📚 Year 11 CIE Statistics Unit Test Mock Paper Analysis | CIE 统计单元测试模拟卷解析
This mock paper analysis is designed to help Year 11 CIE Statistics students review key concepts typically covered in a unit test. Each section presents a typical question, step‑by‑step solution, and tips to avoid common mistakes. Topics range from data representation to probability distributions.
本模拟卷解析旨在帮助学习 CIE 统计的 Year 11 学生复习单元测试中的核心知识点。每个小节给出典型题目、逐步解析以及避坑指南，涵盖从数据表示到概率分布的广泛主题。
1. Histograms and Frequency Density | 直方图与频数密度
A grouped frequency table records the time taken by 50 students to solve a puzzle. Estimate how many students took between 12 and 18 minutes using the histogram concept.
某分组频数表记录了 50 名学生解谜所用的时间。请利用直方图概念估计用时在 12 至 18 分钟之间的学生人数。
| Time, t (minutes) | Frequency |
|---|---|
| 5 ≤ t < 10 | 4 |
| 10 ≤ t < 15 | 12 |
| 15 ≤ t < 20 | 18 |
| 20 ≤ t < 30 | 10 |
| 30 ≤ t < 40 | 6 |
In a histogram the area of each bar is proportional to frequency. We first calculate the frequency density for classes 10-15 and 15-20.
在直方图中，每个条形的面积与频数成正比。我们首先计算区间 10–15 和 15–20 的频数密度。
For 10 ≤ t < 15, class width = 5, frequency density = 12 ÷ 5 = 2.4.
对 10 ≤ t < 15，组距 = 5，频数密度 = 12 ÷ 5 = 2.4。
For 15 ≤ t < 20, class width = 5, frequency density = 18 ÷ 5 = 3.6.
对 15 ≤ t < 20，组距 = 5，频数密度 = 18 ÷ 5 = 3.6。
ts between 12 and 18 minutes, we find the area of the bars covering that interval: from 12–15 (width 3) at density 2.4 gives 3 × 2.4 = 7.2 students; from 15–18 (width 3) at density 3.6 gives 3 × 3.6 = 10.8 students.
为估计 12 到 18 分钟之间的学生人数，我们计算覆盖该区间的条形面积：12–15（宽3）在密度 2.4 下得 3 × 2.4 = 7.2 人；15–18（宽3）在密度 3.6 下得 3 × 3.6 = 10.8 人。
Total estimated students = 7.2 + 10.8 = 18 students.
估计总人数 = 7.2 + 10.8 = 18 人。
2. Cumulative Frequency and Median Estimation | 累积频数曲线与中位数估计
The cumulative frequency table below shows the time taken by 50 students. Use the curve to estimate the median and interquartile range.
下面的累积频数表显示了 50 名学生花费的时间。利用累积频数曲线估计中位数和四分位距。
| Time, t (min) | Cumulative frequency |
|---|---|
| ≤ 10 | 4 |
| ≤ 15 | 16 |
| ≤ 20 | 34 |
| ≤ 30 | 44 |
| ≤ 40 | 50 |
ear interpolation: median = 15 + [(25 - 16)/(34 - 16)] × 5 = 15 + (9/18)×5 = 17.5 minutes.
中位数对应第 25 个值，落在 15 到 20 分钟之间。用线性插值：中位数 = 15 + [(25 - 16)/(34 - 16)] × 5 = 15 + (9/18)×5 = 17.5 分钟。
The lower quartile Q1 is the 12.5th value: Q1 = 10 + [(12.5 - 4)/(16 - 4)] × 5 = 10 + (8.5/12)×5 ≈ 13.54 min.
下四分位数 Q1 是第 12.5 个值：Q1 = 10 + [(12.5 - 4)/(16 - 4)] × 5 = 10 + (8.5/12)×5 ≈ 13.54 分钟。
The upper quartile Q3 is the 37.5th value: Q3 = 20 + [(37.5 - 34)/(44 - 34)] × 10 = 20 + (3.5/10)×10 = 23.5 min.
上四分位数 Q3 是第 37.5 个值：Q3 = 20 + [(37.5 - 34)/(44 - 34)] × 10 = 20 + (3.5/10)×10 = 23.5 分钟。
Interquartile range IQR = 23.5 - 13.54 = 9.96 minutes.
四分位距 IQR = 23.5 - 13.54 = 9.96 分钟。
3. Mean, Variance and Standard Deviation | 均值、方差与标准差
Calculate the mean, variance and standard deviation for the dataset: 12, 15, 18, 21, 22, 25, 28, 30.
计算数据组 12, 15, 18, 21, 22, 25, 28, 30 的均值、方差和标准差。
Sum of values = 12 + 15 + 18 + 21 + 22 + 25 + 28 + 30 = 171. Number of observations n = 8.
数值总和 = 171，观测值个数 n = 8。
Mean μ = 171 ÷ 8 = 21.375
Compute the squared deviations from the mean: (12-21.375)2 + (15-21.375)2 + ... + (30-21.375)2 = (-9.375)2 + (-6.375)2 + (-3.375)2 + (-0.375)2 + (0.625)2 + (3.625)2 + (6.625)2 + (8.625)2.
