📚 Year 9 Cambridge Statistics: Unit Test Mock Paper Walkthrough | 剑桥九年级统计单元测试模拟卷解析
This walkthrough breaks down a full unit test mock paper for Year 9 Cambridge Statistics, covering measures of centre, spread, data representation, probability, and comparing distributions. Each question is explained step by step, highlighting key methods, common pitfalls, and the reasoning behind every answer. Use this to revise and build confidence before your real test.
这篇解析完整拆解了一份针对剑桥九年级统计单元的模拟测试卷，涵盖集中量数、离散程度、数据表示、概率以及分布比较。每道题都逐步讲解，突出关键方法、常见错误和答题逻辑。用它来复习，为真实测试做好充分准备。
1. Mean, Median, Mode and Range | 均值、中位数、众数和极差
The first question gives the test scores of 8 students: 12, 15, 14, 16, 12, 18, 20, 15. We need to find the mean, median, mode, and range.
第一题给出8名学生的测试分数：12, 15, 14, 16, 12, 18, 20, 15。要求计算均值、中位数、众数和极差。
umbers: 12+15+14+16+12+18+20+15 = 122. There are 8 values, so the mean is 122 ÷ 8 = 15.25.
计算均值：总和为12+15+14+16+12+18+20+15 = 122。共8个数据，均值 = 122 ÷ 8 = 15.25。
number of values, the median is the average of the 4th and 5th terms: (15 + 15) ÷ 2 = 15.
中位数：先将分数从小到大排列：12, 12, 14, 15, 15, 16, 18, 20。因为有偶数个数据，中位数取第4和第5个数的平均值：(15 + 15) ÷ 2 = 15。
The mode is the most frequent score. Here 12 and 15 both appear twice, so the data set is bimodal with modes 12 and 15.
众数是出现次数最多的分数。这里12和15都出现了两次，所以数据集是双峰的，众数为12和15。
The range is the difference between the largest and smallest values: 20 – 12 = 8.
极差是最大值与最小值之差：20 – 12 = 8。
A common mistake is forgetting to order the data before finding the median, or using the formula incorrectly for an even count. Always sort first.
常见错误是找中位数前忘记排序，或者在偶数个数据时用错公式。一定要先排序。
2. Grouped Frequency Table: Estimated Mean | 分组频数表：估算均值
This question presents grouped data: 0–10:5, 10–20:12, 20–30:8, 30–40:3. We estimate the mean using midpoints.
这道题给出了分组数据：0–10:5, 10–20:12, 20–30:8, 30–40:3。需要用组中值来估算均值。
oint of each class interval: (0+10)÷2 = 5, (10+20)÷2 = 15, (20+30)÷2 = 25, (30+40)÷2 = 35.
首先求出每个组距的组中值：(0+10)÷2 = 5，(10+20)÷2 = 15，(20+30)÷2 = 25，(30+40)÷2 = 35。
ency: 5×5=25, 15×12=180, 25×8=200, 35×3=105. Sum of these products = 25+180+200+105 = 510.
将每个组中值乘以对应的频数：5×5=25, 15×12=180, 25×8=200, 35×3=105。这些乘积的总和 = 510。
l frequency is 5+12+8+3 = 28. Estimated mean = 510 ÷ 28 ≈ 18.2 (to 3 significant figures).
总频数为 5+12+8+3 = 28。估算均值 = 510 ÷ 28 ≈ 18.2（保留三位有效数字）。
Remember this is an estimate because we assume all values in a class are at the midpoint. The true mean could be slightly different.
注意这只是估算值，因为我们假设每组内的所有数据都集中在组中值上，真实均值可能稍有不同。
3. Bar Chart Construction and Interpretation | 条形图绘制与解读
A survey on favourite sports gave: Football 15, Basketball 10, Volleyball 8, Tennis 7. We draw a bar chart and answer which sport is most popular.
一项关于最喜爱运动的调查结果：足球15人，篮球10人，排球8人，网球7人。我们要绘制条形图并指出哪项运动最受欢迎。
For a bar chart, the categories (sports) go on the horizontal axis, and frequency on the vertical axis. Bars should have equal width and gaps between them.
绘制条形图时，类别（运动）放在横轴，频数放在纵轴。条形宽度应相等，且条与条之间要有空隙。
Draw a vertical scale from 0 to 16, label the axis 'Frequency', and draw bars for each sport up to its frequency. Football's bar is the tallest at 15.
纵轴从0标到16，标注“频数”，然后为每项运动画出对应高度的条形。足球的条形最高，达到15。
From the chart it is clear that Football is the most popular sport. Bar charts are best for comparing discrete categories like this.
从图中可以清楚看出足球是最受欢迎的运动。条形图最适合这样比较离散的类别。
A common error is starting the frequency axis from a number other than 0, which exaggerates differences. Always start your frequency scale at 0.
