Year 7 CCEA Statistics: Unit Test Walkthrough | CCEA 七年级统计单元测试模拟卷解析

📚 Year 7 CCEA Statistics: Unit Test Walkthrough | CCEA 七年级统计单元测试模拟卷解析

Welcome to this step-by-step walkthrough of a full Year 7 CCEA Statistics unit test paper. Each question targets a core skill you need to master: reading pictograms, constructing bar charts, calculating the mean, median, mode and range, working with tally and frequency tables, interpreting line graphs, and comparing data sets. Use these detailed solutions to check your answers, understand common mistakes, and build confidence for your actual test.

欢迎来到这份 CCEA 七年级统计单元测试模拟卷的逐步解析。每一道题都针对你必须掌握的核心技能：阅读象形图、构建条形图、计算平均数、中位数、众数和极差、使用计数和频率表、解读折线图以及比较数据组。请使用这些详细解答来核对答案、理解常见错误并为真正的测试建立信心。

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1. Pictogram Goals | 第1题：进球象形图

The pictogram shows the goals scored by four players in a season. Key: ⚽ = 4 goals. Player A: ⚽⚽⚽; Player B: ⚽⚽; Player C: ⚽⚽⚽⚽⚽; Player D: ⚽⚽⚽⚽.

象形图显示了四名球员在一个赛季中的进球数。图例：⚽ = 4 球。球员 A：⚽⚽⚽；球员 B：⚽⚽；球员 C：⚽⚽⚽⚽⚽；球员 D：⚽⚽⚽⚽。

(a) How many goals did Player C score? First, multiply the number of symbols by the key value: Player C has 5 symbols, so 5 × 4 = 20 goals.

(a) 球员 C 进了多少球？首先，将符号个数乘以图例值：球员 C 有 5 个符号，因此 5 × 4 = 20 球。

(b) How many more goals did Player C score than Player B? Player B has 2 symbols, which means 2 × 4 = 8 goals. The difference is 20 – 8 = 12 goals.

(b) 球员 C 比球员 B 多进多少球？球员 B 有 2 个符号，意味着 2 × 4 = 8 球。相差 20 – 8 = 12 球。

A common mistake is forgetting to multiply by the key, so always check the key before you calculate.

一个常见错误是忘记乘以图例值，所以在计算之前一定要检查图例。

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2. Drawing a Bar Chart | 第2题：绘制条形图

A survey asked 20 students their favourite fruit. The frequency table shows: Apple 6, Banana 5, Orange 4, Grapes 3, Other 2. You need to draw a vertical bar chart with equal gaps and labelled axes.

一项调查询问了 20 名学生最喜欢的水果。频率表显示：苹果 6，香蕉 5，橙子 4，葡萄 3，其他 2。你需要绘制一个垂直条形图，条宽相等且坐标轴有标签。

First, choose a simple scale for the frequency axis. Since the largest value is 6, a scale going up to 7, with each grid line representing 1 student, works well. Draw the vertical axis from 0 to 7 and label it ‘Frequency’. The horizontal axis lists the fruit names. Draw bars of equal width and equal spacing; the height of each bar must exactly match the frequency.

首先，为频率轴选择一个简单的刻度。因为最大数值为 6，所以刻度最高到 7，每个网格线代表 1 个学生，这很合适。画出从 0 到 7 的垂直轴并标记为“频率”。水平轴列出水果名称。画出等宽且等间距的条形；每个条形的高度必须与频率完全吻合。

Always give your chart a clear title, such as ‘Favourite Fruits of 20 Students’. Colour in the bars neatly and make sure they do not touch the next bar.

始终给你的图表一个明确的标题，例如“20 名学生最喜欢的水果”。整齐地给条形上色，并确保每个条形之间不接触。

Accuracy points: if the scale is uneven or bars touch, you will lose marks in the exam.

给分点：如果刻度不均匀或条形相互接触，你会在考试中丢分。

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3. Calculating the Mean | 第3题：计算平均值

The midday temperatures (°C) recorded over 5 days were: 12, 15, 10, 14, 14. Calculate the mean temperature.

记录的五天中午温度（°C）为：12、15、10、14、14。计算平均温度。

Mean = (12 + 15 + 10 + 14 + 14) ÷ 5

平均值 = (12 + 15 + 10 + 14 + 14) ÷ 5

Step 1: Add all the values together. 12 + 15 + 10 + 14 + 14 = 65. Step 2: Divide the total by the number of values. 65 ÷ 5 = 13. So the mean temperature is 13°C.

步骤 1：将所有数值相加。12 + 15 + 10 + 14 + 14 = 65。步骤 2：将总和除以数值的个数。65 ÷ 5 = 13。因此平均温度为 13°C。

The mean gives the central value, but it can be pulled up or down by unusually high or low data points. Always check your addition before dividing.

平均数给出中心值，但它可能会被异常高或异常低的数据点拉高或拉低。在除法前务必检查你的加法。

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4. Finding the Mode | 第4题：寻找众数

A set of shoe sizes from 10 students is recorded: 3, 4, 4, 5, 3, 4, 5, 5, 4, 3. Find the mode.

记录了 10 名学生的鞋码：3、4、4、5、3、4、5、5、4、3。找出众数。

To find the mode, count how many times each value appears. Tally the 3s: there are three (3, 3, 3). The 4s: there are four (4, 4, 4, 4). The 5s: there are three (5, 5, 5). The value that occurs most often is 4, with a frequency of 4. Therefore, the mode is 4.

