📚 Year 10 Cambridge Statistics: Unit Test Mock Exam Analysis | 剑桥10年级统计：单元测试模拟卷解析
This article provides a detailed walkthrough of a mock unit test for Year 10 Cambridge Statistics. Each question targets a core syllabus area—central tendency, cumulative frequency, probability, correlation, sampling, and data representation. Use these step‑by‑step solutions to strengthen both your conceptual understanding and exam confidence.
本文深入解析了一份针对剑桥10年级统计课程的单元模拟测试。每道题都紧扣考纲核心——集中趋势、累积频率、概率、相关性、抽样以及数据展示。通过逐步研习解答过程，你可以夯实概念基础，提升应试能力。
1. Question 1: Mean, Median, Mode, and Range | 问题1：平均数、中位数、众数与范围
Given data set: 12, 15, 14, 10, 18, 14, 12, 16.
给定数据集：12, 15, 14, 10, 18, 14, 12, 16。
Step 1: Calculate the mean. Sum all values: 12 + 15 + 14 + 10 + 18 + 14 + 12 + 16 = 111. Number of values, n = 8. Mean = 111 ÷ 8 = 13.875.
步骤1：计算平均数。将所有数值相加：12 + 15 + 14 + 10 + 18 + 14 + 12 + 16 = 111。数据个数 n = 8。平均数 = 111 ÷ 8 = 13.875。
Step 2: Find the median. Arrange the data in ascending order: 10, 12, 12, 14, 14, 15, 16, 18. With n = 8 (even), the median is the average of the 4th and 5th values: (14 + 14) ÷ 2 = 14.
步骤2：求中位数。将数据按升序排列：10, 12, 12, 14, 14, 15, 16, 18。由于 n = 8（偶数），中位数是第4和第5个值的平均数：(14 + 14) ÷ 2 = 14。
Step 3: Identify the mode. The value 14 appears twice and 12 also appears twice. All other values appear once. The data set is bimodal, with modes 12 and 14.
步骤3：找出众数。数值14出现两次，12也出现两次，其余数值各出现一次。该数据集是双众数的，众数为12和14。
Step 4: Calculate the range. Range = maximum − minimum = 18 − 10 = 8.
步骤4：计算范围。范围 = 最大值 − 最小值 = 18 − 10 = 8。
2. Question 2: Estimating the Mean from a Grouped Frequency Table | 问题2：根据分组频率表估算平均数
The frequency table shows the time (minutes) students spent on homework. Estimate the mean time.
频率表展示了学生做作业所用的时间（分钟）。请估算平均时间。
| Time (min) | Frequency |
|---|---|
| 0–10 | 4 |
| 10–20 | 6 |
| 20–30 | 12 |
| 30–40 | 8 |
| 40–50 | 2 |
Step 1: Find the midpoint (x) of each class. Midpoint = (lower boundary + upper boundary) ÷ 2. The midpoints are 5, 15, 25, 35, 45.
步骤1：计算每个组的中点 (x)。中点 = (组下限 + 组上限) ÷ 2。中点分别为 5, 15, 25, 35, 45。
Step 2: Multiply each midpoint by its frequency (f × x). 5×4=20, 15×6=90, 25×12=300, 35×8=280, 45×2=90. Sum these products: 20+90+300+280+90 = 780.
步骤2：每个中点乘以对应的频数 (f × x)。5×4=20, 15×6=90, 25×12=300, 35×8=280, 45×2=90。将这些乘积相加：20+90+300+280+90 = 780。
Step 3: Divide the total of f × x by the total frequency. Total frequency = 4+6+12+8+2 = 32. Estimated mean = 780 ÷ 32 = 24.375 minutes.
步骤3：用 f × x 的总和除以总频数。总频数 = 4+6+12+8+2 = 32。估算的平均时间 = 780 ÷ 32 = 24.375 分钟。
3. Question 3: Cumulative Frequency – Median and Interquartile Range | 问题3：累积频率——中位数与四分位距
The table below shows the scores of 50 students in a test. Use a cumulative frequency graph to estimate the median and interquartile range.
下表显示了50名学生的测试成绩。请使用累积频率图估算中位数和四分位距。
| Score | Frequency |
|---|---|
| 40–50 | 4 |
| 50–60 | 10 |
| 60–70 | 16 |
| 70–80 | 12 |
| 80–90 | 6 |
| 90–100 | 2 |
Step 1: Construct the cumulative frequency column. Add frequencies successively: 4, 4+10=14, 14+16=30, 30+12=42, 42+6=48, 48+2=50. The cumulative frequencies are 4, 14, 30, 42, 48, 50. Plot points at the upper class boundaries (50, 60, 70, 80, 90, 100) and draw a smooth curve.