常见错误是纵轴不从0开始，这样会夸大差异。务必让频数轴的起点为0。
4. Pie Chart Angles and Usage | 饼图角度计算与适用场景
nis 7), we calculate the sector angle for each sport in a pie chart. Total frequency = 40.
使用相同的调查数据（足球15，篮球10，排球8，网球7），计算饼图中每项运动对应的扇形角度。总频数 = 40。
(10 ÷ 40) × 360° = 90°. Volleyball: (8 ÷ 40) × 360° = 72°. Tennis: (7 ÷ 40) × 360° = 63°.
每个类别的角度 = (类别频数 ÷ 总数) × 360°。足球：(15 ÷ 40) × 360° = 135°；篮球：(10 ÷ 40) × 360° = 90°；排球：(8 ÷ 40) × 360° = 72°；网球：(7 ÷ 40) × 360° = 63°。
e angles sum to 360°: 135°+90°+72°+63° = 360°. This confirms the calculations are correct.
务必检查角度总和是否为360°：135°+90°+72°+63° = 360°。这能确认计算无误。
Pie charts are useful for showing proportions of a whole, but less effective when there are many categories. Here, it clearly shows Football's dominance.
饼图适合展示各部分占整体的比例，但当类别过多时效果较差。这里它清晰地显示了足球的主导地位。
5. Scatter Graphs and Correlation | 散点图与相关性
Data on height (cm) and weight (kg) for 8 students is plotted. The points show a general upward trend. We need to describe the correlation and draw a line of best fit.
8名学生的身高（厘米）和体重（千克）数据被绘成散点图。点的分布呈现总体上升趋势。我们需要描述相关性并画出最佳拟合线。
Describe the correlation: 'There is a strong positive correlation between height and weight. As height increases, weight tends to increase.' Do not say the relationship is perfect unless points lie on a straight line.
描述相关性：“身高与体重之间存在强正相关。随着身高增加，体重也趋于增加。”除非点完全在一条直线上，否则不要说关系是完美的。
To draw the line of best fit, place it so it follows the trend and has roughly equal numbers of points above and below the line. Do not force it through the origin unless the context suggests it.
画最佳拟合线时，让它沿着趋势走，并使线上方和下方的点数大致相等。除非实际情况表明线应经过原点，否则不要强制穿过原点。
Using the line, we can estimate the weight of a student 160 cm tall by reading across from 160 cm on the x-axis to the line, then down to the y-axis.
利用这条线，我们可以估算身高160厘米学生的体重：从横轴上160厘米处垂直向上与拟合线相交，再水平向左读纵轴上的值。
Outliers—points far from the general pattern—should be noted. They can affect the position of the line of best fit.
异常值——距离总体模式很远的点——应当被标出。它们可能会影响最佳拟合线的位置。
6. Basic Probability from a Bag of Coloured Balls | 基础概率：彩色球的抽取
A bag contains 5 red, 3 blue and 2 green balls. One ball is drawn at random. Find P(red), P(not blue) and P(red or green).
一个袋子里有5个红球、3个蓝球和2个绿球。随机抽取一个球。求 P(红)、P(非蓝) 和 P(红或绿)。
= 5 + 3 + 2 = 10. Probability = number of favourable outcomes ÷ total number of outcomes.
球的总数 = 5 + 3 + 2 = 10。概率 = 有利结果的数量 ÷ 所有可能结果的总数。
blue) means the ball is either red or green: 5 + 2 = 7 favourable, so P(not blue) = 7/10.
P(红) = 5/10 = 1/2。P(非蓝) 意味着球是红色或绿色：有利结果数为5+2=7，因此 P(非蓝) = 7/10。
een. These events are mutually exclusive, so we add the probabilities: 5/10 + 2/10 = 7/10.
P(红或绿) 是抽到红球或绿球的概率。这些事件互斥，因此我们把概率相加：5/10 + 2/10 = 7/10。
Notice that P(not blue) and P(red or green) are the same in this context. Always express probabilities as fractions, decimals or percentages as required.
注意在这个情境中 P(非蓝) 和 P(红或绿) 是相等的。务必按要求以分数、小数或百分数表示概率。
7. Tree Diagrams and Sample Spaces | 树状图与样本空间
An experiment consists of tossing a fair coin and then spinning a fair 4-sided spinner labelled 1, 2, 3, 4. Show all possible outcomes and find P(Head and an even number).
一个实验包括抛一枚均匀硬币，然后转动一个标有1,2,3,4的均匀四面转盘。列出所有可能结果，并求 P(正面且偶数)。
Draw a tree diagram: first branch for coin (H, T) each with probability 1/2. From each coin outcome, four branches for the spinner (1,2,3,4) each with probability 1/4.
画出树状图：第一层分支为硬币（正面H、反面T），概率各为1/2。从每个硬币结果出发，第二层为转盘的四个分支（1,2,3,4），概率各为1/4。
Sample space: H1, H2, H3, H4, T1, T2, T3, T4. There are 2 × 4 = 8 equally likely outcomes.
