Year 11 Cambridge Statistics: Mock Unit Test Walkthrough | Year 11 剑桥统计：单元测试模拟卷解析

📚 Year 11 Cambridge Statistics: Mock Unit Test Walkthrough | Year 11 剑桥统计：单元测试模拟卷解析

This article works through a full mock unit test for the Year 11 Cambridge Statistics syllabus. Every question is broken down step by step, highlighting key techniques for handling data, calculating summary statistics, interpreting graphs, and applying probability.

本文完整解析一套针对 Year 11 剑桥统计的单元测试模拟卷。每道题目都逐步拆解，重点演示数据处理、汇总统计量计算、图表解读以及概率应用的关键技巧。

1. Statistical Enquiry Cycle and Data Types | 统计探究循环与数据类型

Question 1 (mock test): A school council wants to investigate students’ lunch habits. They ask: ‘Which year group are you in? How many minutes do you spend eating lunch? Do you buy hot meals, bring a packed lunch, or go home?’ Identify each variable as categorical or numerical. Then suggest one possible follow‑up question that would generate discrete numerical data.

模拟卷第1题：学校学生会想调查学生午餐习惯。他们提问：“你所在的年级？你吃午餐花了多少分钟？你买热餐、自带盒饭还是回家吃？” 判断每个变量是分类变量还是数值变量。然后提出一个能产生离散数值数据的后续问题。

Solution: Year group is categorical (ordinal, since years are ordered groups), time spent eating lunch is continuous numerical (could be measured to any precision), and type of lunch is categorical (nominal). A follow‑up question generating discrete numerical data could be ‘How many days last week did you buy a hot meal?’ – the answer is a whole number count.

解析：年级是分类变量（有序分类，因为年级有顺序），午餐时间是连续数值变量（可以测得无限精确），午餐类型是分类变量（名义分类）。一个能产生离散数值数据的后续问题可以是“上周你有几天买了热餐？”——答案是整数计数。

Understanding the difference between qualitative and quantitative variables, and between discrete and continuous numerical data, is tested in almost every Cambridge Statistics unit test.

区分定性变量与定量变量、离散数值与连续数值数据，几乎是每次剑桥统计单元测试必考的内容。

2. Frequency Tables and Bar Charts | 频数表与条形图

Question 2: The number of siblings for 25 students is recorded: 0, 2, 1, 1, 3, 0, 2, 1, 4, 1, 2, 0, 2, 3, 1, 1, 0, 5, 2, 1, 2, 3, 1, 0, 1. Construct a frequency table and draw a bar chart for this discrete data.

模拟卷第2题：记录了25名学生的兄弟姐妹个数：0, 2, 1, 1, 3, 0, 2, 1, 4, 1, 2, 0, 2, 3, 1, 1, 0, 5, 2, 1, 2, 3, 1, 0, 1。为这组离散数据构建频数表并绘制条形图。

Number of siblings (x)	Tally	Frequency (f)
0	||||	5
1	|||| ||||	9
2	|||| |	6
3	|||	3
4	|	1
5	|	1

The bar chart is drawn with siblings on the horizontal axis and frequency on the vertical axis. Bars are separated to show discrete categories. The heights are 5, 9, 6, 3, 1 and 1 respectively.

条形图的横轴为兄弟姐妹个数，纵轴为频数。条形之间留有间隔，以体现离散分类。各条形高度分别为5、9、6、3、1、1。

When constructing frequency tables on a Cambridge mock, always check that the total frequency equals the sample size. Here 5+9+6+3+1+1 = 25, which matches the number of students.

在剑桥模拟卷中构建频数表时，务必检查总频数是否等于样本容量。这里5+9+6+3+1+1=25，与学生人数一致。

3. Measures of Central Tendency – Mean, Median, Mode | 集中趋势——均值、中位数、众数

Question 3 (continuing from the data in Question 2): Calculate the mean, median and mode for the number of siblings. Explain why the median may be a better measure of central tendency than the mean in this context.

模拟卷第3题（沿用第2题的数据）：计算兄弟姐妹个数的均值、中位数和众数。解释在此背景下中位数为何可能比均值更适合作为集中趋势的度量。

Mean: x̄ = (0×5 + 1×9 + 2×6 + 3×3 + 4×1 + 5×1) ÷ 25 = (0+9+12+9+4+5) ÷ 25 = 39 ÷ 25 = 1.56 siblings.

