IGCSE Edexcel Statistics: Unit Test Mock Paper Walkthrough | IGCSE Edexcel 统计：单元测试模拟卷解析

📚 IGCSE Edexcel Statistics: Unit Test Mock Paper Walkthrough | IGCSE Edexcel 统计：单元测试模拟卷解析

This article provides a detailed walkthrough of a mock unit test for the IGCSE Edexcel Statistics course. Every question is explained step by step, allowing students to identify crucial concepts and avoid common mistakes. The paper covers data classification, averages, charts, probability, cumulative frequency, histograms, moving averages and more, giving you a comprehensive revision aid.

本文详细讲解了一份为 IGCSE Edexcel 统计课程设计的单元测试模拟卷。每道题目都逐步解析，帮助学生掌握核心概念并避免常见错误。试卷涵盖数据分类、平均数、图表、概率、累积频率、直方图、移动平均值等内容，是一份全面的复习资料。

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1. Question 1: Data Classification | 题目1：数据分类

The first question tests your ability to distinguish between primary and secondary data, and to identify discrete and continuous data. You are given four scenarios: (a) Heights of students measured directly by the teacher; (b) Temperatures downloaded from a website; (c) Number of cars counted by a student; (d) Weights of apples picked from a tree.

第一题考查区分一手与二手数据、以及识别离散与连续数据的能力。题目给出四个情景：(a) 老师直接测量的学生身高；(b) 从网上下载的温度记录；(c) 学生统计的过往车辆数；(d) 从树上采摘的苹果重量。

Primary data is data you collect yourself for a specific investigation. Secondary data is obtained from existing sources. Discrete data can only take certain isolated values (often whole numbers), while continuous data can take any value within a range and is usually measured. For (a), the teacher measures the heights, so it is primary data. Height is continuous. (b) The temperatures were recorded by a weather station, so this is secondary data. Temperature is continuous. (c) The student personally counts the cars, making it primary data. The number of cars is discrete. (d) Directly weighing apples gives primary data, and weight is continuous.

一手数据是你为特定调查亲自收集的数据；二手数据来自已有资料。离散数据只能取特定、通常为整数的值，而连续数据可在区间内取任意值，通常是测量得出的。分析：(a) 老师测量身高，是一手数据，身高连续。(b) 温度由气象站记录，是二手数据，温度连续。(c) 学生亲自数车，一手数据，车辆数离散。(d) 直接称重得一手数据，重量连续。

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2. Question 2: Frequency Table and Mean | 题目2：频数分布表与均值

A grouped frequency table shows the heights of 50 plants in cm. Classes: 140 – 150 (frequency 5), 150 – 160 (12), 160 – 170 (20), 170 – 180 (10), 180 – 190 (3). You are asked to estimate the mean height and identify the class containing the median.

一张分组频数表给出了 50 株植物的高度（cm）。组距和频数：140 – 150（5株），150 – 160（12），160 – 170（20），170 – 180（10），180 – 190（3）。要求估计平均高度并指出中位数所在的组。

First, find the midpoint of each class: 145, 155, 165, 175, 185. Multiply each midpoint by its frequency: 145×5=725, 155×12=1860, 165×20=3300, 175×10=1750, 185×3=555. Sum these products: 725+1860+3300+1750+555 = 8190. Total frequency Σf = 50. The estimated mean is 8190 ÷ 50 = 163.8 cm.

先求每组中点：145、155、165、175、185。频数乘中点：145×5=725，155×12=1860，165×20=3300，175×10=1750，185×3=555。总和 = 8190。总频数 Σf = 50。估计均值 = 8190 ÷ 50 = 163.8 cm。

Estimated Mean = Σ(f × midpoint) ÷ Σf = 163.8 cm

For the median, find the position: (50+1) ÷ 2 = 25.5th value. Cumulative frequencies: 5, 17, 37, 47, 50. The 25.5th value lies in the class with cumulative frequency 37, which is 160 – 170 cm. So the median class is 160 ≤ h < 170.

中位数位置为 (50+1) ÷ 2 = 25.5。累积频数：5、17、37、47、50。第 25.5 个值落在累积频数 37 对应的组，即 160 – 170 cm。所以中位数所在组为 160 ≤ h < 170。

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3. Question 3: Bar Charts and Pie Charts | 题目3：条形图与饼图

The favourite colours of 55 students are: Red 15, Blue 25, Green 10, Yellow 5. You need to draw a bar chart and calculate the angles for a pie chart.