样本空间：H1, H2, H3, H4, T1, T2, T3, T4。共有 2 × 4 = 8 个等可能结果。
Favourable outcomes for 'Head and an even number': H2 and H4. That is 2 outcomes. So probability = 2/8 = 1/4.
“正面且偶数”的有利结果：H2 和 H4，共2个。因此概率 = 2/8 = 1/4。
anches: P(H) × P(even) = 1/2 × (2/4) = 1/2 × 1/2 = 1/4. Both methods give the same answer.
或者沿树状图分支相乘：P(H) × P(偶数) = 1/2 × (2/4) = 1/2 × 1/2 = 1/4。两种方法结果相同。
8. Mutually Exclusive and Non‑Mutually Exclusive Events | 互斥事件与非互斥事件
A single card is drawn from a standard 52-card deck. Find P(Heart or Spade) and P(King or Heart). State whether the events are mutually exclusive.
从一副标准52张扑克牌中随机抽一张。求 P(红心或黑桃) 和 P(K或红心)。说明这些事件是否互斥。
e. P(Heart) = 13/52, P(Spade) = 13/52, so P(Heart or Spade) = 13/52 + 13/52 = 26/52 = 1/2.
P(红心或黑桃)：它们互斥，因为一张牌不可能既是红心又是黑桃。P(红心) = 13/52，P(黑桃) = 13/52，所以 P(红心或黑桃) = 13/52 + 13/52 = 26/52 = 1/2。
se the King of Hearts belongs to both categories. Use the formula: P(A or B) = P(A) + P(B) – P(A and B).
对于 P(K或红心)，这些事件不是互斥的，因为红心K同时属于两个类别。使用公式：P(A 或 B) = P(A) + P(B) – P(A 且 B)。
ng) = 4/52, P(Heart) = 13/52, P(King and Heart) = 1/52. So P(King or Heart) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13.
P(K) = 4/52，P(红心) = 13/52，P(K且红心) = 1/52。因此 P(K或红心) = 4/52 + 13/52 – 1/52 = 16/52 = 4/13。
Always check for overlap. If events can occur at the same time, you must subtract the intersection to avoid double counting.
务必检查是否有重叠。如果事件可能同时发生，必须减去交集以避免重复计算。
9. Comparing Distributions Using Median and IQR | 利用中位数和四分位距比较分布
Two classes took the same test. Class A: median 68, IQR 12. Class B: median 70, IQR 20. Compare their performance and consistency.
两个班级参加了同一次测试。A班：中位数68，四分位距12。B班：中位数70，四分位距20。比较他们的表现和稳定性。
The median shows the central tendency. Class B has a slightly higher median, suggesting a higher typical score.
中位数反映集中趋势。B班中位数略高，说明典型成绩更高。
The IQR measures the spread of the middle 50%. A smaller IQR indicates more consistent performance. Class A's IQR of 12 is smaller than Class B's 20, so Class A performed more consistently.
四分位距衡量中间50%数据的分散程度。IQR越小，表现越稳定。A班的IQR为12，小于B班的20，所以A班的成绩更稳定。
We might say: 'Class B achieved a marginally higher median score, but Class A's marks were less spread out, meaning students in Class A scored more uniformly.'
我们可以说：“B班取得了稍高的中位数分数，但A班成绩的离散程度更小，意味着A班学生的分数更均匀。”
When comparing, always reference both a measure of centre and a measure of spread, and write full comparative sentences.
进行比较时，一定要同时引用集中量数和离散量数，并写出完整的比较性语句。
10. Cumulative Frequency and Quartiles | 累积频数与四分位数
A survey recorded daily phone usage (hours) for 60 students in grouped intervals. The cumulative frequency graph is plotted. Use it to estimate the median and quartiles.
一项调查记录了60名学生每天使用手机的时长（小时），数据已分组。绘制了累积频数图。利用该图估算中位数和四分位数。
tal frequency = 60, so median is at cumulative frequency 30. Read across to the time axis.
从累积频数曲线上看，中位数对应总频数一半的位置。总频数 = 60，因此中位数在累积频数30处，水平延伸到时间轴读取数值。
The lower quartile (Q₁) is at cumulative frequency 15 (1/4 of 60), and the upper quartile (Q₃) is at cumulative frequency 45 (3/4 of 60).
下四分位数(Q₁) 位于累积频数15处（60的1/4），上四分位数(Q₃) 位于累积频数45处（60的3/4）。
Interquartile range = Q₃ – Q₁. This estimate helps to understand the spread of the middle half of phone usage.
四分位距 = Q₃ – Q₁。这个估计值有助于了解中间一半学生手机使用时长的分布情况。
Plot the points at the upper class boundaries, not midpoints. Draw a smooth curve through the points. Avoid simply connecting with straight lines, as this would give less accurate readings.
绘制累积频数图时应将点标在组距上界，而不是组中值。用平滑曲线连接各点，避免直接用直线连接，否则读数会不够精确。
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