均值：x̄ = (0×5 + 1×9 + 2×6 + 3×3 + 4×1 + 5×1) ÷ 25 = (0+9+12+9+4+5) ÷ 25 = 39 ÷ 25 = 1.56 个兄弟姐妹。

Median position = (25+1)/2 = 13th value. The ordered list has five 0s, nine 1s, etc. The 13th value falls in the group of 1s, so median = 1. Mode = 1 (highest frequency, 9).

中位数位置 = (25+1)/2 = 第13个值。排序数据中有5个0、9个1……第13个值落在1的分组中，因此中位数 = 1。众数 = 1（频数最高，为9）。

The mean is 1.56, pulled upwards by the few students who have 4 or 5 siblings. The median (1) and mode (1) are not affected by these outliers, so the median gives a more typical value for the majority of students.

均值为1.56，被少数拥有4个或5个兄弟姐妹的学生拉高。中位数（1）和众数（1）不受这些异常值影响，因此中位数给出了更符合大多数学生情况的典型值。

4. Measures of Spread – Range, IQR and Standard Deviation | 离散程度——极差、四分位距与标准差

Question 4: Using the sibling data, find the range, interquartile range (IQR) and estimate the standard deviation. The sum of squares Σ(x – x̄)² is given as 29.36. Interpret what the IQR tells you.

模拟卷第4题：利用兄弟姐妹数据，求极差、四分位距（IQR），并计算标准差。已知离差平方和 Σ(x – x̄)² = 29.36。解读四分位距的含义。

Range = max – min = 5 – 0 = 5 siblings. For IQR: n=25. Q₁ is at position (25+1)/4 = 6.5th value, between the 6th and 7th ordered data points. Both are 1, so Q₁ = 1. Q₃ is at 3(25+1)/4 = 19.5th value, between the 19th and 20th points. The 19th value is 2, the 20th is 2, so Q₃ = 2. Thus IQR = Q₃ – Q₁ = 2 – 1 = 1.

极差 = 最大值 – 最小值 = 5 – 0 = 5 个兄弟姐妹。计算IQR：n=25。Q₁ 位于 (25+1)/4 = 第6.5个值，介于第6和第7个有序数据之间，两者均为1，故 Q₁ = 1。Q₃ 位于 3(25+1)/4 = 第19.5个值，介于第19和第20个之间，第19和第20个值均为2，故 Q₃ = 2。因此 IQR = Q₃ – Q₁ = 2 – 1 = 1。

Standard deviation for a sample: s = √[ Σ(x – x̄)² / (n – 1) ] = √(29.36 / 24) ≈ √1.2233 ≈ 1.106 siblings. The IQR of 1 tells us that the middle 50% of students differ by at most 1 sibling, indicating very low variability around the median.

样本标准差：s = √[ Σ(x – x̄)² / (n – 1) ] = √(29.36 / 24) ≈ √1.2233 ≈ 1.106 个兄弟姐妹。IQR为1说明中间50%的学生兄弟姐妹个数最多相差1，表明中位数周围的变异非常小。

5. Cumulative Frequency and Box Plots | 累积频率与箱线图

Question 5: The waiting times (in minutes) at a bus stop for 50 passengers are summarised in the grouped frequency table below. Draw a cumulative frequency graph, then estimate the median and interquartile range. Finally, construct a box plot.

模拟卷第5题：下表汇总了50名乘客在公交车站的等待时间（分钟）。绘制累积频率图，估计中位数和四分位距，最后构建箱线图。

Waiting time (min)	Frequency
0 ≤ t < 5	8
5 ≤ t < 10	14
10 ≤ t < 15	17
15 ≤ t < 20	9
20 ≤ t < 25	2

We first build a cumulative frequency column: upper class boundaries 5, 10, 15, 20, 25; cumulative frequencies 8, 22, 39, 48, 50. Plot points at (5,8), (10,22), (15,39), (20,48), (25,50) and join with a smooth curve starting at (0,0).

首先建立累积频率列：组上限为5、10、15、20、25；累积频数分别为8、22、39、48、50。在坐标平面上描出点(5,8)、(10,22)、(15,39)、(20,48)、(25,50)，并从(0,0)开始用光滑曲线连接。

From the graph, the median corresponds to the 25th value (n/2 = 25). Reading across at cumulative frequency 25 gives a median of about 10.8 minutes. Q₁ is at 12.5th value ≈ 6.4 min; Q₃ at 37.5th value ≈ 14.2 min. IQR = 14.2 – 6.4 = 7.8 minutes. The box plot uses the five‑number summary: min = 0, Q₁ = 6.4, median = 10.8, Q₃ = 14.2, max = 25.

