Year 11 Edexcel Statistics: Mock Unit Test Walkthrough | Edexcel 11年级统计：单元测试模拟卷解析

📚 Year 11 Edexcel Statistics: Mock Unit Test Walkthrough | Edexcel 11年级统计：单元测试模拟卷解析

This article walks through a mock unit test designed for Year 11 Edexcel Statistics students. The test covers core topics including sampling, data representation, averages, measures of spread, probability, scatter graphs, time series and index numbers. By working through each question, you will deepen your understanding of statistical concepts and develop essential exam techniques. Let us begin the detailed breakdown.

本文详细解析一份为 Edexcel 11年级统计课程设计的模拟单元测试卷。试卷涵盖抽样方法、数据表示、平均数、离散程度、概率、散点图、时间序列和指数等核心主题。通过逐题解析，你将加深对统计概念的理解，并锻炼必要的应试技巧。现在开始详细讲解。

1. Introduction to the Mock Test | 模拟测试介绍

This mock unit test consists of eight questions, each targeting a specific area of the Edexcel Statistics syllabus. The paper is designed to be completed in 60 minutes, mimicking real exam conditions. It assesses both knowledge of statistical methods and the ability to interpret and communicate findings. The following sections provide a full worked solution for each question, highlighting key steps and common pitfalls.

本模拟单元测试包含八道题目，每题针对 Edexcel 统计考纲的一个特定领域。试卷设计为 60 分钟内完成，模拟真实考试环境。它既评价统计方法知识，也考察解读与交流发现的能力。以下各节提供每题完整的解题过程，突出关键步骤和常见错误。

2. Question 1: Sampling Methods | 题目1：抽样方法

A school council wants to investigate Year 11 students’ views on homework. The school secretary provides a list of all 240 Year 11 students arranged by registration group. A random number between 1 and 12 is chosen, and then every 12th student on the list from that starting point is selected for the sample.

学校学生会想调查 11 年级学生对家庭作业的看法。学校秘书提供了一份按注册组排列的 240 名 11 年级学生名单。从 1 到 12 之间随机选取一个起点，然后从该起点开始每隔 12 名学生选取一名进入样本。

Identify the sampling method used. Give one advantage and one disadvantage of this method.

识别所采用的抽样方法，并给出该方法的一个优点和一个缺点。

Worked solution:

This is systematic sampling. The defining feature is selecting members from a list at regular intervals after a random start.

这是系统抽样。其典型特征是在随机起点后按照固定间隔从名单中选取成员。

Advantage: It is quick and easy to apply, and the sample is spread evenly across the entire list, which can reflect the population well if the list is not ordered in a way that introduces bias.

优点：该方法快捷且易于操作，样本均匀分布在整份名单中；若名单排序没有引入偏差，能较好地代表总体。

Disadvantage: If the list has a hidden periodicity (for example, all students in each registration group share a similar characteristic and the interval coincides with group size), the sample may become biased. Also, not every possible sample has an equal chance of being selected.

缺点：若名单具有隐藏的周期性（例如每个注册组的学生具有相似特征，且间隔恰好与组规模一致），样本可能产生偏差。此外，并非所有可能样本都有同等被选中的机会。

3. Question 2: Averages from a Grouped Frequency Table | 题目2：分组频数表求平均值

The table below shows the time, in minutes, that 50 students spent on a mathematics assignment.

下表显示了 50 名学生完成一项数学作业所花费的时间（单位：分钟）。

Time, t (minutes)	Frequency
0 < t ≤ 20	6
20 < t ≤ 40	14
40 < t ≤ 60	18
60 < t ≤ 80	10
80 < t ≤ 100	2

Calculate an estimate for the mean time spent. Explain why your answer is an estimate.

计算所花费时间的平均值估计值，并解释为什么你的答案是估计值。

Step-by-step solution:

First, find the midpoint of each interval. The midpoints are: 10, 30, 50, 70 and 90.

首先，求出每个区间的中点。中点值分别为：10、30、50、70 和 90。

Next, multiply each midpoint by its frequency to obtain the products (fx): 10×6 = 60, 30×14 = 420, 50×18 = 900, 70×10 = 700, 90×2 = 180.

接着，将每个中点与对应的频数相乘，得到 fx 值：10×6 = 60，30×14 = 420，50×18 = 900，70×10 = 700，90×2 = 180。

Sum of frequencies, Σf = 6 + 14 + 18 + 10 + 2 = 50. Sum of fx, Σfx = 60 + 420 + 900 + 700 + 180 = 2260.

频数之和 Σf = 6 + 14 + 18 + 10 + 2 = 50。fx 之和 Σfx = 60 + 420 + 900 + 700 + 180 = 2260。

Estimated mean = Σfx ÷ Σf

x̄ = 2260 ÷ 50 = 45.2 minutes

This is an estimate because we used the midpoint of each class interval to represent all values in that interval; the actual raw data points are not known.

