Year 12 CIE Statistics: Unit Test Mock Paper Walkthrough | Year 12 CIE 统计：单元测试模拟卷解析

📚 Year 12 CIE Statistics: Unit Test Mock Paper Walkthrough | Year 12 CIE 统计：单元测试模拟卷解析

Welcome to this step-by-step walkthrough of a full Year 12 CIE Statistics unit test mock paper. The questions have been designed to reflect the typical style of CIE assessments, covering data representation, measures of central tendency and spread, probability, permutations and combinations, discrete random variables, the binomial distribution, normal approximation, and hypothesis testing. Work through each explanation to consolidate your understanding and exam technique.

欢迎查看这份 Year 12 CIE 统计单元测试模拟卷的详细解析。题目按照 CIE 典型风格设计，涵盖数据表示、集中趋势与离散程度、概率、排列与组合、离散随机变量、二项分布、正态近似以及假设检验。逐一攻克每个解析，巩固你的理解并提升应试技巧。

1. Stem-and-Leaf Diagram, Median and IQR | 茎叶图、中位数与四分位距

Question 1: The times (in minutes) taken by 20 students to complete a puzzle are recorded below. (a) Construct an ordered stem-and-leaf diagram. (b) Calculate the median and the interquartile range (IQR).
Data: 12, 15, 18, 21, 22, 23, 25, 26, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 42, 45.

第1题：记录20名学生完成拼图所需时间（分钟）如下。(a) 画出有序茎叶图。(b) 计算中位数和四分位距 (IQR)。
数据：12, 15, 18, 21, 22, 23, 25, 26, 27, 28, 29, 31, 32, 33, 35, 36, 37, 39, 42, 45。

First, arrange the data in ascending order – it is already ordered. We use the tens digit as the stem and the units digit as the leaf. The stems are 1, 2, 3, and 4. For stem 1, the leaves are 2, 5, 8; for stem 2, leaves 1, 2, 3, 5, 6, 7, 8, 9; for stem 3, leaves 1, 2, 3, 5, 6, 7, 9; and for stem 4, leaves 2, 5. Remember to include a key, e.g. 1 | 2 represents 12 minutes.

首先，将数据按升序排列——该数据已经有序。我们将十位数作为茎，个位数作为叶。茎为1、2、3、4。茎1的叶为2、5、8；茎2的叶为1、2、3、5、6、7、8、9；茎3的叶为1、2、3、5、6、7、9；茎4的叶为2、5。务必包含图例，如 1 | 2 表示12分钟。

The median position for n = 20 is (20+1)/2 = 10.5, so the median is the average of the 10th and 11th values. The 10th value is 28, the 11th is 29, giving a median of 28.5 minutes.

对于 n=20，中位数位置为 (20+1)/2 = 10.5，因此中位数是第10和第11个数值的平均。第10个是28，第11个是29，中位数为28.5分钟。

To find the quartiles, Q1 is at position (20+1)/4 = 5.25, so we interpolate between the 5th (22) and 6th (23) values: Q1 = 22 + 0.25 × (23 – 22) = 22.25. Q3 is at position 3 × (20+1)/4 = 15.75, between the 15th (35) and 16th (36) values: Q3 = 35 + 0.75 × (36 – 35) = 35.75. Hence IQR = 35.75 – 22.25 = 13.5 minutes.

计算四分位数，Q1位于 (20+1)/4 = 5.25 的位置，因此我们在第5个(22)和第6个(23)数值之间插值：Q1 = 22 + 0.25 × (23 – 22) = 22.25。Q3位于 3×(20+1)/4 = 15.75 的位置，在第15个(35)和第16个(36)之间：Q3 = 35 + 0.75 × (36 – 35) = 35.75。因此 IQR = 35.75 – 22.25 = 13.5 分钟。

2. Box Plot and Outlier Detection | 箱线图与异常值检测

Question 2 (using the data from Q1): (a) Using the five-number summary, draw a box plot. (b) Determine if there are any outliers.

第2题（沿用第1题数据）： (a) 根据五数概括法绘制箱线图。(b) 判断是否存在异常值。

The five-number summary is: Minimum = 12, Q1 = 22.25, Median = 28.5, Q3 = 35.75, Maximum = 45. Draw a scale, mark these five positions, construct a box from Q1 to Q3 with a line inside at the median, and then extend whiskers to the minimum and maximum.

五数概括为：最小值=12，Q1=22.25，中位数=28.5，Q3=35.75，最大值=45。画出数轴，标出这五个点，从 Q1 到 Q3 作矩形框，框内居中画中位数线，然后从矩形端引出须线至最小值和最大值。

To check for outliers, compute the fences: Lower fence = Q1 – 1.5 × IQR = 22.25 – 1.5 × 13.5 = 22.25 – 20.25 = 2.0. Upper fence = Q3 + 1.5 × IQR = 35.75 + 20.25 = 56.0. Any value below 2.0 or above 56

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