从图上，中位数对应第25个值（n/2=25）。在累积频率25处读取得到中位数约10.8分钟。Q₁在第12.5个值 ≈ 6.4 分钟；Q₃在第37.5个值 ≈ 14.2 分钟。IQR = 14.2 – 6.4 = 7.8 分钟。箱线图使用五数概括：最小值=0，Q₁=6.4，中位数=10.8，Q₃=14.2，最大值=25。

Always label axes on the cumulative frequency graph and draw dotted lines to show how the quartiles are read. Box plots must be drawn to scale, and whiskers extend to the minimum and maximum values (unless outliers are present).

累积频率图必须标注坐标轴，并用虚线显示四分位数的读取方法。箱线图必须按比例绘制，触须延伸至最小值和最大值（除非存在异常值）。

6. Probability – Simple and Combined Events | 概率——简单与复合事件

Question 6: A bag contains 5 red pens, 3 blue pens and 2 green pens. Two pens are drawn at random without replacement. Work out the probability that at least one pen is red. Express your answer as a fraction in simplest form.

模拟卷第6题：一个袋子里有5支红笔、3支蓝笔和2支绿笔。随机抽取两支笔，不放回。计算至少有一支红笔的概率，并将答案写成最简分数。

Total pens = 10. It is easier to calculate P(no red) = P(both not red). Not red pens = 5 (3 blue + 2 green). P(1st not red) = 5/10 = 1/2. After one non‑red is taken, 4 non‑red out of 9 remain: P(2nd not red) = 4/9. So P(no red) = (5/10) × (4/9) = 20/90 = 2/9. Therefore P(at least one red) = 1 – 2/9 = 7/9.

总笔数 = 10。更容易的方法是计算 P(无红笔) = P(两支都不是红笔)。非红笔共5支（3蓝+2绿）。P(第1支非红) = 5/10 = 1/2。取走一支非红后，剩余9支中有4支非红：P(第2支非红) = 4/9。因此 P(无红笔) = (5/10) × (4/9) = 20/90 = 2/9。于是 P(至少一支红笔) = 1 – 2/9 = 7/9。

Alternatively, a tree diagram shows all branches. Remember that ‘at least one’ problems are often solved fastest using the complement rule: 1 – P(neither).

也可以用树状图展示所有分支。记住，“至少一个”的问题通常用补集规则最快解决：1 – P(都不发生)。

7. Scatter Diagrams and Correlation | 散点图与相关性

Question 7: The table shows the number of hours 7 students spent revising for a test and their test scores. Plot a scatter diagram, describe the correlation, and draw a line of best fit by eye. Use your line to estimate the score for a student who revised for 4.5 hours.

模拟卷第7题：下表给出7名学生复习小时数和测试分数。绘制散点图，描述相关性，并目测画出最佳拟合线。利用你的线估计复习4.5小时的学生的分数。

Hours (x)	Score (y)
1.0	42
2.0	55
2.5	59
3.0	68
4.0	80
5.0	91
5.5	92

When plotted, the points rise from bottom left to top right, showing strong positive correlation. The line of best fit should pass through the ‘centre’ of the points, roughly balancing points above and below. One possible line passes through (2, 55) and (5, 91). Using this line, at x = 4.5 hours the estimated score is about 84.

绘制后，点从左下向右上攀升，显示强正相关。最佳拟合线应穿过点的“中心”，大致使线上方的点和下方的点平衡。可能的一条线经过(2, 55)和(5, 91)。利用这条线，在 x = 4.5 小时处估计分数约为84。

When describing correlation, use the phrases ‘strong positive’, ‘weak negative’, etc., and always refer to the context. Never extrapolate far beyond the data range when using a line of best fit.

描述相关性时，使用“强正相关”、“弱负相关”等术语，并始终结合上下文。在利用最佳拟合线进行预测时，不要过度外推到数据范围之外很远。

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

Question 8: A shop records quarterly ice cream sales (in hundreds of units) over three years. Calculate a four‑point moving average for the first five quarters and comment on the trend.