这是一个估计值，因为我们使用每个组区间的中点来代表该区间内的所有数值；实际原始数据是未知的。

4. Question 3: Interquartile Range and Box Plots | 题目3：四分位距与箱形图

The stem-and-leaf diagram shows the test marks (out of 50) for 19 students.

下列茎叶图显示了 19 名学生的测验分数（满分 50）。

Stem (tens) | Leaf (units)
1 | 4 7
2 | 2 5 6 8 9
3 | 1 3 3 5 7 9
4 | 0 2 4 5 8

Find the median, the lower quartile (Q₁) and the upper quartile (Q₃). Hence, determine the interquartile range (IQR) and draw a box plot for the data.

求中位数、下四分位数 (Q₁) 和上四分位数 (Q₃)。据此确定四分位距 (IQR)，并绘制数据的箱形图。

Worked solution:

First, list all marks in order: 14, 17, 22, 25, 26, 28, 29, 31, 33, 33, 35, 37, 39, 40, 42, 44, 45, 48. There are 19 data values. (Note: the stems give tens, leaves units.)

首先，按顺序列出所有分数：14, 17, 22, 25, 26, 28, 29, 31, 33, 33, 35, 37, 39, 40, 42, 44, 45, 48。共 19 个数据值。（注意：茎表示十位，叶表示个位。）

Median position: (19+1)/2 = 10th value. The 10th value is 33. So median = 33.

中位数位置：(19+1)/2 = 第 10 个值。第 10 个值为 33，因此中位数 = 33。

To find Q₁, consider the lower half (first 9 values): 14, 17, 22, 25, 26, 28, 29, 31, 33. Median of these 9 values is the 5th value, which is 26. So Q₁ = 26.

求 Q₁ 时，考虑下半个数据（前 9 个值）：14, 17, 22, 25, 26, 28, 29, 31, 33。这 9 个值的中位数为第 5 个值，即 26。所以 Q₁ = 26。

For Q₃, take the upper half (last 9 values): 33, 35, 37, 39, 40, 42, 44, 45, 48. Median is the 5th value, 40. Therefore Q₃ = 40.

对于 Q₃，取上半个数据（后 9 个值）：33, 35, 37, 39, 40, 42, 44, 45, 48。中位数为第 5 个值，40。因此 Q₃ = 40。

Interquartile range = Q₃ – Q₁ = 40 – 26 = 14.

四分位距 IQR = Q₃ – Q₁ = 40 – 26 = 14。

To draw the box plot, we need the five-number summary: minimum = 14, Q₁ = 26, median = 33, Q₃ = 40, maximum = 48. The box plot is drawn on a scale from 10 to 50, with a box from 26 to 40 and a vertical line at 33, with whiskers extending to 14 and 48.

绘制箱形图需要五数概括：最小值 = 14，Q₁ = 26，中位数 = 33，Q₃ = 40，最大值 = 48。在 10 到 50 的尺度上绘制，箱子从 26 到 40，中间在 33 处画一条竖线，须线延伸至 14 和 48。

5. Question 4: Probability and Two-way Tables | 题目4：概率与双向表

80 students were asked whether they prefer football or basketball. The results are summarised in the two-way table.

80 名学生被询问他们喜欢足球还是篮球。结果汇总在双向表中。

	Football	Basketball	Total
Boys	28	12	40
Girls	18	22	40
Total	46	34	80

A student is chosen at random. Find the probability that the student:

随机选择一名学生。求以下概率：

(a) prefers basketball.
(b) is a girl and prefers football.
(c) prefers football given that the student is a boy.
(d) is a boy given that the student prefers basketball.

(a) 喜欢篮球；(b) 是女生且喜欢足球；(c) 已知是男生，喜欢足球的概率；(d) 已知喜欢篮球，是男生的概率。

Solutions:

(a) P(Basketball) = Total who prefer basketball / Total students = 34/80 = 17/40.

(a) P(篮球) = 喜欢篮球的总人数 ÷ 总人数 = 34/80 = 17/40。

(b) P(Girl and Football) = 18/80 = 9/40.

(b) P(女生且足球) = 18/80 = 9/40。

(d) P(Boy | Basketball) = Number of boys who prefer basketball / Total who prefer basketball = 12/34 = 6/17.

(d) P(男生 | 篮球) = 喜欢篮球的男生人数 ÷ 喜欢篮球的总人数 = 12/34 = 6/17。

6. Question 5: Probability Tree Diagrams | 题目5：概率树图

A bag contains 5 red counters and 3 blue counters. Two counters are drawn at random without replacement. Construct a probability tree diagram and use it to find the probability that:

一个袋子里有 5 个红色筹码和 3 个蓝色筹码。随机抽取两个筹码，不放回。构建概率树图，并用它求以下事件的概率：

(i) both counters are red.
(ii) the counters are of different colours.

