📚 KS3 OCR Statistics: Unit Test Mock Paper Walkthrough | KS3 OCR 统计：单元测试模拟卷解析
This article provides a step-by-step guide through a mock unit test for the KS3 OCR Statistics topic. It covers a variety of question styles you might encounter, including interpreting bar charts and pie charts, calculating averages, working with probability, filling two-way tables, and designing a data collection method. Each question is followed by detailed solutions in both English and Chinese, helping you build confidence and exam technique.
本文逐步解析一套KS3 OCR统计单元测试模拟卷，内容涵盖柱状图与饼图解读、平均数计算、概率问题、双向表补全以及数据收集方法设计等常见题型。每道题均附有中英双语详细解答，帮助你巩固知识点，提升应试技巧。
1. Question 1: Bar Chart Interpretation | 问题1：柱状图解读
The bar chart below shows the number of goals scored by a football team each month from January to June.
下表展示了一支足球队在1月至6月每月的进球数。
| Month | 月份 | Goals | 进球数 |
|---|---|
| January | 1月 | 3 |
| February | 2月 | 5 |
| March | 3月 | 4 |
| April | 4月 | 6 |
| May | 5月 | 2 |
| June | 6月 | 4 |
Questions:
(a) Which month had the highest number of goals?
(b) Calculate the total number of goals scored over the 6 months.
(c) Find the mean number of goals per month.
问题：
(a) 哪个月的进球数最多？
(b) 计算这6个月的总进球数。
(c) 求每月平均进球数。
Part (a): Identify the maximum. Compare all the values: 3, 5, 4, 6, 2, 4. April has 6 goals, which is the highest.
第(a)部分：找出最大值。比较所有数值：3、5、4、6、2、4。4月有6个进球，是最高的。
Part (b): Total goals. Add all the monthly goals together.
第(b)部分：总进球数。将所有月份的进球数相加。
Total = 3 + 5 + 4 + 6 + 2 + 4 = 24 goals
总进球数 = 3 + 5 + 4 + 6 + 2 + 4 = 24 个
Part (c): Mean. Mean = total sum ÷ number of months.
第(c)部分：平均值。平均值 = 总和 ÷ 月数。
Mean = 24 ÷ 6 = 4 goals per month
平均值 = 24 ÷ 6 = 4 个进球/月
Always check that the mean lies within the range of the data (between 2 and 6). Here it does.
一定要检查平均值是否落在数据的范围内（2到6之间）。本例中确实在范围内。
2. Question 2: Pie Chart Interpretation | 问题2：饼图解读
A survey asked 200 students which sport they preferred. The results are shown in the pie chart with the following percentages: Football 40%, Cricket 25%, Tennis 20%, and Others 15%.
一项调查询问了200名学生最喜欢的运动。结果如饼图所示：足球40%，板球25%，网球20%，其他15%。
Questions:
(a) How many students chose Cricket?
(b) What fraction of the students chose Tennis? Give your answer in its simplest form.
(c) What is the ratio of Football to Tennis supporters?
问题：
(a) 有多少名学生选择了板球？
(b) 选择网球的学生占多少比例？请用最简分数表示。
(c) 足球支持者与网球支持者的人数比是多少？
Part (a): Cricket calculation. Cricket represents 25% of 200 students. 25% is equivalent to ¼.
第(a)部分：板球计算。板球占200名学生的25%。25%相当于¼。
Number of Cricket fans = 25% of 200 = 0.25 × 200 = 50 students
板球爱好者人数 = 200的25% = 0.25 × 200 = 50 人
Part (b): Tennis fraction. Tennis is 20%. As a fraction this is 20/100.
第(b)部分：网球比例。网球占20%。写成分数是20/100。
20/100 simplifies to 1/5 (divide numerator and denominator by 20)
20/100 化简为 1/5（分子分母同除以20）
Part (c): Ratio Football to Tennis. Football = 40%, Tennis = 20%. The ratio of percentages is 40 : 20.
第(c)部分：足球与网球之比。足球=40%，网球=20%。百分比之比为40 : 20。
40 : 20 simplifies to 2 : 1 (divide both by 20)
40 : 20 化简为 2 : 1（两者同除以20）
In terms of actual numbers: Football 80, Tennis 40 also gives 80 : 40 = 2 : 1. Always simplify ratios.
按实际人数：足球80人，网球40人，比值为80 : 40 = 2 : 1。比值一定要化到最简。
3. Question 3: Mean, Median, Mode and Range | 问题3：平均数、中位数、众数与极差
A student recorded the number of books read by 8 friends: 7, 8, 7, 9, 10, 7, 6, 8. Use this data set to answer the following.
一位学生记录了8名朋友的阅读书本数：7, 8, 7, 9, 10, 7, 6, 8。请根据这组数据回答以下问题。
(a) Find the mode. | (a) 求众数。
The mode is the number that appears most often. The data sorted: 6, 7, 7, 7, 8, 8, 9, 10. The number 7 appears three times, more than any other.
众数是出现次数最多的数。将数据排序：6, 7, 7, 7, 8, 8, 9, 10。数字7出现了三次，比其他的都多。
Mode = 7
众数 = 7
(b) Find the median. | (b) 求中位数。
The median is the middle value when the data is in order. Since there are 8 numbers (an even count), the median is the mean of the 4th and 5th values.
