📚 Year 9 AQA Statistics: Unit Test Mock Paper Walkthrough | Year 9 AQA 统计：单元测试模拟卷解析
Welcome to the full walkthrough of a Year 9 AQA Statistics unit test mock paper. This resource covers key topics including averages, charts, sampling, probability, stem-and-leaf diagrams and more. Each question is presented with a detailed, step-by-step solution to help you understand the methods and boost your confidence for real assessments. Work through the questions first and then check your answers with the explanations.
欢迎阅读 Year 9 AQA 统计单元测试模拟卷的完整解析。本资料涵盖平均数、图表、抽样、概率、茎叶图等关键主题。每道题都配有详细的分步解答，帮助你理解方法并提升真实考试的自信心。先尝试自己作答，再用解析核对答案。
1. Q1: Mean, Median, Mode and Range | 问题1：平均数、中位数、众数和范围
The following data shows the number of books read by ten students in one month: 3, 5, 2, 7, 5, 4, 5, 6, 3, 8. Calculate the mean, median, mode and range.
以下数据显示了十名学生在一个月内阅读的书籍数量：3, 5, 2, 7, 5, 4, 5, 6, 3, 8。计算平均数、中位数、众数和范围。
To handle the data clearly, always start by ordering the values from smallest to largest: 2, 3, 3, 4, 5, 5, 5, 6, 7, 8.
为了清晰处理数据，首先将数值从小到大排序：2, 3, 3, 4, 5, 5, 5, 6, 7, 8。
Mean = (Sum of all values) ÷ (Number of values) = (3+5+2+7+5+4+5+6+3+8) ÷ 10 = 48 ÷ 10 = 4.8.
平均数 = (所有数值之和) ÷ (数值个数) = (3+5+2+7+5+4+5+6+3+8) ÷ 10 = 48 ÷ 10 = 4.8。
Median: with ten values (an even number), the median is the mean of the 5th and 6th values in the ordered list. The 5th value is 5 and the 6th value is 5, so the median is (5+5) ÷ 2 = 5.
中位数：有十个数值（偶数个），中位数是排序列表中第5个和第6个数值的平均数。第5个数值是5，第6个数值是5，因此中位数为 (5+5) ÷ 2 = 5。
Mode: the value that appears most often. Here 5 appears three times, more than any other number, so the mode is 5.
众数：出现次数最多的数值。此处5出现了三次，多于其他任何数值，因此众数为5。
Range = Maximum value − Minimum value = 8 − 2 = 6.
范围 = 最大值 − 最小值 = 8 − 2 = 6。
2. Q2: Reading and Interpreting Bar Charts | 问题2：解读条形图
A survey asked 40 students about their favourite ice cream flavour. The results are shown in the table below.
一项调查询问了40名学生最喜欢的冰淇淋口味。结果如下表所示。
| Flavour | 口味 | Frequency | 频数 |
|---|---|
| Chocolate | 巧克力 | 12 |
| Vanilla | 香草 | 18 |
| Strawberry | 草莓 | 6 |
| Mint | 薄荷 | 4 |
(a) How many students chose Vanilla? (b) Which flavour is the mode? (c) How many more students chose Vanilla than Strawberry? (d) If the data were shown on a bar chart, which axis would show the frequency?
(a) 有多少学生选择了香草？ (b) 哪种口味是众数？ (c) 选择香草的学生比选择草莓的学生多多少人？ (d) 如果用条形图显示数据，哪一个轴显示频数？
Total frequency = 12 + 18 + 6 + 4 = 40, which matches the number of students surveyed.
总频数 = 12 + 18 + 6 + 4 = 40，与接受调查的学生人数相符。
(a) 18 students chose Vanilla. (b) The mode is Vanilla, because it has the highest frequency (18).
(a) 18名学生选择了香草。 (b) 众数是香草，因为它的频数最高 (18)。
(c) Difference = Vanilla − Strawberry = 18 − 6 = 12 students. (d) In a bar chart, the vertical axis (y-axis) normally shows the frequency, while the horizontal axis (x-axis) shows the categories.
