📚 KS3 Edexcel Statistics: Unit Test Mock Paper Solutions | KS3 Edexcel 统计：单元测试模拟卷解析
Welcome to this detailed walkthrough of a KS3 Edexcel Statistics unit test mock paper. We will work through ten typical exam-style questions, covering key topics such as averages, charts, probability and data comparison. Each question is followed by a step-by-step solution with clear explanations, designed to help you master the essential skills assessed at Key Stage 3.
欢迎来到KS3 Edexcel统计单元测试模拟卷的详细解析。我们将逐一分析十道典型的考试风格题目，涵盖平均数、图表、概率和数据比较等关键主题。每道题都附有分步解答和清晰的解释，旨在帮助你掌握KS3阶段评估的核心技能。
1. Finding the Mean and Range | 计算平均值与范围
A student recorded the time (in minutes) they spent on homework each day for a week: 30, 25, 45, 35, 40, 50, 55. Calculate the mean time and the range.
一名学生记录了一周内每天做作业的时间（分钟）：30, 25, 45, 35, 40, 50, 55。计算平均时间和范围。
+ 25 + 45 + 35 + 40 + 50 + 55 = 280 minutes. There are 7 values, so divide the total by 7.
要求平均值，首先将所有数值相加。总和为30 + 25 + 45 + 35 + 40 + 50 + 55 = 280分钟。因为有7个数值，所以将总和除以7。
Mean = Total sum ÷ Number of values = 280 ÷ 7 = 40 minutes
The mean homework time is 40 minutes. This represents a typical daily amount, but individual days vary.
平均作业时间为40分钟。这代表了一个典型的每日时长，但每天之间会有波动。
The range is the difference between the largest and smallest values. Here the largest is 55 and the smallest is 25.
范围是最大值与最小值之间的差。这里最大值是55，最小值是25。
Range = 55 – 25 = 30 minutes
A range of 30 minutes shows there is quite a bit of variation in the time spent on homework during the week.
30分钟的范围表明一周内做作业的时间存在相当大的差异。
2. Median and Mode from a List | 从列表求中位数和众数
The test scores of 9 students are: 18, 22, 18, 25, 30, 22, 18, 27, 24. Find the median and mode.
9名学生的测试分数为：18, 22, 18, 25, 30, 22, 18, 27, 24。求中位数和众数。
First, order the numbers from smallest to largest: 18, 18, 18, 22, 22, 24, 25, 27, 30.
首先，将数字从小到大排列：18, 18, 18, 22, 22, 24, 25, 27, 30。
the median is the 5th value in the ordered list, because the position is (9 + 1) ÷ 2 = 5.
因为有9个数据，中位数是排序后列表中的第5个值，因为位置是 (9 + 1) ÷ 2 = 5。
The 5th value is 22, so the median score is 22. This means half the students scored below 22 and half above.
第5个值是22，因此中位数分数为22。这意味着一半学生的分数低于22，一半高于22。
The mode is the number that appears most frequently. 18 appears three times, which is more than any other value.
众数是出现频率最高的数字。18出现了三次，比其他任何数值都多。
Therefore the mode is 18. There can be more than one mode if several values tie for the highest frequency, but here only 18 repeats most.
因此众数是18。如果有多个值以相同的最高频率并列，则可以有多个众数，但这里只有18重复次数最多。
3. Interpreting a Bar Chart | 解读条形图
The bar chart below shows quarterly sales (in £) for a small shop: Q1 = £2000, Q2 = £3500, Q3 = £2800, Q4 = £4000. Use the information to answer the questions.
下面的条形图显示了一家小店的季度销售额（英镑）：第一季度 = £2000，第二季度 = £3500，第三季度 = £2800，第四季度 = £4000。请根据信息回答问题。
Which quarter had the highest sales? The tallest bar is Quarter 4, so the highest sales are in Q4 with £4000.
哪个季度销售额最高？最高的条形是第四季度，因此销售额最高的是Q4，为£4000。
Calculate the total sales for the year. Add all four quarters: £2000 + £3500 + £2800 + £4000 = £12,300.
计算全年总销售额。将四个季度的销售额相加：£2000 + £3500 + £2800 + £4000 = £12,300。
What is the range of the quarterly sales? The highest is £4000 and the lowest is £2000, so range = £4000 – £2000 = £2000.