中位数是排序后中间的值。因为有8个数据（偶数个），中位数是第4和第5个数的平均值。
Ordered: 6, 7, 7, 7, 8, 8, 9, 10 → Median = (7 + 8) ÷ 2 = 7.5
排序：6, 7, 7, 7, 8, 8, 9, 10 → 中位数 = (7 + 8) ÷ 2 = 7.5
(c) Calculate the mean. | (c) 计算平均数。
First find the total sum: 7+8+7+9+10+7+6+8 = 62. Then divide by 8.
先求总和：7+8+7+9+10+7+6+8 = 62。然后除以8。
Mean = 62 ÷ 8 = 7.75
平均数 = 62 ÷ 8 = 7.75
(d) Find the range. | (d) 求极差。
The range is the difference between the largest and smallest values.
极差是最大值与最小值之差。
Range = 10 – 6 = 4
极差 = 10 – 6 = 4
4. Question 4: Line Graph and Interpolation | 问题4：折线图与插值估算
The table below shows the temperature recorded at a school weather station on a sunny morning.
下表显示了学校气象站在一个晴朗上午记录的每小时温度。
| Time | 时间 | Temperature (°C) | 温度 (°C) |
|---|---|
| 08:00 | 12 |
| 09:00 | 14 |
| 10:00 | 15 |
| 11:00 | 17 |
| 12:00 | 19 |
Questions:
(a) Plot a line graph of temperature against time.
(b) At what time was the temperature highest?
(c) Estimate the temperature at 10:30.
问题：
(a) 绘制温度随时间变化的折线图。
(b) 什么时间温度最高？
(c) 估算10:30时的温度。
Part (a): Plotting. Use time on the horizontal axis (x-axis) and temperature on the vertical axis (y-axis). Plot the points and join them with straight lines. (A sketch would show a rising trend.)
第(a)部分：绘图。 横轴（x轴）为时间，纵轴（y轴）为温度。描出各点并用直线连接。（草图应呈现上升趋势。）
Part (b): Highest temperature. Look at the table: the maximum temperature is 19°C at 12:00.
第(b)部分：最高温度。 观察表格：最高温度为19°C，出现在12:00。
Part (c): Estimating at 10:30. 10:30 is halfway between 10:00 (15°C) and 11:00 (17°C). Assuming a steady increase, the temperature would be about halfway between 15 and 17.
第(c)部分：估算10:30温度。 10:30是10:00（15°C）和11:00（17°C）的中间时刻。假设匀速上升，温度大约在15和17的中间。
Estimate = (15 + 17) ÷ 2 = 16°C
估算值 = (15 + 17) ÷ 2 = 16°C
This is called interpolation. The line graph makes such estimates visually clear.
这称为插值。折线图使得这类估算在视觉上非常直观。
5. Question 5: Basic Probability | 问题5：基础概率
A bag contains 3 red marbles, 5 blue marbles and 2 green marbles. All marbles are identical except for colour. One marble is taken at random.
一个袋子里有3个红色弹珠、5个蓝色弹珠和2个绿色弹珠。除颜色外，所有弹珠完全相同。随机取出一个弹珠。
Questions:
(a) Find the probability that the marble is blue.
(b) Find the probability that the marble is not red.
问题：
(a) 求取出蓝色弹珠的概率。
(b) 求取出非红色弹珠的概率。
Part (a): Total marbles = 3 + 5 + 2 = 10. Number of favourable outcomes (blue) = 5.
第(a)部分：总弹珠数 = 3 + 5 + 2 = 10。有利结果（蓝色）的数量 = 5。
P(blue) = 5/10 = 1/2 or 0.5
P(蓝色) = 5/10 = 1/2 或 0.5
Part (b): ‘Not red’ means the marble is either blue or green. Number of non-red marbles = 5 + 2 = 7.
第(b)部分：“非红色”指蓝色或绿色弹珠。非红色弹珠的数量 = 5 + 2 = 7。
P(not red) = 7/10 or 0.7
P(非红色) = 7/10 或 0.7
Alternatively, P(not red) = 1 – P(red) = 1 – 3/10 = 7/10. Both methods should give the same answer.
另外，P(非红色) = 1 – P(红色) = 1 – 3/10 = 7/10。两种方法的答案应一致。
6. Question 6: Two-Way Table | 问题6：双向表
A school surveyed 100 students (40 boys and 60 girls) about their favourite leisure activity: Football, Reading, or Gaming. The results are partly filled in.
某学校对100名学生（男生40名，女生60名）进行了最喜欢的休闲活动调查：足球、阅读或游戏。部分结果已填入表格。
| Football | 足球 | Reading | 阅读 | Gaming | 游戏 | Total | 总计 | |
|---|---|---|---|---|
| Boys | 男生 | 20 | 10 | ? | 40 |
| Girls | 女生 | 15 | 30 | ? | 60 |
| Total | 总计 | 35 | 40 | ? | 100 |
Questions:
(a) Complete the missing values in the table.
(b) What is the probability that a randomly chosen student prefers reading?
(c) Given that a student chosen at random prefers football, what is the probability the student is a girl?
问题：
(a) 补全表格中的缺失值。
(b) 随机选择一名学生，喜欢阅读的概率是多少？
(c) 已知随机选择的学生喜欢足球，求这名学生是女生的概率。
Part (a): For boys, Gaming = 40 – (20+10) = 10. For girls, Gaming = 60 – (15+30) = 15. Then the total Gaming column = 10 + 15 = 25. The completed table is shown below.
第(a)部分： 男生游戏人数 = 40 – (20+10) = 10。女生游戏人数 = 60 – (15+30) = 15。游戏总人数列 = 10 + 15 = 25。完整表格如下。
| Football | Reading | Gaming | Total | |
|---|---|---|---|---|
| Boys | 20 | 10 | 10 |
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