要找到众数，计算每个数值出现的次数。数一下 3：有三个（3、3、3）。4：有四个（4、4、4、4）。5：有三个（5、5、5）。出现次数最多的数值是 4，频率为 4。因此众数是 4。

If two values appear most often with the same frequency, the data set would have two modes (bimodal). In this case, only 4 appears most often.

如果有两个数值以相同的最高频率出现，那么数据集就有两个众数（双峰）。在这个例子中，只有 4 出现最频繁。

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5. Determining the Median | 第5题：确定中位数

Using the same shoe size data, work out the median. First, arrange the numbers in ascending order: 3, 3, 3, 4, 4, 4, 4, 5, 5, 5.

使用相同的鞋码数据计算中位数。首先，将数字按升序排列：3、3、3、4、4、4、4、5、5、5。

There are 10 numbers (an even count). The median is found by taking the two middle numbers and calculating their mean. The 5th value is 4 and the 6th value is also 4. So the median = (4 + 4) ÷ 2 = 4.

一共有 10 个数字（偶数个）。中位数需要取中间的两个数并计算它们的平均值。第 5 个值是 4，第 6 个值也是 4。因此中位数 = (4 + 4) ÷ 2 = 4。

If there were an odd number of values, you would simply pick the middle one. Always remember to order the data first, or the median will be wrong.

如果数值个数是奇数，你只需取正中间的那个值。请务必记住先对数据进行排序，否则中位数会出错。

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6. Working out the Range | 第6题：计算极差

The range shows how spread out the data is. For the shoe size data, find the range.

极差显示数据的离散程度。对于鞋码数据，求极差。

Range = Maximum value – Minimum value

极差 = 最大值 – 最小值

The largest shoe size is 5, and the smallest is 3. Range = 5 – 3 = 2. A small range like this tells us the shoe sizes are quite consistent across the group.

最大的鞋码是 5，最小的是 3。极差 = 5 – 3 = 2。像这样较小的极差告诉我们该组学生的鞋码很接近。

The range is affected by extremes; one very high or low value can make the range huge, even if most values are close together.

极差受极端值的影响；即使大多数值都很接近，一个非常高或非常低的值也能让极差变得很大。

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7. Choosing the Best Average | 第7题：选择最佳平均数

The test scores of a class are: 10, 11, 12, 12, 13, 14, 15, 16, 40. Which average, the mean or the median, better represents the typical score? Explain your choice.

一个班级的考试分数为：10、11、12、12、13、14、15、16、40。哪种平均数——均值还是中位数——更能代表典型分数？解释你的选择。

First, find the mean: sum = 10+11+12+12+13+14+15+16+40 = 143. 143 ÷ 9 ≈ 15.9. The mean is pulled up to about 16 by the unusually high score of 40.

首先求均值：总和 = 10+11+12+12+13+14+15+16+40 = 143。143 ÷ 9 ≈ 15.9。均值被异常高分 40 拉高到大约 16。

Now find the median. Order the data: 10, 11, 12, 12, 13, 14, 15, 16, 40. The middle value (5th) is 13. The median is 13, which is much closer to where most scores actually lie.

再求中位数。排序：10、11、12、12、13、14、15、16、40。中间值（第 5 个）是 13。中位数是 13，这更接近大多数分数的实际情况。

Because the extreme score 40 distorts the mean but not the median, the median gives a fairer picture of a typical student’s performance. In general, use the median when you have outliers.

因为极端分数 40 扭曲了均值但没有影响中位数，中位数能更公正地反映典型学生的成绩。一般来说，当数据有离群值时，应使用中位数。

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8. Completing a Tally and Frequency Table | 第8题：完成计数与频率表

The raw data below shows the pets owned by 15 students: Cat, Dog, Dog, Fish, Cat, Rabbit, Cat, Dog, Fish, Cat, Cat, Dog, Rabbit, Dog, Cat. Create a frequency table with tallies.

以下原始数据显示了 15 名学生养的宠物：猫、狗、狗、鱼、猫、兔、猫、狗、鱼、猫、猫、狗、兔、狗、猫。创建一个带计数标记的频率表。

We go through the list and add a tally mark for each pet. For Cat: we count one by one and draw a tally mark for the fifth across the previous four, giving a group of five. The completed frequency table should look like this:

我们逐一浏览列表，并为每个宠物添加计数标记。对于猫：我们一一点数，在数到第五个时划一条斜线穿过前面四个，形成一个“五”的标记组。完成的频率表应该像这样：

Pet 宠物	Tally 计数	Frequency 频率
Cat	IIII	6
Dog	IIII	5
Fish	II	2
Rabbit	II	2

Check that the total frequency adds up to 15, which matches the number of data entries. Tally charts help you avoid counting the same item twice and keep your work organised.

检查总频率加起来是否为 15，这与数据条目的数量相符。计数图能帮助你避免重复计数同一项，并使你的工作井井有条。

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9. Reading a Line Graph | 第9题：阅读折线图

A line graph shows the outside temperature recorded every 2 hours from 08:00 to 18:00. The temperatures plotted are: 08:00 – 10°C, 10:00 – 12°C, 12:00 – 16°C, 14:00 – 18°C, 16:00 – 17°C, 18:00 – 14°C.

一张折线图显示了从 08:00 到 18:00 每两小时记录一次的室外温度。绘制的温度点为：08:00 – 10°C，10:00 – 12°C，12:00 – 16°C，14:00 – 18°C，16:00 – 17°C，18:00 – 14°C。

(a) What was the temperature at 12:00? By reading the

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