模拟卷第8题：一家商店记录了三年内每季度冰淇淋销量（单位：百件）。计算前五个季度的四点移动平均，并评论趋势。

Quarter	Sales
Y1 Q1	23
Y1 Q2	45
Y1 Q3	62
Y1 Q4	28
Y2 Q1	27
Y2 Q2	49

Four‑point moving averages are placed between quarters. For the first average: (23+45+62+28) ÷ 4 = 158 ÷ 4 = 39.5, centred between Q2 and Q3 of Y1. Second: (45+62+28+27) ÷ 4 = 162 ÷ 4 = 40.5, between Y1 Q3 and Y1 Q4. Third: (62+28+27+49) ÷ 4 = 166 ÷ 4 = 41.5, between Y1 Q4 and Y2 Q1. The moving averages show a slight upward trend, with sales gradually increasing over time after smoothing out seasonal fluctuations.

四点移动平均放置在两个季度中间。第一个平均：(23+45+62+28) ÷ 4 = 158 ÷ 4 = 39.5，位于Y1的第2季度和第3季度之间。第二个：(45+62+28+27) ÷ 4 = 162 ÷ 4 = 40.5，位于Y1 Q3与Y1 Q4之间。第三个：(62+28+27+49) ÷ 4 = 166 ÷ 4 = 41.5，位于Y1 Q4与Y2 Q1之间。移动平均显示轻微上升趋势，在消除季节性波动后，销量随时间逐渐增长。

Remember that a 4‑point moving average smooths out quarterly variations, allowing us to see the underlying trend. Always label the time axis clearly and plot the centred moving averages on the same graph as the original series.

记住，四点移动平均可以平滑季度波动，让我们看到潜在趋势。务必清楚地标注时间轴，并将中心化后的移动平均与原序列画在同一张图上。

9. Using Statistical Diagrams to Compare Distributions | 利用统计图比较分布

Question 9: Parallel box plots are drawn for the test scores of two classes, A and B. The five‑number summary for class A: min=32, Q₁=54, median=67, Q₃=80, max=98. For class B: min=45, Q₁=60, median=72, Q₃=82, max=96. Compare the central tendency and spread of the two classes.

模拟卷第9题：为A班和B班的测试分数绘制并列箱线图。A班五数概括：最小值=32，Q₁=54，中位数=67，Q₃=80，最大值=98。B班：最小值=45，Q₁=60，中位数=72，Q₃=82，最大值=96。比较两个班的集中趋势和离散程度。

From the summaries, class B has a higher median (72) than class A (67), so on average B performed better. The interquartile range for class A is 80 – 54 = 26, while for class B it is 82 – 60 = 22, showing that the middle 50% of scores in class B are less spread out. However, class A has a much lower minimum (32) and a slightly higher maximum, so its range is larger (98 – 32 = 66 vs 96 – 45 = 51). This suggests class A contains a wider range of abilities, including some very low achievers.

由数据概括可知，B班中位数（72）高于A班（67），因此平均而言B班表现更好。A班的四分位距为80 – 54 = 26，B班为82 – 60 = 22，表明B班中间50%的成绩离散程度更小。然而，A班的最小值低得多（32），最大值略高，因此极差更大（98 – 32 = 66 vs 96 – 45 = 51）。这说明A班学生能力差异更大，包含一些成绩非常低的学生。

When comparing distributions, always point to specific statistics – median for centre, IQR/range for spread – and refer to the context. Use phrases like ‘higher median indicates better typical performance’ and ‘smaller IQR shows more consistent results’.

在比较分布时，始终要引用具体的统计量——中位数用于描述中心，IQR或极差用于描述离散——并结合上下文。使用“较高的中位数表明典型表现更好”和“较小的四分位距显示成绩更一致”等表述。

10. Common Mistakes and How to Avoid Them | 常见错误及如何避免

Mock tests often reveal small but costly errors. A typical mistake is confusing the formulas for population standard deviation and sample standard deviation: always divide by (n – 1) for a sample. Another error is misreading cumulative frequency graphs – remember to use the upper class boundaries, not midpoints. When calculating probability without replacement, students frequently forget to reduce the denominator after the first pick. Also, when drawing a line of best fit, it must be a single straight line, not a curve, and should not be forced through the origin unless the context justifies it.

模拟测试往往会暴露虽小却代价高昂的错误。一个典型错误是混淆总体标准差和样本标准差的公式：对于样本，永远除以 (n – 1)。另一个错误是误读累积频率图——记住使用组上限

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