(c) 差值 = 香草 − 草莓 = 18 − 6 = 12名学生。 (d) 在条形图中，纵轴 (y轴) 通常显示频数，横轴 (x轴) 显示类别。
3. Q3: Scatter Graphs and Correlation | 问题3：散点图与相关性
The table shows the number of hours six students spent revising for a test and their scores.
下表显示了六名学生为一次测验复习的小时数及其分数。
| Hours | 小时数 | 2 | 3 | 5 | 1 | 4 | 6 |
|---|---|---|---|---|---|---|
| Score | 分数 | 50 | 60 | 80 | 40 | 70 | 90 |
(a) Plot the points on a scatter graph. (b) Describe the correlation shown. (c) Draw a line of best fit and use it to estimate the score for a student who revised for 3.5 hours.
(a) 在散点图上描出各点。 (b) 描述所示的相关性。 (c) 画一条最佳拟合线，并用它估计复习了3.5小时的学生的分数。
(a) The points to plot are (2, 50), (3, 60), (5, 80), (1, 40), (4, 70), (6, 90). Once plotted, they rise from the bottom left towards the top right. (b) The graph shows a strong positive correlation: as the number of revision hours increases, the test score tends to increase.
(a) 需要描点的坐标为 (2, 50), (3, 60), (5, 80), (1, 40), (4, 70), (6, 90)。描点后，这些点从左下方向右上方上升。 (b) 图表显示强正相关：随着复习小时数的增加，测验分数倾向于提高。
(c) A line of best fit should pass through the ‘centre’ of the points, with roughly equal numbers of points above and below it. Using the line, when hours = 3.5, the estimated score is about 65. (Accept answers in the range 63-67 depending on the line drawn.)
(c) 最佳拟合线应穿过数据点的“中心”，使线两侧的点数大致相等。利用这条线，当复习小时数为3.5时，估计分数约为65分。（根据所画直线，答案在63-67之间均可接受。）
4. Q4: Sampling Methods | 问题4：抽样方法
A school has 600 students in Years 7 to 10. The numbers in each year are: Year 7: 200, Year 8: 150, Year 9: 150, Year 10: 100. The headteacher wants to survey a stratified sample of 60 students. (a) Explain why stratified sampling might be better than simple random sampling here. (b) Calculate how many students should be selected from each year.
一所学校有600名7至10年级的学生。各年级人数为：7年级200人，8年级150人，9年级150人，10年级100人。校长希望抽取一个60名学生的分层样本。 (a) 解释为什么此时分层抽样可能比简单随机抽样更好。 (b) 计算每个年级应抽取多少名学生。
(a) Stratified sampling ensures that each year group is represented in the sample in proportion to its size in the population. A simple random sample might, by chance, miss a year group completely or overrepresent a small group. This would make the sample less representative.
(a) 分层抽样能确保每个年级在样本中的比例与其在总体中的比例一致。而简单随机样本可能偶然会完全遗漏某个年级，或使小型年级层比例过高，这会使样本代表性变差。
(b) The total population is 600, and we want a sample of 60, so the sampling fraction is 60/600 = 1/10. Number from Year 7 = 200 × (1/10) = 20. Year 8 = 150 × (1/10) = 15. Year 9 = 150 × (1/10) = 15. Year 10 = 100 × (1/10) = 10. Check: 20+15+15+10 = 60.
(b) 总体为600人，样本量为60，因此抽样比例为 60/600 = 1/10。7年级抽取人数 = 200 × (1/10) = 20人。8年级 = 150 × (1/10) = 15人。9年级 = 150 × (1/10) = 15人。10年级 = 100 × (1/10) = 10人。验证：20+15+15+10 = 60。
5. Q5: Basic Probability | 问题5：基本概率
A bag contains 3 red balls, 5 blue balls and 2 green balls. One ball is taken at random. Find the probability that the ball is: (a) red, (b) blue or green, (c) not red.
