Year 10 Cambridge Statistics: Unit Test Mock Paper Analysis | 剑桥 Year 10 统计学：单元测试模拟卷解析

📚 Year 10 Cambridge Statistics: Unit Test Mock Paper Analysis | 剑桥 Year 10 统计学：单元测试模拟卷解析

This mock paper has been designed to mirror the style and content of a typical Year 10 Cambridge Statistics unit test. It consists of 8 questions that progress from basic terminology to more demanding data interpretation and probability reasoning. Each section below presents one question, followed by a detailed bilingual breakdown of the solution, highlighting common errors and reinforcing the underlying statistical concepts.

这份模拟卷模拟了典型的剑桥 Year 10 统计学单元测试的风格和内容。试卷包含 8 道题目，从基础术语逐步递进到更有挑战的数据解释和概率推理。下方每一节先呈现一道题目，再以中英双语详细拆解解题过程，突出常见错误并强化背后的统计概念。

1. Classifying Data Types | 数据类型的分类

Question: State whether each of the following is qualitative or quantitative. For quantitative data, further classify it as discrete or continuous. (a) The colour of a car. (b) The number of siblings a student has. (c) The time taken to complete a puzzle.

English explanation: Qualitative data describe qualities or categories that cannot be measured with numbers in a meaningful arithmetic way. ‘Colour of a car’ is a label, so it is qualitative. Quantitative data are numerical. The ‘number of siblings’ can only take whole numbers (you can’t have 2.3 siblings), making it discrete quantitative. The ‘time taken to complete a puzzle’ can be measured to any degree of precision, so it is continuous quantitative.

中文解析：定性数据描述的是无法用数字进行有意义算术运算的性质或类别。“汽车颜色”只是一个标签，因此是定性数据。定量数据是数值型数据。“兄弟姐妹的数量”只能取整数值（不可能有 2.3 个兄弟姐妹），因此是离散定量数据。“完成拼图所需的时间”可以测量到任意精度，因此是连续定量数据。

2. Constructing a Frequency Distribution Table | 构建频数分布表

Question: The heights (in cm) of 20 students are recorded: 158, 162, 158, 165, 170, 162, 158, 167, 162, 165, 170, 158, 165, 162, 167, 170, 162, 165, 158, 170. Construct a frequency distribution table and calculate the mode.

English explanation: First, list the distinct data values in ascending order: 158, 162, 165, 167, 170. Then tally the occurrences. 158 appears 5 times, 162 appears 6 times, 165 appears 4 times, 167 appears 2 times, and 170 appears 3 times. The mode is the value with the highest frequency, which is 162 cm.

中文解析：首先，将不同的数据值按升序排列：158、162、165、167、170。然后统计频数：158 出现 5 次，162 出现 6 次，165 出现 4 次，167 出现 2 次，170 出现 3 次。众数是频数最高的值，即 162 厘米。

3. Pie Chart Angle Calculations | 饼图角度计算

Question: In a survey of 180 people, 45 prefer tea, 90 prefer coffee, and the rest prefer juice. Calculate the angle for each sector in a pie chart representing this data.

English explanation: The total angle in a pie chart is 360°. The fraction for tea is 45/180 = 1/4, so the angle is (1/4) × 360° = 90°. Coffee: 90/180 = 1/2, angle = 180°. Juice: the remaining people = 180 – 45 – 90 = 45, fraction = 45/180 = 1/4, angle = 90°. Always check that the angles sum to 360°.

中文解析：饼图的总角度为 360°。茶的比例为 45/180 = 1/4，因此角度为 (1/4) × 360° = 90°。咖啡：90/180 = 1/2，角度为 180°。果汁：剩余人数 = 180 – 45 – 90 = 45，比例为 45/180 = 1/4，角度为 90°。务必检验角度之和为 360°。

4. Mean, Median, and Mode from a Frequency Table | 根据频数表求平均数、中位数与众数

Question: The table shows the number of pets owned by 30 families.

Number of pets	Frequency
0	4
1	8
2	10
3	6
4	2

Calculate the mean, median, and mode.

English explanation: To find the mean, multiply each value by its frequency and sum: (0×4)+(1×8)+(2×10)+(3×6)+(4×2) = 0+8+20+18+8 = 54. Divide by total frequency 30: mean = 54/30 = 1.8 pets. The mode is the value with the highest frequency: 2 pets (frequency 10). For the median, the total number of data points is 30, so the median lies between the 15th and 16th values. The cumulative frequency reaches 4 (0 pets), then 4+8=12 (1 pet), then 12+10=22 (2 pets); hence the 15th and 16th values are both 2, so median = 2.

中文解析：计算平均数时，将每个值乘以其频数并求和：(0×4)+(1×8)+(2×10)+(3×6)+(4×2) = 0+8+20+18+8 = 54。除以总频数 30：平均数 = 54/30 = 1.8 只宠物。众数是频数最高的值：2 只宠物（频数 10）。中位数：数据总数为 30，中位数位于第 15 和第 16 个值之间。累积频数：4（0 只宠物），然后是 4+8=12（1 只宠物），接着 12+10=22（2 只宠物）；因此第 15 和第 16 个值都是 2，故中位数 = 2。

5. Range and Interquartile Range | 极差与四分位距

Question: Find the range and interquartile range (IQR) of the following data set: 12, 7, 9, 15, 8, 5, 14, 11, 10.

