Year 10 CIE Statistics Unit Test Mock Paper Analysis | 十年级CIE统计单元测试模拟卷解析

📚 Year 10 CIE Statistics Unit Test Mock Paper Analysis | 十年级CIE统计单元测试模拟卷解析

This mock paper is designed to help Year 10 CIE Statistics students consolidate key topics in descriptive statistics and introductory probability. Each question is broken down into clear, step-by-step solutions highlighting common pitfalls and essential exam techniques.

本模拟卷旨在帮助十年级CIE统计学生巩固描述性统计与概率初步的核心内容。每道题都配有清晰的逐步解析，突出常见错误与关键的应试策略。

1. Overview of the Mock Test | 模拟卷概述

The unit test targets the first half of the CIE Statistics syllabus, including data presentation, measures of central tendency and spread, cumulative frequency, and basic probability. The 10-question paper is designed to be completed in 45 minutes and reflects the style of past-paper items.

本次单元测试覆盖CIE统计课程的前半部分，包括数据展示、集中趋势与离散程度度量、累积频率以及基础概率。这份包含10道题的试卷需在45分钟内完成，题目风格贴近真题。

Question types: computational, diagrammatic, and interpretive
题型：计算题、图示题和解读题
Covered topics: mean, median, mode, range, frequency tables, bar charts, stem-and-leaf diagrams, quartiles, box plots, cumulative frequency, simple probability, combined events, tree diagrams, and expectation
涵盖主题：均值、中位数、众数、极差、频数表、条形图、茎叶图、四分位数、箱线图、累积频率、简单概率、组合事件、树状图以及期望值

2. Question 1: Calculating Mean, Median, Mode and Range | 第1题：计算均值、中位数、众数与极差

Data set: 45, 50, 55, 60, 60, 65, 70, 75, 80, 90

数据集：45, 50, 55, 60, 60, 65, 70, 75, 80, 90

Solution:

解答：

Step 1: Mean — Sum all values: 45 + 50 + 55 + 60 + 60 + 65 + 70 + 75 + 80 + 90 = 650. Number of values n = 10. Mean = 650/10 = 65.

步骤1：均值 — 所有数值相加：45 + 50 + 55 + 60 + 60 + 65 + 70 + 75 + 80 + 90 = 650。数据个数 n = 10。均值 = 650/10 = 65。

Step 2: Median — Ordered data already given. n is even, so median = average of the 5th and 6th values: (60 + 65)/2 = 62.5.

步骤2：中位数 — 已排序。n为偶数，中位数取第5和第6个值的平均：(60 + 65)/2 = 62.5。

Step 3: Mode — The value that appears most often is 60 (appears twice).

步骤3：众数 — 出现次数最多的数是60（出现两次）。

Step 4: Range = maximum − minimum = 90 − 45 = 45.

步骤4：极差 = 最大值 − 最小值 = 90 − 45 = 45。

3. Question 2: Mean from a Frequency Table | 第2题：频数表求均值

Table:

表格：

Score x	0	1	2	3	4
Frequency f	2	5	8	3	2

Step 1: Multiply each score by its frequency to get fx: 0×2=0, 1×5=5, 2×8=16, 3×3=9, 4×2=8.

步骤1：将每个得分乘以其频数，得 fx：0×2=0, 1×5=5, 2×8=16, 3×3=9, 4×2=8。

Step 2: Sum of fx = 0+5+16+9+8 = 38. Sum of frequencies = 2+5+8+3+2 = 20.

步骤2：Σfx = 0+5+16+9+8 = 38。Σf = 2+5+8+3+2 = 20。

Step 3: Mean = Σfx / Σf = 38/20 = 1.9.

步骤3：均值 = Σfx/Σf = 38/20 = 1.9。

Common mistake: forgetting to sum frequencies correctly or dividing by the number of categories instead of total frequency.

常见错误：忘记正确累加频数，或除以类别数而非总频数。

4. Question 3: Bar Chart Interpretation | 第3题：条形图解读

A survey asked 40 students their favourite colour: Red (10), Blue (12), Green (8), Yellow (6), Other (4). Draw a bar chart and give one key feature.

一项调查询问了40名学生最喜欢的颜色：红色(10)、蓝色(12)、绿色(8)、黄色(6)、其他(4)。画出条形图并给出一个关键特征。

Drawing guidance:

绘图指导：

Use equal-width bars with gaps between them. Label the horizontal axis ‘Colour’ and the vertical axis ‘Frequency’. The height of each bar corresponds to the frequency. The vertical scale should start at 0.

