, you will find eight carefully designed questions that cover the key topics from your unit test, including averages (mean, median, mode), range, frequency tables, bar charts, pie charts, and basic probability. Each question is broken down step by step, with clear explanations to help you understand the reasoning and avoid common mistakes.

欢迎阅读这份 Year 7 WJEC 统计单元测试模拟卷的详细解析。本解析包含了八道精心设计的题目，涵盖了单元测试中的核心考点，包括平均数、中位数、众数、极差、频率表、条形图、饼图和基础概率。每道题都进行了逐步拆解，并配有清晰的解释，帮助你理解解题思路，避免常见错误。

1. Mock Test Structure | 模拟卷结构

The mock test consists of eight questions, each targeting a specific skill from the WJEC Year 7 Statistics curriculum. The questions progress from straightforward calculations to those requiring interpretation and reasoning. You are advised to spend about 40 minutes on this set of questions, mimicking a real test environment. The topics covered are: calculating the mean, finding the median, identifying the mode, working out the range, choosing an appropriate average when outliers are present, organising data into a frequency table and drawing a bar chart, interpreting information from a pie chart, and solving a simple probability problem.

本模拟卷包含八道题目，每道题都针对WJEC七年级统计课程中的一项特定技能。题目从直接的运算逐步过渡到需要解释和推理的题型。建议你在40分钟内完成这组问题，以模拟真实考试环境。涉及的考点包括：计算平均数、求中位数、识别众数、计算极差、在存在异常值时选择合适的平均数、将数据整理成频率表并绘制条形图、解读饼图中的信息以及解决一个简单的概率问题。

2. Question 1: Calculating the Mean | 问题1：计算平均数

Question: The ages of five students are 10, 12, 11, 10, 13. Calculate the mean age.

题目：五名学生的年龄分别是10, 12, 11, 10, 13。计算平均年龄。

To find the mean, first add all the values together. The sum of the ages is 10 + 12 + 11 + 10 + 13 = 56. Next, count how many values there are; here n = 5. Finally, divide the sum by the number of values.

要求平均数，首先将所有数值相加。年龄的总和为 10 + 12 + 11 + 10 + 13 = 56。然后，数出一共有多少个数值；这里 n = 5。最后，用总和除以数值的个数。

Mean = 56 ÷ 5 = 11.2

The mean age of the five students is 11.2 years. Always remember to include the unit (years) in your final answer when the question refers to a measured quantity. If the result is a decimal, it is perfectly acceptable to leave it as 11.2 rather than rounding unnecessarily.

五名学生的平均年龄是11.2岁。当题目涉及带有单位的量时，请一定记得在最终答案中写上单位（岁）。如果结果是小数，直接保留为11.2即可，不必进行不必要的四舍五入。

3. Question 2: Finding the Median | 问题2：求中位数

Question: Find the median of the following test scores: 7, 9, 5, 8, 6, 10, 4.

题目：求出以下测验成绩的中位数：7, 9, 5, 8, 6, 10, 4。

The median is the middle value when the data is arranged in order. Begin by listing the scores from smallest to largest: 4, 5, 6, 7, 8, 9, 10. Since there are 7 numbers (an odd amount), the median is simply the (7+1)/2 = 4th value. Count along: the 4th value is 7.

中位数是将数据按顺序排列后位于中间位置的那个值。首先将这些成绩从小到大列出：4, 5, 6, 7, 8, 9, 10。因为有7个数值（奇数个），中位数就是第 (7+1)/2 = 4 个数值。数到第四个值，即为7。

If the dataset had an even number of values, we would take the average of the two middle numbers. But in this case, the median test score is 7. A common mistake is forgetting to order the values first, which might lead to picking ‘8’ incorrectly.