55 名学生最喜欢的颜色：红色 15 人，蓝色 25 人，绿色 10 人，黄色 5 人。题目要求绘制条形图并计算饼图中各扇形的圆心角。

For the bar chart, each colour is a category on the horizontal axis. The vertical axis must be scaled appropriately, say up to 30. Draw four bars with heights equal to the frequencies. Remember to label both axes and give the chart a title. Bars should be of equal width with gaps between them.

条形图中，横轴为颜色类别，纵轴需适当缩放（例如最大至 30）。绘制四根柱，高度等于频数。务必标记坐标轴并给出标题，柱宽相等且柱间留有间隙。

For a pie chart, the total frequency is 55. The angle for each colour is (frequency ÷ 55) × 360°. Red: (15 ÷ 55) × 360° ≈ 98.2°; Blue: (25 ÷ 55) × 360° ≈ 163.6°; Green: (10 ÷ 55) × 360° ≈ 65.5°; Yellow: (5 ÷ 55) × 360° ≈ 32.7°. Check: approximately 98.2 + 163.6 + 65.5 + 32.7 = 360°.

饼图中，总频数为 55。各颜色角度 = (频数÷55) × 360°。红色：(15÷55)×360°≈98.2°；蓝色：(25÷55)×360°≈163.6°；绿色：(10÷55)×360°≈65.5°；黄色：(5÷55)×360°≈32.7°。总和约 360°，正确。

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4. Question 4: Box Plots and Comparisons | 题目4：箱线图与比较

Two sets of summary statistics are given for test scores. Set A: Min 2, Q1 5, Median 8, Q3 12, Max 18. Set B: Min 4, Q1 6, Median 9, Q3 11, Max 15. Draw box plots and compare the distributions.

给出了两组考试分数的五数概括。A组：最小值2，下四分位数5，中位数8，上四分位数12，最大值18。B组：最小值4，下四分位数6，中位数9，上四分位数11，最大值15。要求绘制箱线图并比较分布。

Draw a scale from 0 to 20. For Set A, mark the five numbers with small vertical lines. Draw the box from Q1 (5) to Q3 (12) with a line at the median (8). Extend whiskers to the min (2) and max (18). Repeat for Set B: box from 6 to 11 with median at 9, whiskers to 4 and 15. Place the second box plot immediately below the first for easy comparison.

先绘制从 0 到 20 的刻度。对于 A 组，用小竖线标出五个数值。从 Q1 5 到 Q3 12 画矩形，并在中位数 8 处画线。触须延伸到最小值 2 和最大值 18。B 组相同：矩形从 6 到 11，中位数 9，触须至 4 和 15。将第二张箱线图画在正下方以便比较。

Comparing: Set A has a wider spread (range 16) than Set B (range 11). The interquartile range for A is 12 – 5 = 7, while for B it is 11 – 6 = 5, so A is more variable in the middle 50%. Set B has a higher median, suggesting overall performance was better. Set A is slightly positively skewed (right-skewed) because the whisker from Q3 to Max is longer than from Min to Q1, while Set B appears more symmetric.

比较：A 组的离散程度更大（极差 16），B 组极差仅为 11。A 组的四分位距为 12 – 5 = 7，B 组为 5，说明 A 组中间 50% 的数据更分散。B 组中位数更高，表明整体成绩较好。A 组略显正偏态（右偏），因为 Q3 到最大值的触须较长；而 B 组更趋近对称。

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5. Question 5: Probability Calculations | 题目5：概率计算

A bag contains 3 red, 5 blue and 2 green marbles. A marble is taken at random. Find: (a) P(blue); (b) P(red or green); (c) if the first marble is blue and not replaced, find the probability that a second marble is also blue.

袋中有 3 个红球、5 个蓝球和 2 个绿球。随机抽出一个。求：(a) 抽到蓝球的概率；(b) 抽到红球或绿球的概率；(c) 若先抽到一个蓝球且不放回，再抽一个仍是蓝球的概率。

Total marbles = 3 + 5 + 2 = 10. (a) P(blue) = number of blue ÷ total = 5/10 = 1/2. (b) Red or green: these are mutually exclusive, so add their probabilities = (3/10) + (2/10) = 5/10 = 1/2. Alternatively, P(not blue) = 1 – 1/2 = 1/2. (c) After removing one blue, there are now 4 blue left out of 9 marbles. So P(second blue) = 4/9.