使用等宽的条形，条形之间留有空隙。横轴标记为’颜色’，纵轴标记为’频数’。每个条形的高度对应其频数。纵轴刻度必须从0开始。

Key feature: Blue is the modal category as it has the highest bar.

关键特征：蓝色是众数类别，因其条形最高。

Avoid common errors such as using a broken scale without justification or drawing bars that touch.

避免常见错误，如无正当理由使用断裂刻度，或条形之间紧贴。

5. Question 4: Stem-and-Leaf Diagram | 第4题：茎叶图

Data: 23, 25, 27, 30, 31, 33, 33, 35, 36, 40, 42

数据：23, 25, 27, 30, 31, 33, 33, 35, 36, 40, 42

Construct a stem-and-leaf diagram using a key, then find the median and mode.

制作茎叶图并附上图例，然后找出中位数和众数。

Stem | Leaf
2    | 3 5 7
3    | 0 1 3 3 5 6
4    | 0 2
Key: 2|3 means 23

Median: n = 11, so 6th value = 33. Mode = 33 (appears twice).

中位数：n = 11，第6个值是33。众数 = 33（出现两次）。

The leaves must be sorted and in a single vertical column. Always include a key and ensure equal spacing for legibility.

叶必须排序并呈单列垂直排列。务必添加图例并保持等间距以确保可读性。

6. Question 5: Quartiles and Box-and-Whisker Plot | 第5题：四分位数和箱线图

Using the data from Q4 (23, 25, 27, 30, 31, 33, 33, 35, 36, 40, 42), find Q₁, Q₂, Q₃, and draw a box plot.

利用第4题的数据（23, 25, 27, 30, 31, 33, 33, 35, 36, 40, 42），求 Q₁、Q₂、Q₃ 并绘制箱线图。

Ordered data: n = 11. Q₂ (median) = 6th value = 33. Lower half (first 5 values): 23, 25, 27, 30, 31 → Q₁ = median of lower half = 27. Upper half (last 5 values): 33, 35, 36, 40, 42 → Q₃ = 36. Minimum = 23, maximum = 42. IQR = 36 – 27 = 9.

排序数据：n = 11。Q₂（中位数）= 第6个值 = 33。下半部分（前5个值）：23,25,27,30,31 → Q₁ = 下半部分的中位数 = 27。上半部分（后5个值）：33,35,36,40,42 → Q₃ = 36。最小值 = 23，最大值 = 42。四分位距 IQR = 36 – 27 = 9。

Box plot: draw a number line, place vertical lines at min, Q₁, median, Q₃, max. Box from Q₁ to Q₃ with a line at median. Whiskers extend to min and max. Check for outliers: none here as 1.5×IQR limits are not breached.

箱线图：在数轴上画出表示最小值、Q₁、中位数、Q₃、最大值的竖线。箱体从 Q₁ 到 Q₃，箱内中线表示中位数。触须延伸至最小值和最大值。检查异常值：此处未超出 1.5×IQR 界限。

7. Question 6: Cumulative Frequency Curve | 第6题：累积频率曲线

The table shows the lengths of 40 leaves.

表格显示了40片叶子的长度。

Length (mm)	1−10	11−20	21−30	31−40	41−50
Frequency	3	7	14	10	6

Complete the cumulative frequency table and draw a cumulative frequency curve, then estimate the median and interquartile range.

完成累积频率表，绘制累积频率曲线，然后估算中位数和四分位距。

Cumulative frequencies: ≤10: 3, ≤20: 3+7=10, ≤30: 10+14=24, ≤40: 24+10=34, ≤50: 34+6=40. Plot points at upper class boundaries (10.5, 3), (20.5, 10), (30.5, 24), (40.5, 34), (50.5, 40). Join with a smooth curve.

累积频率：≤10: 3, ≤20: 3+7=10, ≤30: 10+14=24, ≤40: 24+10=34, ≤50: 34+6=40。描点于上组界：(10.5,3), (20.5,10), (30.5,24), (40.5,34), (50.5,40)。用光滑曲线连接。

Median from curve: go to cumulative frequency 20 (50% of 40), draw horizontal line to curve, then vertical to axis → approximately 28 mm. Q₁ at 10 → ~20 mm, Q₃ at 30 → ~35 mm. IQR ≈ 35 − 20 = 15 mm.