如果数据集包含偶数个数值，我们则需要取中间两个数的平均数。但在本题中，测验成绩的中位数就是7。一个常见的错误是忘记先将数据排序，这可能会使人错误地选中“8”。

4. Question 3: Identifying the Mode | 问题3：找出众数

Question: A student records the colours of cars passing the school gate: red, blue, red, green, blue, red, white, blue, red. What is the mode?

题目：一名学生记录了经过校门口的汽车颜色：红、蓝、红、绿、蓝、红、白、蓝、红。众数是什么？

The mode is the value that appears most frequently. To find it, tally the frequency of each colour: Red appears 4 times, Blue appears 3 times, Green appears once, and White appears once. Because red has the highest frequency (4), it is the mode.

众数是出现频率最高的那个值。要找到众数，可以统计每种颜色出现的次数：红色出现4次，蓝色出现3次，绿色出现1次，白色出现1次。由于红色的出现频率最高（4次），因此众数是红色。

Note that ‘red’ is a categorical variable, so the answer is the label itself, not the number 4. If two colours had tied for the highest frequency, the dataset would have two modes (bimodal). In WJEC Year 7, you may also be asked to find the mode from a frequency table.

请注意，“红色”属于分类变量，因此答案就是该标签本身，而不是数字4。如果有两种颜色的出现频率并列最高，那么数据集就会有两个众数（双众数）。在WJEC七年级的课程中，你也可能需要从频率表中找出众数。

5. Question 4: Working Out the Range | 问题4：计算极差

Question: The monthly rainfall (in mm) in a town over six months was: 45, 52, 38, 61, 47, 55. Calculate the range.

题目：某城镇六个月的月降雨量（单位：毫米）分别为：45, 52, 38, 61, 47, 55。计算极差。

The range measures the spread of the data and is found by subtracting the smallest value from the largest value. First, identify the maximum rainfall, which is 61 mm, and the minimum, which is 38 mm. Then subtract: Range = 61 – 38 = 23 mm.

极差用于衡量数据的分散程度，计算方法是用最大值减去最小值。首先找出最大降雨量61毫米和最小降雨量38毫米。然后相减：极差 = 61 – 38 = 23 毫米。

The range tells us that the rainfall varied by 23 mm across those six months. A larger range indicates more variability. Remember to include the unit ‘mm’ in your final answer, as the original data had a unit. It is also good practice to double-check that you have correctly spotted the highest and lowest values, especially if the data is not sorted.

极差告诉我们，在这六个月中，降雨量相差了23毫米。极差越大，说明数据的波动性越大。记得在最终答案中加上单位“毫米”，因为原始数据带有单位。此外，最好再检查一遍自己是否准确找出了最大值和最小值，尤其是在数据未排序的情况下。

6. Question 5: Choosing the Right Average (Outliers) | 问题5：选择合适的平均数（异常值）

Question: The salaries of employees in a small company are: £18,000, £19,000, £20,000, £21,000, £22,000, and £150,000 (the CEO). Explain which measure of central tendency (mean or median) best represents the typical salary, and give a reason.

题目：一家小公司员工的年薪如下：£18,000, £19,000, £20,000, £21,000, £22,000 和 £150,000（CEO）。解释哪种集中趋势指标（平均数还是中位数）最能代表典型的薪酬水平，并说明理由。

First, let us calculate the mean. The sum is £18,000 + £19,000 + £20,000 + £21,000 + £22,000 + £150,000 = £250,000. Divided by 6 employees, the mean salary is £41,667. This value is heavily distorted by the CEO’s salary, which is an outlier. Most employees earn much less than £41,667.

首先来计算平均数。总和为 £18,000 + £19,000 + £20,000 + £21,000 + £22,000 + £150,000 = £250,000。除以6名员工，平均薪酬为 £41,667。这个数值因CEO的高薪（异常值）而被严重拉高，大多数员工的收入远远低于 £41,667。

The median, on the other hand, is the middle value of the ordered list: £20,000 and £21,000 are the two middle values, so the median is £20,500. This figure better reflects the earnings of a typical employee because the outlier does not affect the median. Therefore, the median is the more representative average.