季度销售额的范围是多少？最高为£4000，最低为£2000，因此范围 = £4000 – £2000 = £2000。
Bar charts are useful for comparing categories quickly. Always read the scale carefully and label axes correctly.
条形图便于快速比较各个类别。务必仔细阅读刻度，并正确标记坐标轴。
4. Pie Chart Angle Calculations | 饼图角度计算
A survey of 200 students asked how they travel to school. The results were: Walk 80, Bus 60, Car 50, Bicycle 10. Calculate the angle for each sector in a pie chart.
一项针对200名学生的调查询问了他们的上学方式。结果是：步行80人，公交60人，汽车50人，自行车10人。计算饼图中各扇区的角度。
a pie chart, the total angle is 360° and each sector angle is (frequency ÷ total) × 360°.
在饼图中，总角度为360°，每个扇区的角度为 （频数 ÷ 总数）× 360°。
(60 ÷ 200) × 360° = 108°. Car: (50 ÷ 200) × 360° = 90°. Bicycle: (10 ÷ 200) × 360° = 18°.
步行： (80 ÷ 200) × 360° = 0.4 × 360° = 144°。同样，公交： (60 ÷ 200) × 360° = 108°。汽车： (50 ÷ 200) × 360° = 90°。自行车： (10 ÷ 200) × 360° = 18°。
s sum to 360°: 144° + 108° + 90° + 18° = 360°. This confirms the calculations are correct.
务必检查所有角度之和是否为360°：144° + 108° + 90° + 18° = 360°。这证实了计算是正确的。
Using a protractor, you can now draw the pie chart accurately. Label each sector with the method and percentage if required.
现在你可以使用量角器精确绘制饼图。如果需要，标记每个扇区及其对应的方式和百分比。
5. Scatter Graphs and Correlation | 散点图与相关性
A scatter graph shows the maths scores and science scores of 15 students. The points generally rise from left to right. One student scored 70 in maths.
散点图显示了15名学生的数学成绩和科学成绩。这些点总体上从左到右呈上升趋势。一名学生的数学成绩为70分。
Describe the correlation. Because the points go uphill, there is a positive correlation. As maths scores increase, science scores also tend to increase.
描述相关性。由于散点呈上升趋势，因此存在正相关。随着数学成绩的提高，科学成绩通常也会提高。
We draw a line of best fit, which passes through the middle of the points with roughly equal numbers above and below it.
我们画一条最佳拟合线，它穿过点的中间，大致使线上方和下方的点数相等。
To estimate the science score for a maths score of 70, find 70 on the horizontal axis, go up to the line, then read across to the vertical axis.
要估算数学成绩为70分时的科学成绩，在横轴上找到70，向上延伸到直线，然后平移到纵轴读取数值。
Suppose the line passes through 68 on the science axis. The estimated science score is 68. This is an interpolation because 70 lies within the data range.
假设直线在科学轴上经过68。则估计的科学成绩为68。这是一个内插值，因为70位于数据范围内。
Scatter graphs help us see relationships, but correlation does not imply causation. Always be cautious when making predictions outside the data range (extrapolation).
散点图帮助我们观察关系，但相关性并不意味因果关系。在数据范围之外进行预测（外推）时，务必谨慎。
6. Basic Probability | 基础概率
A bag contains 4 red counters, 3 blue counters and 2 green counters. One counter is taken at random. Find the probabilities of the following events.
一个袋子里有4个红色筹码、3个蓝色筹码和2个绿色筹码。随机取出一个筹码。求以下事件的概率。
robability is calculated as (number of favourable outcomes) ÷ (total number of possible outcomes).
筹码总数 = 4 + 3 + 2 = 9。概率的计算方法是 （有利结果的数量）÷ （所有可能结果的总数）。
ed) = 4/9. Probability of blue: P(blue) = 3/9 = 1/3. Probability of green: P(green) = 2/9.
红色概率：P(红) = 4/9。蓝色概率：P(蓝) = 3/9 = 1/3。绿色概率：P(绿) = 2/9。
Probability of not green: this is the complement of getting green, so P(not green) = 1 – P(green) = 1 – 2/9 = 7/9.