总数 = 3 + 5 + 2 = 10。(a) P(蓝) = 5/10 = 1/2。(b) 红或绿：互斥事件，概率相加 = 3/10 + 2/10 = 5/10 = 1/2；或用 1 – 1/2 = 1/2。(c) 取出一个蓝球后，剩下 4 个蓝球，共 9 个球，因此 P(第二个蓝球) = 4/9。

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

Data shows the number of hours (x) a student revises and the test score (y). Pairs: (2, 45), (5, 55), (1, 30), (6, 70), (4, 50), (3, 40). Plot a scatter graph, describe the correlation, and draw a line of best fit. Use it to estimate the score for 7 hours of revision.

数据给出了学生复习时间（x, 小时）与测试成绩（y）的配对：(2, 45), (5, 55), (1, 30), (6, 70), (4, 50), (3, 40)。绘制散点图，描述相关性并画一条最佳拟合线。据其估计复习 7 小时对应的分数。

Plot each point on axes labelled “Hours of revision” (x, 0-7) and “Test score” (y, 0-80). The points show an upward trend, indicating positive correlation. The line of best fit should pass through the mean point (x̄, ȳ). Means: x̄ = (2+5+1+6+4+3)/6 = 21/6 = 3.5; ȳ = (45+55+30+70+50+40)/6 = 290/6 ≈ 48.3. Draw a straight line through (3.5, 48.3) that balances points above and below. Extend the line to x=7. Reading across gives an estimated score of about 78-80.

在横轴为“复习时间”（0-7）、纵轴为“测试成绩”（0-80）的坐标系上描点。各点呈上升趋势，表明存在正相关。最佳拟合线应通过均值点 (x̄, ȳ)。均值：x̄ ≈ 3.5，ȳ ≈ 48.3。画一条穿过均值点并大致平分两侧散点的直线。延伸到 x=7 处，读取 y 值约为 78-80。

Best fit line: Score ≈ 8.6 × Hours + 18 (approx.) giving 78 for 7 hours.

It is vital to describe the correlation as “positive” and note that the relationship appears linear. Avoid using the line to predict too far beyond the given data range (extrapolation can be unreliable).

描述相关性时务必使用“正相关”，并指出关系近似线性。避免超出数据范围进行过多外推，否则预测将不可靠。

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7. Question 7: Cumulative Frequency and Quartiles | 题目7：累积频率与四分位数

The mass of 80 newborn babies (kg) is grouped. Classes: 2.0-2.5 (f=6), 2.5-3.0 (f=14), 3.0-3.5 (f=24), 3.5-4.0 (f=22), 4.0-4.5 (f=10), 4.5-5.0 (f=4). Construct a cumulative frequency table, draw the curve and estimate the median and interquartile range.

80 名新生儿的体重（kg）分组数据：组距 2.0-2.5（频数 6），2.5-3.0（14），3.0-3.5（24），3.5-4.0（22），4.0-4.5（10），4.5-5.0（4）。建立累积频率表，绘制曲线并估算中位数和四分位距。

Use upper class boundaries for plotting: 2.5, 3.0, 3.5, 4.0, 4.5, 5.0. Cumulative frequencies: 6, 20 (6+14), 44 (20+24), 66 (44+22), 76 (66+10), 80 (76+4). Plot points (upper boundary, cumulative frequency). Join with a smooth curve. For median, locate the 40th value (½ of 80). Draw a horizontal line from 40 to the curve, then down to the mass axis — approximately 3.35 kg. For Q1 (20th value) ≈ 2.95 kg, Q3 (60th value) ≈ 3.85 kg. IQR = 3.85 – 2.95 = 0.90 kg.

画累积频率曲线时使用组的上界：2.5, 3.0, 3.5, 4.0, 4.5, 5.0。累积频率：6, 20, 44, 66, 76, 80。描点（上界，累积频率）并用光滑曲线连接。中位数对应第 40 个值，从 40 画水平线交曲线再向下读数，约为 3.35 kg。下四分位数 Q1（第 20 个值）≈ 2.95 kg，上四分位数 Q3（第 60 个值）≈ 3.85 kg。四分位距 IQR = 0.90 kg。

Median ≈ 3.35 kg, Q1 ≈ 2.95 kg, Q3 ≈ 3.85 kg, IQR = 0.90 kg

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8. Question 8: Histograms with Unequal Class Widths | 题目8：不等组距直方图

The time (minutes) students spent on a quiz: 0-10 (14 students), 10-15 (12), 15-20 (8), 20-30 (10), 30-50 (6). Draw a histogram and comment on the shape.