相比之下，中位数是排序后列表的中间值：中间两个值是£20,000 和 £21,000，因此中位数为 £20,500。这个数值更能反映一名普通员工的收入水平，因为异常值不会影响中位数。因此，中位数是更具代表性的平均数。

In WJEC exams, you must always explain your choice by referring to the effect of the outlier. A classic sentence frame is: ‘The median is a better average because it is not pulled up/down by the extreme value.’

在WJEC考试中，你必须通过指出异常值的影响来解释你的选择。一个经典的答题句式是：“中位数更适合作为平均数，因为它不会被极值拉高或拉低。”

7. Question 6: Frequency Table and Bar Chart | 问题6：频率表和条形图

Question: The following list shows the number of books read by 20 students in a month: 3, 2, 4, 3, 5, 2, 1, 3, 4, 2, 3, 5, 4, 2, 3, 1, 4, 3, 2, 5. (a) Complete the frequency table. (b) Draw a fully labelled bar chart.

题目：以下列表显示了20名学生在一个月内阅读的书籍数量：3, 2, 4, 3, 5, 2, 1, 3, 4, 2, 3, 5, 4, 2, 3, 1, 4, 3, 2, 5。(a) 完成频率表。(b) 绘制一张带有完整标签的条形图。

To complete the frequency table, tally each occurrence. The distinct values are 1, 2, 3, 4, and 5. Count carefully:

Books = 1: appears 2 times
Books = 2: appears 5 times
Books = 3: appears 6 times
Books = 4: appears 4 times
Books = 5: appears 3 times

要完成频率表，需要统计每个数值出现的次数。这里不同的取值为1, 2, 3, 4, 5。仔细数：

书籍数量 = 1：出现2次
书籍数量 = 2：出现5次
书籍数量 = 3：出现6次
书籍数量 = 4：出现4次
书籍数量 = 5：出现3次

The completed frequency table looks like this:

完成的频率表如下所示：

Number of books	Frequency
1	2
2	5
3	6
4	4
5	3

For the bar chart, draw a horizontal axis labelled ‘Number of books’ with categories 1 to 5. The vertical axis is ‘Frequency’ and should be scaled from 0 to 6, with regular intervals. Draw bars of equal width, with heights corresponding to the frequencies: bar for 1 reaches 2, bar for 2 reaches 5, bar for 3 reaches 6, bar for 4 reaches 4, bar for 5 reaches 3. Leave equal gaps between the bars and give the chart a clear title, such as ‘Bar chart showing number of books read by students’.

绘制条形图时，横轴标为“书籍数量”，类别为1到5。纵轴为“频率”，刻度从0到6，间隔均匀。绘制等宽的条形，其高度对应频率：1的条形高为2，2的条形高为5，3的条形高为6，4的条形高为4，5的条形高为3。条形之间留出相等的间隙，并给图表加上清晰的标题，如“显示学生阅读书籍数量的条形图”。

8. Question 7: Interpreting a Pie Chart | 问题7：解读饼图

Question: A pie chart shows how 360 pupils travel to school. The ‘Bus’ sector has an angle of 120°. How many pupils travel by bus?

题目：一个饼图显示了360名学生的上学交通方式。“公交车”扇区的角度为120°。有多少名学生乘坐公交车上学？

In a pie chart, the full circle (360°) represents the total frequency, which is 360 pupils. Therefore, each degree represents 360 ÷ 360 = 1 pupil. For a sector of 120°, the number of pupils is simply 120 × 1 = 120. Alternatively, use the proportional method: (120/360) × 360 = 120.