English explanation: First, sort the data in ascending order: 5, 7, 8, 9, 10, 11, 12, 14, 15. The range = maximum – minimum = 15 – 5 = 10. For IQR, locate the lower quartile (Q₁) and upper quartile (Q₃). With 9 data points, the median (Q₂) is the 5th value: 10. The lower half is 5, 7, 8, 9; Q₁ is the median of this half, i.e., (7+8)/2 = 7.5. The upper half is 11, 12, 14, 15; Q₃ = (12+14)/2 = 13. IQR = Q₃ – Q₁ = 13 – 7.5 = 5.5.

中文解析：首先将数据按升序排列：5, 7, 8, 9, 10, 11, 12, 14, 15。极差 = 最大值 – 最小值 = 15 – 5 = 10。对于四分位距，先找出下四分位数（Q₁）和上四分位数（Q₃）。9 个数据点，中位数（Q₂）为第 5 个值：10。下半部分数据为 5, 7, 8, 9；Q₁ 为这部分的中位数，即 (7+8)/2 = 7.5。上半部分数据为 11, 12, 14, 15；Q₃ = (12+14)/2 = 13。IQR = Q₃ – Q₁ = 13 – 7.5 = 5.5。

6. Drawing and Interpreting a Box Plot | 箱线图的绘制与解读

Question: Using the five-number summary from the data in question 5 (min = 5, Q₁ = 7.5, median = 10, Q₃ = 13, max = 15), draw a box plot and comment on the skewness.

English explanation: A box plot consists of a box from Q₁ to Q₃ with a vertical line at the median. Whiskers extend to the minimum and maximum (assuming no outliers). Here, the box spans from 7.5 to 13 on the scale, with median at 10. The left whisker (5 to 7.5) is slightly shorter than the right whisker (13 to 15), and the median is closer to Q₁ than to Q₃. This suggests a slight positive (right) skew, although with such a small data set the skew is mild.

中文解析：箱线图由一个从 Q₁ 到 Q₃ 的矩形组成，并在中位数位置画一条竖线。触须延伸至最小值和最大值（假设无异常值）。此处，矩形在刻度上从 7.5 到 13，中位数位于 10。左侧触须（5 至 7.5）略短于右侧触须（13 至 15），且中位数更靠近 Q₁ 而非 Q₃。这表明数据略有正偏（右偏），但由于数据集较小，偏态程度轻微。

7. Basic Probability and Expected Frequency | 基础概率与期望频数

Question: A fair six-sided die is rolled 300 times. (a) What is the probability of rolling a number greater than 4? (b) How many times would you expect to roll a number greater than 4 in 300 rolls?

English explanation: Numbers greater than 4 on a die are 5 and 6, so two favourable outcomes out of six. Probability = 2/6 = 1/3. The expected frequency is the probability multiplied by the number of trials: (1/3) × 300 = 100 times.

中文解析：骰子上大于 4 的点数为 5 和 6，共两种有利结果，总可能结果为 6 种。概率 = 2/6 = 1/3。期望频数为概率乘以试验次数：(1/3) × 300 = 100 次。

8. Tree Diagrams and the Multiplication Rule | 树状图与乘法法则

Question: A bag contains 4 red and 6 blue marbles. Two marbles are drawn at random without replacement. (a) Draw a tree diagram to represent all possible outcomes. (b) Calculate the probability that both marbles drawn are red. (c) Calculate the probability that the two marbles are different colours.

English explanation: Part (a) – the tree has first branch: red 4/10, blue 6/10. Second branch: if red first, remaining are 3 red, 6 blue; probabilities: red 3/9, blue 6/9. If blue first, remaining are 4 red, 5 blue; probabilities: red 4/9, blue 5/9. (b) Probability both red = (4/10) × (3/9) = 12/90 = 2/15. (c) Different colours: (red then blue) + (blue then red) = (4/10 × 6/9) + (6/10 × 4/9) = 24/90 + 24/90 = 48/90 = 8/15. Always check that all final probabilities sum to 1.

中文解析：第 (a) 部分：树状图第一层分支：红色 4/10，蓝色 6/10。第二层分支：若第一次抽到红色，剩余 3 红 6 蓝；概率：红色 3/9，蓝色 6/9。若第一次抽到蓝色，剩余 4 红 5 蓝；概率：红色 4/9，蓝色 5/9。(b) 两次均为红色的概率 = (4/10) × (3/9) = 12/90 = 2/15。(c) 颜色不同的概率 = (先红后蓝) + (先蓝后红) = (4/10 × 6/9) + (6/10 × 4/9) = 24/90 + 24/90 = 48/90 = 8/15。务必检验所有最终概率之和为 1。

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Year 10 Cambridge Statistics: Unit Test Mock Paper Analysis | 剑桥 Year 10 统计学：单元测试模拟卷解析