步骤1：构建累积频率列。逐次累加频数：4, 4+10=14, 14+16=30, 30+12=42, 42+6=48, 48+2=50。累积频率为 4, 14, 30, 42, 48, 50。以各组上限（50, 60, 70, 80, 90, 100）为横坐标描点，并绘制平滑曲线。
Step 2: Estimate the median. The median position is at half the total frequency: 50 ÷ 2 = 25. Draw a horizontal line from 25 on the cumulative frequency axis to the curve, then vertically down to the score axis. From the curve, median ≈ 66.
步骤2：估算中位数。中位数的位置在总频数的一半处：50 ÷ 2 = 25。从累积频率轴的25处画水平线与曲线相交，再垂直向下读取分数轴。由曲线可得，中位数 ≈ 66。
Step 3: Estimate the lower quartile (Q1) and upper quartile (Q3). Q1 position = 50 ÷ 4 = 12.5 → Q1 ≈ 58. Q3 position = 3×50 ÷ 4 = 37.5 → Q3 ≈ 76.
步骤3：估算下四分位数（Q1）和上四分位数（Q3）。Q1 位置 = 50 ÷ 4 = 12.5 → Q1 ≈ 58。Q3 位置 = 3×50 ÷ 4 = 37.5 → Q3 ≈ 76。
Step 4: Calculate the interquartile range (IQR). IQR = Q3 − Q1 = 76 − 58 = 18.
步骤4：计算四分位距（IQR）。四分位距 = Q3 − Q1 = 76 − 58 = 18。
4. Question 4: Box-and-Whisker Plot | 问题4：盒形图
Using the five‑number summary from Question 3 (minimum = 40, Q1 = 58, median = 66, Q3 = 76, maximum = 100), draw and interpret the box plot.
使用问题3中的五数概括（最小值 = 40，Q1 = 58，中位数 = 66，Q3 = 76，最大值 = 100），绘制并解读盒形图。
Step 1: Sketch the box plot. Draw a scale on the horizontal axis. Mark the minimum (40) and maximum (100) with short vertical lines. Draw a box from Q1 (58) to Q3 (76) and a vertical line inside the box at the median (66). Connect the box to the extreme values with whiskers.
步骤1：绘制盒形图。在横轴上标出刻度。用短竖线标出最小值（40）和最大值（100）。从 Q1（58）到 Q3（76）画一个矩形盒，在盒内中位数（66）处画一条竖线。用须线将盒子与极值连接。
Step 2: Interpret the box plot. The box shows the middle 50% of scores, which range from 58 to 76. The median (66) is closer to Q1, indicating a slight positive skew. The right whisker is longer than the left, confirming a longer tail towards higher scores.
步骤2：解读盒形图。盒子表示中间50%的分数，范围从58到76。中位数（66）更靠近 Q1，显示分布轻微正偏。右须线比左须线长，证实高分段有更长的尾部。
5. Question 5: Basic Probability with Mutually Exclusive Events | 问题5：互斥事件的基本概率
A bag contains 3 red balls, 2 blue balls, and 5 green balls. One ball is selected at random. Find the probability that the ball is (a) red or blue; (b) not green.
一个袋子中装有3个红球、2个蓝球和5个绿球。随机抽取一个球。求该球是 (a) 红色或蓝色的概率；(b) 不是绿色的概率。
Step 1: Determine total number of outcomes. Total balls = 3 + 2 + 5 = 10.
步骤1：确定结果总数。球的总数 = 3 + 2 + 5 = 10。
Step 2: Part (a) – red or blue. These events are mutually exclusive. Number of favourable outcomes = 3 + 2 = 5. P(red or blue) = 5 ÷ 10 = ½ = 0.5.
步骤2：(a) 部分——红色或蓝色。这些事件互斥。有利结果数 = 3 + 2 = 5。P(红色或蓝色) = 5 ÷ 10 = ½ = 0.5。
Step 3: Part (b) – not green. ‘Not green’ means red or blue, so favourable = 5. Alternatively, P(not green) = 1 − P(green) = 1 − (5/10) = 0.5.
步骤3：(b) 部分——不是绿色。“不是绿色”意味着红色或蓝色，有利结果 = 5。或者用互补事件计算：P(不是绿色) = 1 − P(绿色) = 1 − (5/10) = 0.5。
6. Question 6: Scatter Plot and Correlation | 问题6：散点图与相关性
The table shows the values of two variables, x and y. Plot a scatter diagram and describe the correlation. Estimate the value of y when x = 4.5 using a line of best fit.
下表给出了两个变量 x 和 y 的数值。绘制散点图并描述其相关性。利用最佳拟合线估计 x = 4.5 时的 y 值。
| x | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|---|---|
| y | 2 | 4 | 5 | 7 | 8 | 10 | 12 |
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