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。求取出的球是以下的概率： (a) 红色， (b) 蓝色或绿色， (c) 不是红色。
Total number of balls = 3 + 5 + 2 = 10. Probability is always calculated as (number of favourable outcomes) / (total number of outcomes).
球的总数 = 3 + 5 + 2 = 10。概率总是用 (有利结果的数量) / (可能结果的总数) 来计算。
(a) P(red) = number of red balls / total = 3/10. (b) ‘Blue or green’ means we count the blue and green balls together: 5 + 2 = 7. So P(blue or green) = 7/10. (c) ‘Not red’ is the complement of ‘red’. P(not red) = 1 − P(red) = 1 − 3/10 = 7/10, which matches the blue and green combined probability.
(a) P(红色) = 红球数量 / 总数 = 3/10。 (b) “蓝色或绿色”意味着将蓝球和绿球算在一起：5 + 2 = 7。因此 P(蓝色或绿色) = 7/10。 (c) “不是红色”是“红色”的互补事件。P(不是红色) = 1 − P(红色) = 1 − 3/10 = 7/10，与蓝、绿球的组合概率一致。
6. Q6: Listing Outcomes with Two Dice | 问题6：列出两个骰子的结果
Two fair six-sided dice are rolled. (a) List all the possible outcomes where the sum of the two numbers is 7. (b) Hence, or otherwise, find the probability that the sum is 7.
掷两个公平的六面骰子。 (a) 列出两个数字之和为7的所有可能结果。 (b) 由此，或通过其他方法，求出和为7的概率。
When two dice are rolled, there are 6 × 6 = 36 equally likely outcomes. We can systematically list pairs where the sum is 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). Note that (1,6) is different from (6,1).
掷两颗骰子时，共有 6 × 6 = 36 个等可能的结果。我们可以系统地列出和为7的数对：(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)。请注意 (1,6) 和 (6,1) 是不同的。
There are 6 outcomes that give a sum of 7. Therefore, P(sum of 7) = number of favourable outcomes / total outcomes = 6/36. This simplifies to 1/6.
共有6种结果使得和为7。因此，P(和为7) = 有利结果数 / 总结果数 = 6/36，化简为 1/6。
7. Q7: Stem-and-Leaf Diagrams | 问题7：茎叶图
An incomplete stem-and-leaf diagram shows the scores of 10 students in a spelling test. The key is: 2 | 3 means 23. Complete the ordered stem-and-leaf diagram, write down the actual scores and then find the median and the range.
一个未完成的茎叶图显示了10名学生在拼写测验中的分数。图例：2 | 3 表示23。完成有序的茎叶图，写出实际分数，然后求出中位数和范围。
Given unordered leaves: Stem 2: 3, 8, 1, 5 ; Stem 3: 2, 0, 9 ; Stem 4: 1, 4, 2. First, order the leaves within each stem: Stem 2 becomes 1, 3, 5, 8 ; Stem 3 becomes 0, 2, 9 ; Stem 4 becomes 1, 2, 4. The scores are: 21, 23, 25, 28, 30, 32, 39, 41, 42, 44.
给定的无序叶：茎2的叶：3, 8, 1, 5；茎3的叶：2, 0, 9；茎4的叶：1, 4, 2。首先，将每个茎上的叶排序：茎2变为1, 3, 5, 8；茎3变为0, 2, 9；茎4变为1, 2, 4。分数为：21, 23, 25, 28, 30, 32, 39, 41, 42, 44。
There are 10 scores. Median is the mean of the 5th and 6th values: 5th score is 30, 6th is 32; median = (30+32) ÷ 2 = 31. Range = highest (44) − lowest (21) = 23.
共有10个分数。中位数是第5个和第6个值的平均数：第5个分数是30，第6个是32；中位数 = (30+32) ÷ 2 = 31。范围 = 最大值 (44) − 最小值 (21) = 23。
8. Q8: Finding a Missing Value Using the Mean | 问题8：利用平均数求缺失值
The mean of five numbers is 20. Four of the numbers are 18, 22, 25 and 15. Work out the missing number.
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