📚 Year 8 Cambridge Statistics Mock Test Walkthrough | 剑桥8年级统计：单元测试模拟卷解析
Welcome to this detailed walkthrough of a Year 8 Cambridge Statistics mock test. Designed to mirror the style and content of a real unit assessment, this paper covers the core topics: calculating averages and range, working with frequency tables, interpreting graphs, and basic probability. Use this article to check your answers, understand common mistakes, and build confidence for your actual test.
欢迎阅读这篇剑桥8年级统计模拟卷的详细解析。这份试卷模拟了真实单元评估的风格和内容，涵盖了核心主题：计算平均数与极差、处理频率表、解读图表以及基础概率。通过本文，你可以核对答案、理解常见错误，并为实际测试树立信心。
1. Mean, Median, Mode & Range | 平均数、中位数、众数和极差
Question: The numbers below show the points scored by a netball team in 9 matches. Find the mean, median, mode and range.
12, 18, 15, 12, 20, 15, 14, 12, 17
题目：以下数字显示了一支无挡板篮球队在9场比赛中的得分。求平均数、中位数、众数和极差。
12, 18, 15, 12, 20, 15, 14, 12, 17
To find the mean, first add all values: 12+18+15+12+20+15+14+12+17 = 135. There are 9 matches, so divide by 9: 135 ÷ 9 = 15. The mean score is 15.
求平均数，先求和：12+18+15+12+20+15+14+12+17 = 135。共9场比赛，除以9得 135 ÷ 9 = 15。平均得分为15。
For the median, arrange the data in order: 12, 12, 12, 14, 15, 15, 17, 18, 20. The middle (5th) value is 15, so the median is 15.
将数据按顺序排列：12, 12, 12, 14, 15, 15, 17, 18, 20。中间（第5个）值是15，因此中位数为15。
The mode is the most frequent number. Here 12 appears three times, more than any other number. Thus the mode is 12.
众数是出现次数最多的数值。12出现了三次，多于其他任何数，所以众数是12。
Range = maximum − minimum = 20 − 12 = 8. So the range is 8.
极差 = 最大值 − 最小值 = 20 − 12 = 8。极差为8。
2. Mean from a Frequency Table | 根据频率表求平均数
Question: The frequency table shows the number of books read by 30 students in a month. Calculate the mean number of books.
题目：频率表显示了30名学生一个月内阅读的书籍数量。计算平均读书量。
| Books (x) | Frequency (f) |
|---|---|
| 1 | 4 |
| 2 | 7 |
| 3 | 10 |
| 4 | 6 |
| 5 | 3 |
First, add a column for ‘fx’ by multiplying each x by its frequency: 1×4=4, 2×7=14, 3×10=30, 4×6=24, 5×3=15. Sum these: 4+14+30+24+15 = 87. Total frequency is 30. Mean = total fx ÷ total f = 87 ÷ 30 = 2.9 books.
首先，增加列 ‘fx’，用每个 x 乘以其频率：1×4=4, 2×7=14, 3×10=30, 4×6=24, 5×3=15。求和：87。总频率为30。平均数 = 87 ÷ 30 = 2.9 本书。
3. Interpreting a Bar Chart | 解读条形图
Question: The bar chart (not shown here) displays the favourite fruit of Year 8 students: apples 22, bananas 35, oranges 18, grapes 25. Which fruit is the most popular? How many more students chose bananas than oranges?
题目：条形图（此处未显示）展示了8年级学生最喜欢的水果：苹果22人，香蕉35人，橙子18人，葡萄25人。哪种水果最受欢迎？选择香蕉的学生比选择橙子的多多少人？
Look at the heights of the bars: bananas have the highest frequency (35), so they are the most popular. To find the difference: 35 (bananas) − 18 (oranges) = 17 more students prefer bananas.
观察条形的高度：香蕉的频率最高（35），因此最受欢迎。求两者差值：35 − 18 = 17，因此喜欢香蕉的学生比喜欢橙子的多17人。
4. Pie Chart Angles | 饼图角度计算
Question: A survey asked 60 students how they travel to school. The results: Walk 24, Bus 15, Car 12, Bike 9. Work out the angle for each sector in a pie chart.