计算各值与均值的偏差平方和：(-9.375)2 + (-6.375)2 + ... + (8.625)2。
40.640625 + 11.390625 + 0.140625 + 0.390625 + 13.140625 + 43.890625 + 74.390625 = 271.875.
平方和 40.640625 + 11.390625 + 0.140625 + 0.390625 + 13.140625 + 43.890625 + 74.390625 = 271.875。
Variance σ2 = 271.875 ÷ 8 = 33.984375
Standard deviation σ = √33.984375 ≈ 5.83
4. Box Plots and Outlier Detection | 箱线图与异常值检测
A five‑number summary gives: minimum = 3, Q1 = 8, median = 12, Q3 = 15, maximum = 20. Determine if there are any outliers and sketch the box plot.
一组五数概括为：最小值 3、Q1 = 8、中位数 12、Q3 = 15、最大值 20。判断是否存在异常值并绘制箱线图。
Interquartile range IQR = 15 - 8 = 7.
四分位距 IQR = 15 - 8 = 7。
Lower fence = Q1 - 1.5 × IQR = 8 - 10.5 = -2.5. Upper fence = Q3 + 1.5 × IQR = 15 + 10.5 = 25.5.
下边界 = Q1 - 1.5 × 7 = 8 - 10.5 = -2.5；上边界 = Q3 + 1.5 × 7 = 15 + 10.5 = 25.5。
All data values lie between -2.5 and 25.5, so no outliers are present.
所有数据值均在 -2.5 与 25.5 之间，因此没有异常值。
The box is drawn from Q1 to Q3 with a line at the median. Whiskers extend to the minimum and maximum. No outliers are marked.
箱体从 Q1 画至 Q3，并在中位数处画一条线。须线延伸至最小值和最大值。无需标记异常值。
5. Basic Probability and Tree Diagrams | 基础概率与树形图
A bag contains 3 red and 5 blue balls. Two balls are drawn at random without replacement. Find the probability that both balls are the same colour.
一个袋子装有 3 个红球和 5 个蓝球。随机不放回地抽取两个球，求两球颜色相同的概率。
diagram: first draw P(Red) = 3/8, P(Blue) = 5/8. After one red is taken, second draw P(Red|Red) = 2/7, P(Blue|Red) = 5/7. After one blue, P(Red|Blue) = 3/7, P(Blue|Blue) = 4/7.
画出树形图：第一次抽取 P(红) = 3/8，P(蓝) = 5/8。若第一次为红，第二次 P(红|红) = 2/7，P(蓝|红) = 5/7；若第一次为蓝，第二次 P(红|蓝) = 3/7，P(蓝|蓝) = 4/7。
robability both red = (3/8) × (2/7) = 6/56. Probability both blue = (5/8) × (4/7) = 20/56.
两球全红的概率 = (3/8) × (2/7) = 6/56。两球全蓝的概率 = (5/8) × (4/7) = 20/56。
P(same colour) = 6/56 + 20/56 = 26/56 = 13/28
6. Conditional Probability and Venn Diagrams | 条件概率与韦恩图
For two events A and B, P(A) = 0.4, P(B) = 0.5 and P(A ∪ B) = 0.7. Find P(A|B) and illustrate with a Venn diagram.
对于两个事件 A 和 B，P(A) = 0.4，P(B) = 0.5，P(A ∪ B) = 0.7。求 P(A|B) 并用韦恩图表示。
First find the intersection: P(A ∩ B) = P(A) + P(B) - P(A ∪ B) = 0.4 + 0.5 - 0.7 = 0.2.
先求交集：P(A ∩ B) = P(A) + P(B) - P(A ∪ B) = 0.4 + 0.5 - 0.7 = 0.2。
Conditional probability P(A|B) = P(A ∩ B) / P(B) = 0.2 / 0.5 = 0.4.
条件概率 P(A|B) = P(A ∩ B) / P(B) = 0.2 / 0.5 = 0.4。
In the Venn diagram, the overlap region is 0.2; the remaining part of A is 0.2, the remaining part of B is 0.3, and outside is 0.3.
在韦恩图中，重叠区域为 0.2；A 的剩余部分为 0.2，B 的剩余部分为 0.3，外部为 0.3。
7. Discrete Random Variables and Expectation | 离散随机变量与期望值
The probability distribution of a discrete random variable X is given below. Find E(X) and Var(X).
离散随机变量 X 的概率分布如下。求 E(X) 和 Var(X)。
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X=x) | 0.2 | 0.3 | 0.4 | 0.1 |
E(X) = Σ x·P(X=x) = 1×0.2 + 2×0.3 + 3×0.4 + 4×0.1 = 0.2 + 0.6 + 1.2 + 0.4 = 2.4
For variance we compute E(X2) = 12×0.2 + 22×0.3 + 32×0.4 + 42×0.1 = 0.2 + 1.2 + 3.6 + 1.6 = 6.6.
方差计算需
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