在饼图中，整个圆（360°）代表总频数，即360名学生。因此，每1度代表 360 ÷ 360 = 1 名学生。对于120°的扇区，学生人数就是 120 × 1 = 120。也可以使用比例法计算：(120/360) × 360 = 120。

Many students forget that the total angle is always 360° and mistakenly try to use 100%. Remember, pie charts are based on angles, not percentages, unless stated otherwise. If the total frequency were different, say 180 pupils, then the multiplier would be (180/360) = 0.5 pupils per degree. In this case, the numbers conveniently matched so 1° = 1 pupil, but it is crucial to show the working.

许多学生会忘记总角度始终是360°，而错误地尝试使用百分比。请记住，除非另有说明，饼图是基于角度而非百分比绘制的。如果总频数不同，例如是180名学生，那么每度对应的学生数就变为 180/360 = 0.5。在本题中，数字恰好对应使得 1° = 1 名学生，但展示计算过程仍然至关重要。

9. Question 8: Basic Probability | 问题8：基础概率

Question: A bag contains 3 red counters, 5 blue counters, and 2 green counters. One counter is chosen at random. What is the probability that the counter is (a) blue, (b) not red?

题目：一个袋子里装有3个红色筹码、5个蓝色筹码和2个绿色筹码。随机抽取一个筹码。抽到 (a) 蓝色和 (b) 不是红色的概率分别是多少？

Probability is calculated as (number of favourable outcomes) / (total number of possible outcomes). First, find the total number of counters: 3 + 5 + 2 = 10. For part (a), the number of blue counters is 5, so P(blue) = 5/10, which simplifies to 1/2.

概率的计算公式为：（有利结果的数量）/（所有可能结果的总数）。首先，求出筹码的总数：3 + 5 + 2 = 10。对于 (a) 部分，蓝色筹码的数量是5，因此 P(蓝色) = 5/10，化简后为 1/2。

For part (b), ‘not red’ means the counter could be blue or green. The number of favourable outcomes is 5 (blue) + 2 (green) = 7. Therefore, P(not red) = 7/10. You could also use the complement rule: P(not red) = 1 – P(red) = 1 – (3/10) = 7/10. Both methods are valid.

对于 (b) 部分，“不是红色”意味着筹码可以是蓝色或绿色。有利结果的数量为 5（蓝）+ 2（绿）= 7。因此，P(不是红色) = 7/10。你也可以使用补集法则：P(不是红色) = 1 – P(红色) = 1 – (3/10) = 7/10。两种方法均正确。

In WJEC assessments, you should give your final probability as a fraction in its simplest form, although a decimal or percentage is sometimes acceptable. Always check that your fraction is less than or equal to 1; a probability greater than 1 indicates an error.

在WJEC评估中，最终的概率应写成最简分数形式，尽管有时也接受小数或百分数。记得检查你的分数是否小于或等于1；概率大于1则说明有误。

10. Key Revision Tips | 备考重点提示

Before your test, ensure you can confidently calculate the mean, median, mode, and range from both a raw list and a frequency table. Practice ordering data accurately for the median, and always check whether you need to find the middle of two numbers. When constructing charts, label axes, use a ruler, and provide a title. For probability, express answers as simplified fractions and use the complement rule to save time. Most importantly, read each question carefully—particularly those asking you to ‘explain’ your choice of average. Use precise vocabulary like ‘outlier’ and ‘spread’ to access full marks.

在测验之前，请确保你能够自信地从原始数据列表和频率表中计算平均数、中位数、众数和极差。练习准确排序数据以求得中位数，并时常检查是否需要取中间两个数的平均值。在绘制图表时，记得标注坐标轴、使用直尺并给出标题。处理概率问题时，将答案写成最简分数，并善用补集法则以节省时间。最为重要的是，仔细阅读每一道题——尤其是那些要求你“解释”选择某种平均数的题目。使用诸如“异常值”和“分散程度”等精确词汇，有助于你取得满分。

Published by TutorHao | Statistics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Year 7 WJEC Statistics: Unit Test Mock Paper Analysis | Year 7 WJEC 统计：单元测试模拟卷解析