非绿色的概率：这是取出绿色的补集，因此 P(非绿) = 1 – P(绿) = 1 – 2/9 = 7/9。
All probabilities are between 0 and 1. A probability of 0 means impossible; 1 means certain. The sum of all possible outcomes is 1.
所有概率都介于0和1之间。概率为0意味着不可能发生；为1意味着必然发生。所有可能结果的概率之和为1。
7. Averages from a Frequency Table | 频数表中的平均数
pets × 3 households, 1 pet × 8, 2 pets × 6, 3 pets × 3. Calculate the mean number of pets.
下表显示了20户家庭拥有的宠物数量：0只宠物 × 3户，1只 × 8户，2只 × 6户，3只 × 3户。计算平均宠物数量。
y multiplying each number of pets by its frequency: (0 × 3) + (1 × 8) + (2 × 6) + (3 × 3).
首先，通过将每种宠物数量乘以其频数来计算宠物总数：(0 × 3) + (1 × 8) + (2 × 6) + (3 × 3)。
Total pets = 0 + 8 + 12 + 9 = 29
The total frequency is 3 + 8 + 6 + 3 = 20 households. Then divide: mean = 29 ÷ 20.
总频数为 3 + 8 + 6 + 3 = 20户家庭。然后相除：平均值 = 29 ÷ 20。
Mean = 1.45 pets
This is an appropriate average for discrete data. The mean does not have to be a whole number, even though you cannot have a fraction of a pet – it represents an average across the group.
这是适用于离散数据的平均数。尽管现实中不可能有几分之几只宠物，但平均值不必是整数——它代表了整组的平均水平。
ve frequency and identify the most common value (mode = 1 because frequency 8 is highest).
要通过表格找中位数和众数，可使用累积频数并找出最常见的值（众数为1，因为频数8最高）。
8. Comparing Data Sets | 数据集的比较
Two classes took the same maths test. Class A had a mean score of 65 and a range of 30. Class B had a mean score of 70 and a range of 15. Compare their performance.
两个班级参加了同样的数学测验。A班的平均分为65，范围为30。B班的平均分为70，范围为15。比较他们的表现。
The mean tells us about the typical score. Class B has a higher mean (70 > 65), so on average, Class B performed better than Class A.
平均值可以告诉我们典型分数。B班的平均分更高（70 > 65），因此平均而言，B班的表现优于A班。
The range describes the spread of scores. Class B has a smaller range (15) compared to Class A (30), which means scores in Class B are more consistent and less spread out.
范围描述了分数的分散程度。B班的范围（15）比A班（30）小，这意味着B班的分数更加集中，波动较小。
In contrast, Class A's large range suggests that some students scored very high and some very low, so the performance is more varied.
相比之下，A班较大的范围表明一些学生得分很高，另一些得分很低，因此成绩差异更大。
When comparing, always use both an average and a measure of spread to give a full picture. The mode could also be used if suits the data.
进行比较时，始终结合平均数及分散程度的度量，以给出全面的描述。如果数据合适，也可以使用众数。
9. Stem and Leaf Diagrams | 茎叶图
The heights (in cm) of 8 students are: 132, 145, 138, 151, 147, 133, 142, 150. Draw an ordered stem and leaf diagram and find the median.
8名学生的身高（厘米）为：132, 145, 138, 151, 147, 133, 142, 150。绘制有序茎叶图并求中位数。
First, order the data: 132, 133, 138, 142, 145, 147, 150, 151. Choose stems as the tens and leaves as the units. Stem 13 represents 130-139.
首先，将数据排序：132, 133, 138, 142, 145, 147, 150, 151。以十位数作茎，个位数作叶。茎13代表130-139。
The ordered stem and leaf:
13 | 2 3 8
14 | 2 5 7
15 | 0 1
Key: 13|2 = 132 cm
有序茎叶图：
13 | 2 3 8
14 | 2 5 7
15 | 0 1
图例：13|2 = 132 厘米
With 8 values, the median lies between the 4th and 5th values. From the diagram, the 4th is 142 (1st in 14 stem) and 5th is 145 (2nd in 14 stem).
有8个数据，中位数位于第4和第5个值的中间。从图中可知，第4个是142（茎14的第一个），第5个
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