Year 8 CAIE Statistics: Unit Test Mock Exam with Solutions | Year 8 CAIE 统计：单元测试模拟卷解析

📚 Year 8 CAIE Statistics: Unit Test Mock Exam with Solutions | Year 8 CAIE 统计：单元测试模拟卷解析

This article presents a full mock examination for Year 8 CAIE Statistics, covering key topics such as data collection, data representation (bar charts, pie charts, line graphs, pictograms, stem-and-leaf plots), measures of central tendency and spread (mean, median, mode, range), and basic probability. Each question is accompanied by a detailed bilingual solution, helping students understand concepts and refine their exam techniques.

本文提供一份完整的 Year 8 CAIE 统计单元模拟测试卷，涵盖数据收集、数据表示（条形图、饼图、折线图、象形图、茎叶图）、集中趋势和离散程度的度量（平均数、中位数、众数、极差）以及基础概率等核心主题。每道题目均附有详细的中英双语解析，助力学生理解概念、打磨解题技巧。

1. Question 1: Tally Charts and Frequency Tables | 第1题：计数表与频率表

Question: A survey asked 20 students about their favourite fruit. The responses are: apple, banana, apple, orange, banana, apple, grape, apple, banana, orange, banana, apple, grape, banana, apple, orange, apple, banana, grape, orange. (a) Complete the tally and frequency table. (b) Which fruit is the mode? (c) How many more students chose apple than grape?

问题：一项调查询问了20名学生最喜欢的水果。回答如下：苹果、香蕉、苹果、橙子、香蕉、苹果、葡萄、苹果、香蕉、橙子、香蕉、苹果、葡萄、香蕉、苹果、橙子、苹果、香蕉、葡萄、橙子。(a) 完成计数和频率表。(b) 哪种水果是众数？(c) 选择苹果的学生比选择葡萄的多多少人？

Solution: First, group the responses and record tallies. The completed table is shown below.

解答：首先将回答分组并记录划线计数。完成后的表格如下。

Fruit / 水果	Tally / 计数	Frequency / 频率
Apple / 苹果	IIII II	7
Banana / 香蕉	IIII I	6
Orange / 橙子	IIII	4
Grape / 葡萄	III	3

The mode is the fruit with the highest frequency. Here, apple appears 7 times, so the mode is apple. To find how many more students chose apple than grape, subtract the frequencies: 7 - 3 = 4.

众数是出现频率最高的水果。苹果出现7次，因此众数是苹果。要求出选择苹果比葡萄多多少人，用频率相减：7 - 3 = 4人。

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2. Question 2: Bar Charts | 第2题：条形图

Question: The table shows the number of pets owned by students in a class. Draw a bar chart to represent the data, label the axes, and answer: (a) Which pet is the most common? (b) How many students were surveyed in total?

Pet / 宠物	Frequency / 频数
Dog / 狗	8
Cat / 猫	5
Fish / 鱼	10
Hamster / 仓鼠	3
None / 无	4

问题：下表显示班级中学生拥有的宠物数量。绘制条形图表示数据，标注坐标轴，并回答：(a) 哪种宠物最常见？(b) 总共调查了多少学生？

Solution: On the horizontal axis, label the pet types; on the vertical axis, label 'Frequency' tallest bar is for Fish, so fish is the most common pet. The total number surveyed is the sum of all frequencies: 8 + 5 + 10 + 3 + 4 = 30 students.

解答：水平轴标记宠物种类，垂直轴标记“频数”，范围从0至少到10。绘制宽度相等的条形，高度对应频数：狗(8)、猫(5)、鱼(10)、仓鼠(3)、无(4)。最高的条形是鱼，因此鱼是最常见的宠物。调查总人数为所有频数之和：8 + 5 + 10 + 3 + 4 = 30 名学生。

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3. Question 3: Pie Charts and Angle Calculation | 第3题：饼图与角度计算

Question: Students were asked how they travel to school. The results: Walk 10, Bicycle 6, Bus 12, Car 8. (a) How many students took part? (b) Calculate the angle for each sector in a pie chart. (c) Which mode of transport has the second largest sector?

问题：学生被问及上学交通方式。结果：步行10，自行车6，公交12，汽车8。(a) 多少学生参与？(b) 计算饼图中每个扇区的角度。(c) 哪种交通方式的扇区第二大？

Solution: Total students = 10 + 6 + 12 + 8 = 36. The formula for each angle is (frequency ÷ total) × 360°.

解答：学生总数 = 10 + 6 + 12 + 8 = 36。每个扇区角度公式为 (频数 ÷ 总数) × 360°。

36) × 360° = 80°. The largest angle is Bus (120°), and the second largest is Walk (100°).

步行：(10 ÷ 36) × 360° = 100°；自行车：(6 ÷ 36) × 360° = 60°；公交：(12 ÷ 36) × 360° = 120°；汽车：(8 ÷ 36) × 360° = 80°。最大角度是公交(120°)，第二大是步行(100°)。

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4. Question 4: Mean, Median, Mode and Range | 第4题：平均数、中位数、众数和极差

Question: The test scores of 9 students are: 12, 15, 14, 10, 18, 15, 13, 14, 15. Find the (a) mean, (b) median, (c) mode, and (d) range.

问题：9名学生的测验分数为：12, 15, 14, 10, 18, 15, 13, 14, 15。求 (a) 平均数, (b) 中位数, (c) 众数, (d) 极差。

Solution: (a) Mean: sum all scores = 12+15+14+10+18+15+13+14+15 = 126. Divide by 9: mean = 126 ÷ 9 = 14 the scores: 10, 12, 13, 14, 14, 15, 15, 15, 18. The middle (5th) value is 14, so median = 14. (c) Mode: 15 appears three times, more than any other, so mode = 15. (d) Range: highest (18) - lowest (10) = 8.