题目：一项调查询问了60名学生的上学交通方式。结果：步行24人，公交车15人，私家车12人，自行车9人。计算饼图中每个扇形的角度。
Total frequency = 60. The whole pie chart is 360°. The angle for Walk: (24/60) × 360 = 0.4 × 360 = 144°. For Bus: (15/60) × 360 = ¼ × 360 = 90°. For Car: (12/60) × 360 = 0.2 × 360 = 72°. For Bike: (9/60) × 360 = 0.15 × 360 = 54°. Check: 144+90+72+54 = 360°.
总频率为60。整个饼图为360°。步行扇形角度：(24/60) × 360 = 144°。公交车：(15/60) × 360 = 90°。私家车：(12/60) × 360 = 72°。自行车：(9/60) × 360 = 54°。验证总和为360°。
5. Scatter Graphs & Correlation | 散点图与相关性
Question: A scatter graph plots ‘hours spent revising’ against ‘test score %’. The points show an upward trend from bottom left to top right. Describe the type of correlation. What does this tell you?
题目：散点图绘制了“复习时间”与“测试分数百分比”的关系。各点呈现出从左下到右上的上升趋势。描述相关性的类型，并说明这说明了什么。
The pattern shows a positive correlation: as the number of hours revising increases, the test score tends to increase. This suggests that more revision is linked to higher scores, but it does not prove cause and effect.
该模式呈现正相关：随着复习时间的增加，测试分数往往也增加。这表明更多的复习与更高的分数相关联，但不能证明因果关系。
6. Probability Scales & Simple Events | 概率尺度与简单事件
Question: A bag contains 3 red, 2 blue and 5 green counters. One counter is picked at random. Mark on a probability scale the chance of picking: (a) a red counter, (b) a yellow counter, (c) a green counter.
题目：一个袋子装有3个红色、2个蓝色和5个绿色筹码。随机抽取一个。在概率尺度上标出抽到以下筹码的概率：(a) 红色，(b) 黄色，(c) 绿色。
Total counters = 3+2+5 = 10. Probability (red) = 3/10 = 0.3, which lies between ‘unlikely’ and ‘even chance’. Probability (yellow) = 0, so it is ‘impossible’. Probability (green) = 5/10 = ½ = 0.5, an ‘even chance’.
筹码总数 = 10。红色概率 = 3/10 = 0.3，位于“不太可能”与“等可能性”之间。黄色概率 = 0，为“不可能”。绿色概率 = 5/10 = ½ = 0.5，属于“等可能性”。
7. Experimental Probability & Expectation | 实验概率与期望值
Question: A fair six-sided die is rolled 120 times. How many times would you expect to get a 3?
题目：一枚公平的六面骰子投掷120次。你期望出现3的次数是多少？
The theoretical probability of rolling a 3 is 1/6. Expected number = probability × number of trials = (1/6) × 120 = 20. So you would expect to roll a 3 about 20 times.
掷出3的理论概率为1/6。期望次数 = 概率 × 试验次数 = (1/6) × 120 = 20。因此预计约20次。
Question: In an experiment, Emma spun a spinner 50 times and got ‘blue’ 18 times. What is the experimental probability of blue?
问题：在实验中，Emma旋转了50次转盘，得到“蓝色”18次。蓝色的实验概率是多少？
Experimental probability = number of successful outcomes / total trials = 18/50 = 0.36 or 36%. This may differ from the theoretical probability.
实验概率 = 成功次数 / 总次数 = 18/50 = 0.36（或36%）。这可能与理论概率不同。
8. Stem-and-Leaf Diagrams | 茎叶图
Question: The stem-and-leaf diagram shows the ages of people at a cinema: Stem 1 | 2 4 7; Stem 2 | 0 3 5 5 8; Stem 3 | 1 1 2; Key: 1|2 means 12. Find the median age and the range.
题目：茎叶图显示了电影院观众的年龄：茎1 | 2 4 7；茎2 | 0 3 5 5 8；茎3 | 1 1 2；关键：1|2表示12。求年龄的中位数和极差。
List all ages in order: 12, 14, 17, 20, 23, 25, 25, 28, 31, 31, 32. There are 11 values. The median is the 6th value, which is 25. Range = 32 − 12 = 20.
按顺序列出所有年龄：12,14,17,20,23,25,25,28,31,31,32。共11个值。中位数为第6个值，即
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