学生完成测验所用时间（分钟）：0-10（14人），10-15（12），15-20（8），20-30（10），30-50（6）。绘制直方图并评论分布形状。

In a histogram, area represents frequency. Since class widths differ, calculate frequency density: frequency density = frequency ÷ class width. For 0-10: width = 10, density = 14 ÷ 10 = 1.4. 10-15: width = 5, density = 12 ÷ 5 = 2.4. 15-20: width = 5, density = 8 ÷ 5 = 1.6. 20-30: width = 10, density = 10 ÷ 10 = 1.0. 30-50: width = 20, density = 6 ÷ 20 = 0.3. Plot frequency density on the vertical axis. The bars have no gaps. The histogram peaks in the 10-15 minute interval and is positively skewed (right-skewed), with a long tail for higher times.

在直方图中，面积代表频数。因组距不同，需计算频数密度：频数密度 = 频数 ÷ 组距。0-10：宽度 10，密度 14÷10=1.4。10-15：宽度 5，密度 12÷5=2.4。15-20：密度 8÷5=1.6。20-30：密度 10÷10=1.0。30-50：密度 6÷20=0.3。纵轴标为频数密度，矩形间无间隙。直方图在 10-15 分钟处达到峰值，呈正偏态（右偏），右侧有长尾。

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9. Question 9: Moving Averages and Time Series | 题目9：移动平均值与时间序列

Quarterly sales (£1000s) for a shop: Q1-8, Q2-12, Q3-15, Q4-10, Q1-9, Q2-14, Q3-17, Q4-11. Calculate the four-point moving averages and comment on the trend.

某商店季度销售额（千英镑）：Q1-8, Q2-12, Q3-15, Q4-10, Q1-9, Q2-14, Q3-17, Q4-11。计算四点移动平均值并评论趋势。

Four-point moving averages smooth out seasonal variation. For the first four quarters, sum = 8+12+15+10 = 45, moving total = 45, but the average is placed between Q2 and Q3. Average = 45 ÷ 4 = 11.25. Next, drop Q1, add next Q1 (9): 12+15+10+9 = 46, average = 46 ÷ 4 = 11.5. Continue: 15+10+9+14 = 48, average = 12.0; 10+9+14+17 = 50, average = 12.5; 9+14+17+11 = 51, average = 12.75. The moving averages are 11.25, 11.5, 12.0, 12.5, 12.75. Plot these at the midpoints of the time intervals. The trend shows a clear increase over time, suggesting growing sales.

四点移动平均值可消除季节性波动。前四个季度总和 = 8+12+15+10 = 45，移动平均 = 45÷4 = 11.25，时间点位于 Q2 与 Q3 之间。接着，去掉 Q1，加入下一个 Q1 值 9：12+15+10+9 = 46，平均 = 11.5。继续：15+10+9+14=48，平均=12.0；10+9+14+17=50，平均=12.5；9+14+17+11=51，平均=12.75。这些移动平均值（11.25, 11.5, 12.0, 12.5, 12.75）绘制在对应时间段中点。趋势明显上升，表明销售持续增长。

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10. Question 10: Sampling Bias and Inference | 题目10：抽样偏差与推断

A school wants to know the average time students spend on homework. Describe two potential sources of bias: asking only students from a single class, and using a questionnaire placed on a social media group. Suggest how to obtain a more representative sample.

某学校想了解学生花在家庭作业上的平均时间。指出两种潜在的偏差来源：仅询问一个班级的学生，以及在社交媒体群组中发放问卷。给出如何获取更具代表性样本的建议。

If only one class is surveyed, the sample is not representative because, for instance, that class may be a top set with higher study hours. This is selection bias. A social media questionnaire may lead to voluntary response bias, as only students who are highly active online or strongly opinionated tend to respond. Furthermore, some students without access to the platform are excluded.

若仅调查一个班级，样本不具代表性，比如该班可能是重点班，学习时间更长。这是选择偏差。社交媒体问卷可能导致自愿响应偏差，只有非常活跃或持有强烈意见的学生才会参与。此外，无法使用该平台的学生会被排除在外。

To improve, use stratified random sampling: divide the student body into year groups or ability levels, then randomly select participants from each group in proportion to the total. Ensure the questionnaire is distributed in a way that reaches all selected individuals, such as during a tutor period. This reduces bias and gives results that better reflect the whole school.

改进方法：使用分层随机抽样。将全体学生按年级或能力水平分层，按比例从各层随机选取参与者。确保问卷能以覆盖所有选中间个体的方式发放，例如在辅导课时间。这样可减少偏差，使结果更能代表全校。

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