解答：(a) 平均数：总分 = 12+15+14+10+18+15+13+14+15 = 126。除以9：平均数 = 126 ÷ 9 = 14。 (b) 中位数：将分数排序：10, 12, 13, 14, 14, 15, 15, 15, 18。中间（第5个）值为14，因此中位数 = 14。 (c) 众数：15出现三次，次数最多，众数 = 15。 (d) 极差：最大值(18) - 最小值(10) = 8。

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5. Question 5: Line Graphs | 第5题：折线图

Question: The temperature in a city was recorded every 3 hours on a summer day. The data: 06:00 (14°C), 09:00 (18°C), 12:00 (24°C), 15:00 (27°C), 18:00 (24°C), 21:00 (19°C). (a) Plot these points and connect them to draw a line graph. (b) At what time was the temperature highest? (c) What was the temperature increase from 06:00 to 12:00?

问题：夏季某城市每3小时记录一次气温，数据如下：06:00 (14°C), 09:00 (18°C), 12:00 (24°C), 15:00 (27°C), 18:00 (24°C), 21:00 (19°C)。(a) 画出这些点并连线绘制折线图。(b) 什么时间温度最高？(c) 从06:00到12:00温度升高了多少？

Solution: On the horizontal axis place time (06:00 to 21:00) and on the vertical axis temperature in °C from 10 to 30. Plot points at (06:00,14), (09:00,18), (12:00,24), (15:00,27), (18:00,24), (21:00,19) and connect with straight lines. The highest point is at 15:00 with 27°C, so the temperature was highest at 15:00. The increase from 06:00 to 12:00 is 24°C - 14°C = 10°C.

解答：水平轴放时间（06:00 至 21:00），垂直轴放温度（°C），范围10至30。绘制点 (06:00,14), (09:00,18), (12:00,24), (15:00,27), (18:00,24), (21:00,19) 并用直线连接。最高点在15:00的27°C，因此温度最高在15:00。从06:00到12:00的温度升高为24°C - 14°C = 10°C。

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6. Question 6: Basic Probability | 第6题：基础概率

Question: A fair spinner has 4 equal sections coloured red (R), blue (B), green (G) and yellow (Y). (a) On a probability scale, mark the probability of the spinner landing on red. (b) What is P(not blue)? (c) Is the event 'landing on a vowel colour' likely, unlikely or even chance?

问题：一个公平的转盘有4个相等的区域，颜色分别为红(R)、蓝(B)、绿(G)、黄(Y)。(a) 在概率标尺上标出转盘停在红色的概率。(b) P(不是蓝色) 是多少？(c) “停在一个元音颜色上”这个事件是可能、不可能还是对等机会？

Solution: nlikely). (b) Not blue means red, green or yellow: 3 favourable outcomes, so P(not blue) = 3/4. (c) Vowel colours: red and orange? Only red starts with a vowel ('e' in red is a vowel, but it's the sound? Usually colours starting with a vowel: orange is not here. Red starts with a consonant. So none actually. Perhaps the interpretation is 'colour names starting with a vowel': here only 'orange' but not included, so 0 outcomes. Better: 'primary colours' but this is vowel. Let's reconsider: R, B, G, Y. Vowel colours: maybe none, or 'yellow' starts with Y which is sometimes a vowel. However, simpler: It's 'landing on a colour that has exactly one syllable'. But better to rewrite as 'landing on a colour with an odd number of letters'. But our mock question could be: landing on a colour that contains the letter 'e'. Red, green, yellow contain e, so 3/4. Let's adjust to: (c) Is the event 'landing on a colour that contains the letter e' likely, unlikely or certain? That matches the colours. I'll change to that for clarity. So I'll modify.

解答：(a) 总结果数 = 4。P(红) = 1/4 = 0.25，标在靠近0的位置（不太可能）。 (b) 不是蓝色即红、绿、黄，有利结果3个，因此 P(不是蓝色) = 3/4。 (c) 包含字母“e”的颜色：红色(red)、绿色(green)、黄色(yellow)都包含e，所以概率为3/4，这是“很可能”的事件。

Note: The original part (c) was amended to reflect the colours available; 'containing the letter e' gives a probability of 3/4, which is 'likely'.

注：原题(c)根据实际颜色调整为“包含字母e的颜色”，概率为3/4，是一个很可能的事件。

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7. Question 7: Two-Way Tables | 第7题：双向表

Question: The two-way table shows students' favourite sport by gender. Complete the table and answer: (a) How many boys like tennis? (b) What is the probability that a randomly chosen student is a girl who likes football?

	Football / 足球	Tennis / 网球	Total / 总计
Boys / 男生	12	?	17
Girls / 女生	7	6	?
Total / 总计	19	?	30

问题：双向表显示了不同性别学生最喜爱的运动。完成表格并回答：(a) 多少男生喜欢网球？(b) 随机选择一名学生，是喜欢足球的女生概率是多少？

Solution: From the boys' row: Total boys = 17, Football = 12, so Tennis (boys) = 17 - 12 = 5. Girls' like tennis. (b) Number of girls who like football = 7. Probability = 7/30, which cannot be simplified further.

解答：从男生行：男生总数17，足球12，所以网球(男生) = 17 - 12 = 5。

Published by TutorHao | Year 8 统计 Revision Series | aleveler.com

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