从曲线读中位数：纵轴累积频率20处作水平线交曲线，再作垂线至横轴 → 约28 mm。Q₁ 取10 → ~20 mm，Q₃ 取30 → ~35 mm。IQR ≈ 35 − 20 = 15 mm。

Always use graph paper and plot points carefully. Do not connect the first point to the origin unless there is a true zero frequency.

务必使用坐标纸，谨慎描点。除非数据从零频数开始，否则不要将第一个点与原点相连。

8. Question 7: Simple Probability | 第7题：简单概率

A bag contains 3 red, 4 blue and 5 green balls. One ball is picked at random. Find: (a) P(red); (b) P(not blue).

一个袋子里装有3个红球、4个蓝球和5个绿球。随机抽取一球。求：(a) P(红球)；(b) P(非蓝球)。

Total balls = 3 + 4 + 5 = 12.

总球数 = 3 + 4 + 5 = 12。

(a) P(red) = number of red / total = 3/12 = 1/4.

(a) P(红) = 红球数 / 总数 = 3/12 = 1/4。

(b) P(not blue) = 1 – P(blue) = 1 – 4/12 = 1 – 1/3 = 2/3. Alternatively, count favourable: red + green = 3+5 = 8, so 8/12 = 2/3.

(b) P(非蓝) = 1 – P(蓝) = 1 – 4/12 = 1 – 1/3 = 2/3。或直接计数有利结果：红+绿 = 3+5 = 8，即 8/12 = 2/3。

Always simplify fractions and ensure the probability is between 0 and 1.

记得约分并确保概率值在0到1之间。

9. Question 8: Mutually Exclusive Events | 第8题：互斥事件

In a single roll of a fair six-sided die, let A be the event ‘rolling an even number’ and B be ‘rolling a number less than 3’. Determine if A and B are mutually exclusive, and find P(A ∪ B).

掷一枚均匀的六面骰子一次，设事件A为“掷出偶数”，事件B为“掷出小于3的数”。判断A与B是否互斥，并求 P(A ∪ B)。

A = {2,4,6}, B = {1,2}. Since A ∩ B = {2} (non-empty), A and B are not mutually exclusive.

A = {2,4,6}，B = {1,2}。因为 A ∩ B = {2}（非空），所以A与B不是互斥事件。

P(A) = 3/6 = 1/2, P(B) = 2/6 = 1/3, P(A ∩ B) = 1/6.

P(A) = 3/6 = 1/2，P(B) = 2/6 = 1/3，P(A ∩ B) = 1/6。

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 1/2 + 1/3 − 1/6 = (3+2−1)/6 = 4/6 = 2/3.

P(A ∪ B) = P(A) + P(B) − P(A ∩ B) = 1/2 + 1/3 − 1/6 = (3+2−1)/6 = 4/6 = 2/3。

The key is to correctly identify the intersection. Mutually exclusive events would have P(A ∩ B) = 0, so then P(A ∪ B) = P(A) + P(B).

关键在于正确找出交集。互斥事件的 P(A ∩ B) = 0，此时 P(A ∪ B) = P(A) + P(B)。

10. Question 9: Tree Diagrams and Combined Events | 第9题：树状图与组合事件

A jar has 3 red and 5 blue marbles. Two marbles are drawn without replacement. Draw a tree diagram and find the probability that both are red, and the probability of at least one red.

一个罐子里有3个红球和5个蓝球。无放回地抽取两次。画树状图，并求两次都抽到红球的概率，以及至少抽到一个红球的概率。

Tree diagram (first draw, second draw):

树状图（第一次抽取，第二次抽取）：

           / Red (2/7)   → RR: (3/8)×(2/7) = 6/56 = 3/28
Red (3/8) —
           \ Blue (5/7)  → RB: (3/8)×(5/7) = 15/56

           / Red (3/7)   → BR: (5/8)×(3/7) = 15/56
Blue (5/8)—
           \ Blue (4/7)  → BB: (5/8)×(4/7) = 20/56 = 5/14

P(both red) = 3/28.

P(两次红) = 3/28。

P(at least one red) = 1 − P(no red) = 1 − P(BB) = 1 − 5/14 = 9/14. Alternatively, sum RR + RB +

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Year 10 CIE Statistics Unit Test Mock Paper Analysis | 十年级CIE统计单元测试模拟卷解析