(i) 两个筹码均为红色；(ii) 两个筹码颜色不同。

Worked solution:

First draw: P(Red) = 5/8, P(Blue) = 3/8. For the second draw, the probabilities change depending on the first outcome.

第一次抽取：P(红) = 5/8，P(蓝) = 3/8。第二次抽取时，概率取决于第一次的结果。

If first is Red, remaining: 4 Red, 3 Blue, total 7. So P(Red|Red) = 4/7, P(Blue|Red) = 3/7.

如果第一次是红色，剩余：4 红，3 蓝，共 7 个。因此 P(红|红) = 4/7，P(蓝|红) = 3/7。

If first is Blue, remaining: 5 Red, 2 Blue, total 7. So P(Red|Blue) = 5/7, P(Blue|Blue) = 2/7.

如果第一次是蓝色，剩余：5 红，2 蓝，共 7 个。因此 P(红|蓝) = 5/7，P(蓝|蓝) = 2/7。

(i) P(both red) = P(Red) × P(Red|Red) = (5/8) × (4/7) = 20/56 = 5/14.

(i) P(两个红) = P(红) × P(红|红) = (5/8) × (4/7) = 20/56 = 5/14。

(ii) Different colours can happen in two ways: (Red then Blue) or (Blue then Red).

(ii) 颜色不同可以通过两种方式发生：（红然后蓝）或（蓝然后红）。

P(Red then Blue) = (5/8) × (3/7) = 15/56. P(Blue then Red) = (3/8) × (5/7) = 15/56. Total = 15/56 + 15/56 = 30/56 = 15/28.

P(红然后蓝) = (5/8) × (3/7) = 15/56。P(蓝然后红) = (3/8) × (5/7) = 15/56。总和 = 15/56 + 15/56 = 30/56 = 15/28。

7. Question 6: Scatter Graphs and Correlation | 题目6：散点图与相关性

The table gives the hours of revision and the marks achieved by 8 students in a statistics test.

下表给出了 8 名学生在统计测验中的复习小时数与取得的成绩。

Revision (hours)	1.5	2.0	3.0	4.5	5.0	6.0	7.5	8.0
Mark	38	42	51	55	63	72	78	85

Plot a scatter graph. Describe the correlation and interpret what it means. Identify any potential outlier.

绘制散点图。描述相关性并解释其含义。识别任何可能的离群值。

Response guidance:

When plotting, put revision hours on the horizontal axis and test mark on the vertical axis. The points show a clear upward pattern.

绘图时，将复习小时数放在横轴，测验分数放在纵轴。数据点呈现明显的上升趋势。

This indicates positive correlation: as revision hours increase, the test mark tends to increase. This suggests that spending more time revising is associated with higher achievement in the test.

这表明存在正相关：随着复习时间增加，测验分数往往也增加。这意味着投入更多时间复习与取得更高成绩有关联。

Looking at the data, the point (1.5, 38) and the point (8.0, 85) follow the trend well. However, the point (6.0, 72) appears slightly below the general pattern, but still reasonable. Without a formal test, there is no strong outlier; the scatter is fairly linear.

观察数据，(1.5, 38) 和 (8.0, 85) 这两点很好地符合趋势。然而，点 (6.0, 72) 看上去略低于总体形态，但仍然合理。在没有正式检验的情况下，没有明显的离群值；散点大致呈线性。

8. Question 7: Time Series and Moving Averages | 题目7：时间序列与移动平均

The number of daily visitors to a museum was recorded for the first 8 days of a month.

某家博物馆记录了某个月前 8 天的每日游客人数。

Day	1	2	3	4	5	6	7	8
Visitors	120	132	125	141	138	150	149	155

Calculate a 3-point moving average for the data. Plot the original time series and the moving averages on the same axes, and use the trend to predict the number of visitors on day 9.

计算该数据的 3 点移动平均值。在同一坐标轴上绘制原始时间序列和移动平均值，并利用趋势预测第 9 天的游客人数。

Worked solution:

A 3-point moving average for day 2 is calculated as (120 + 132 + 125) ÷ 3 = 125.7 (to 1 d.p.). Continue for days 3 to 7.

第 2 天的 3 点移动平均值计算为 (120 + 132 + 125) ÷ 3 = 125.7（保留一位小数）。依此计算至第 7 天。

Day	Moving Average
2	125.7
3	132.7
4	134.7
5	143.0
6	145.7
7	151.3

Plotting the moving averages shows a clear upward trend. By extending

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Year 11 Edexcel Statistics: Mock Unit Test Walkthrough | Edexcel 11年级统计：单元测试模拟卷解析