IGCSE Cambridge Statistics: Unit Test Mock Paper Walkthrough | IGCSE 剑桥统计：单元测试模拟卷解析

📚 IGCSE Cambridge Statistics: Unit Test Mock Paper Walkthrough | IGCSE 剑桥统计：单元测试模拟卷解析

This mock paper walkthrough covers the most common Cambridge IGCSE Statistics unit test topics, including averages, frequency tables, cumulative frequency, probability, scatter diagrams, histograms, time series, box plots, index numbers and sampling. Each question below is followed by a clear worked solution so you can check your method and learn the typical marking points.

本模拟卷解析涵盖剑桥 IGCSE 统计单元测试最常见的考点，包括平均数、频数表、累积频数、概率、散点图、直方图、时间序列、箱线图、指数和抽样。下面的每道题都配有清晰的解题步骤，方便你核对方法并掌握典型得分点。

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1. Question 1: Mean, Median, Mode and Range | 第1题：平均数、中位数、众数和极差

The data set is 7, 9, 11, 13, 15, 15, 18, 21, 24. Find the mean, median, mode and range.

数据集为 7、9、11、13、15、15、18、21、24。求平均数、中位数、众数和极差。

lue, also 15. The range is the largest value minus the smallest value, giving 24 - 7 = 17.

首先将数据从小到大排列。平均数是所有数值之和除以数据个数，因此 Σx = 133，n = 9。中位数是第 5 个数据，因为 (9 + 1) ÷ 2 = 5，结果为 15。众数是出现次数最多的数，也是 15。极差是最大值减最小值，即 24 - 7 = 17。

Mean = 133 ÷ 9 = 14.8, Median = 15, Mode = 15, Range = 17

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2. Question 2: Frequency Table and Estimated Mean | 第2题：频数表与估计平均数

y 7, 20 ≤ t < 30 has frequency 9, and 30 ≤ t < 40 has frequency 5. Estimate the mean time.

该分组频数表显示了 25 名学生完成拼图所用的时间：0 ≤ t < 10 的频数为 4，10 ≤ t < 20 的频数为 7，20 ≤ t < 30 的频数为 9，30 ≤ t < 40 的频数为 5。估计平均时间。

15 + 9 × 25 + 5 × 35 = 20 + 105 + 225 + 175 = 525. Then divide by the total frequency 25.

用每个组距的中点作为代表值：5、15、25 和 35。将每个中点乘以相应频数后相加：4 × 5 + 7 × 15 + 9 × 25 + 5 × 35 = 20 + 105 + 225 + 175 = 525。然后除以总频数 25。

Estimated mean = 525 ÷ 25 = 21 minutes

The modal class is 20 ≤ t < 30 because it has the largest frequency 9.

众数所在组是 20 ≤ t < 30，因为该组频数最大，为 9。

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3. Question 3: Cumulative Frequency, Median and Interquartile Range | 第3题：累积频数、中位数与四分位距

The cumulative frequency table for the heights of 50 students is: less than 150 cm: 0, less than 155 cm: 8, less than 160 cm: 20, less than 165 cm: 35, less than 170 cm: 42, less than 175 cm: 50. Draw a cumulative frequency curve and estimate the median and interquartile range.

50 名学生身高的累积频数表为：低于 150 cm：0，低于 155 cm：8，低于 160 cm：20，低于 165 cm：35，低于 170 cm：42，低于 175 cm：50。绘制累积频数曲线，并估计中位数和四分位距。

0 ÷ 2 = 25. Reading from the curve gives approximately 162 cm. The lower quartile is at 25% of the total frequency, 12.5, giving about 156 cm. The upper quartile is at 75% 37.5, giving about 168 cm. The interquartile range is upper quartile minus lower quartile.

将组距上界与累积频数描点，并用光滑曲线连接。中位数对应总频数的一半，即 50 ÷ 2 = 25，从曲线上读得约为 162 cm。下四分位数对应总频数的 25%，即 12.5，约为 156 cm。上四分位数对应 75%，即 37.5，约为 168 cm。四分位距等于上四分位数减下四分位数。

Median ≈ 162 cm, IQR = 168 - 156 = 12 cm

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4. Question 4: Probability from a Two-Way Table | 第4题：双列表格中的概率

Eighty students each choose either History or Geography. The table shows: 45 boys, 35 girls; 20 boys choose History, 25 boys choose Geography, 14 girls choose History, 21 girls choose Geography. A student is selected at random. Find P(boy), P(girl and Geography), P(Geography), and P(boy given that the student chose History).

80 名学生每人选择历史或地理。表格显示：45 名男生，35 名女生；20 名男生选择历史，25 名男生选择地理；14 名女生选择历史，21 名女生选择地理。随机选择一名学生。求 P(男生)、P(女生且地理)、P(地理) 以及 P(男生 | 选择历史)。

21, so P(girl and Geography) = 21 ÷ 80 = 0.2625. The total choosing Geography is 25 + 21 =

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