📚 IGCSE WJEC Statistics: Unit Test Mock Paper Walkthrough | IGCSE WJEC 统计：单元测试模拟卷解析
This walkthrough covers the most common question styles in a WJEC IGCSE Statistics unit test. For each section, a short mock question is followed by the method and answer. Work through every example before checking the solution to get the most from this revision resource.
本文解析 WJEC IGCSE 统计单元测试中最常见的题型。每一节先给出一个简短的模拟题，再给出方法和答案。在查看解答前先自己尝试每一道例题，可以最大化这份复习资料的价值。
1. Statistical investigation and data collection | 统计调查与数据收集
A WJEC unit test often begins by asking you to identify the population, sample and sampling frame in a practical context. You must also explain one advantage of using a sample instead of a census.
WJEC 单元测试通常首先要求你在实际情境中识别总体、样本和抽样框。你还必须解释使用样本而不是普查的一个优点。
Mock question: A school wants to estimate the average number of hours Year 11 students exercise each week. It selects 80 students from the Year 11 register. State the population, the sampling frame and one advantage of sampling.
模拟题：一所学校想估计 11 年级学生每周平均锻炼小时数。它从 11 年级名册中抽取了 80 名学生。请说明总体、抽样框以及抽样的一个优点。
Answer: Population = all Year 11 students at the school. Sampling frame = the Year 11 register or list of names. Advantage of sampling: it is quicker and cheaper than asking every student, and it still gives a reliable estimate if the sample is representative.
答案：总体 = 该校所有 11 年级学生。抽样框 = 11 年级名册或名单。抽样的优点：与询问每名学生相比，抽样更快捷、成本更低，而且如果样本具有代表性，仍能给出可靠的估计。
2. Types of data | 数据类型
You need to classify data as qualitative or quantitative, and then as discrete or continuous where relevant. Ordinal data are qualitative but have a natural order, such as satisfaction ratings.
你需要将数据分为定性数据或定量数据，并在相关时进一步分为离散数据或连续数据。有序数据属于定性数据，但具有自然顺序，例如满意度评分。
Mock question: Classify each of the following: shoe size, hair colour, exam mark out of 50, height in cm, and a rating from 1 to 5.
模拟题：对以下各项进行分类：鞋码、头发颜色、满分 50 分的考试分数、身高（厘米）以及 1 到 5 的评分。
Answer: shoe size is discrete quantitative data because it takes fixed numerical values; hair colour is nominal qualitative data; exam mark is discrete quantitative data; height is continuous quantitative data; rating 1 to 5 is ordinal qualitative data.
答案：鞋码是离散定量数据，因为它取固定的数值；头发颜色是名义定性数据；考试分数是离散定量数据；身高是连续定量数据；1 到 5 的评分是有序定性数据。
3. Frequency tables and grouped data | 频数分布表与分组数据
When raw data are large or spread out, we group them into equal class intervals. A frequency table must show clear class boundaries, tallies if needed, and the total frequency.
当原始数据较多或分布较广时，我们把它们分成等距的组。频数表必须显示清晰的组界，必要时显示计数符号，并给出总频数。
Mock question: Twenty students scored these marks out of 100: 56, 61, 64, 59, 70, 72, 68, 65, 74, 78, 81, 76, 69, 63, 60, 77, 82, 79, 71, 66. Using groups 50-59, 60-69, 70-79, 80-89, construct a frequency table and state the modal class.
模拟题：20 名学生的百分制成绩如下：56、61、64、59、70、72、68、65、74、78、81、76、69、63、60、77、82、79、71、66。使用 50-59、60-69、70-79、80-89 分组，构建频数表并指出众数类。
Answer: 50-59 has frequency 2; 60-69 has 7; 70-79 has 8; 80-89 has 3. Total frequency = 20. The modal class is 70-79 because it has the highest frequency.
答案：50-59 的频数为 2；60-69 为 7；70-79 为 8；80-89 为 3。总频数 = 20。众数类是 70-79，因为它的频数最高。
4. Mean, median and mode | 平均数、中位数与众数
The mean is calculated from a frequency table using Σfx ÷ Σf, where x is the class midpoint for grouped data. The median is the middle value when data are ordered. For n values, use position (n + 1) ÷ 2.
对于频数表，平均数用 Σfx ÷ Σf 计算，其中 x 是分组数据的组中点。中位数是将数据排序后的中间值。对于 n 个数据，位置为 (n + 1) ÷ 2。
Mock question: For the frequency table above, estimate the mean using midpoints and find the median class. Use midpoints 54.5, 64.5, 74.5, 84.5.
模拟题：对于上面的频数表，使用组中点估计平均数，并找出中位数所在类。使用组中点 54.5、64.5、74.5、84.5。
Answer: Σfx = (2×54.5)+(7×64.5)+(8×74.5)+(3×84.5) = 109 + 451.5 + 596 + 253.5 = 1410. Mean = 1410 ÷ 20 = 70.5 marks. The median position is (20+1)÷2 = 10.5, so the median lies in the 70-79 class.
答案：Σfx = (2×54.5)+(7×64.5)+(8×74.5)+(3×84.5) = 109 + 451.5 + 596 + 253.5 = 1410。平均数 = 1410 ÷ 20 = 70.5 分。中位数位置为 (20+1)÷2 = 10.5，因此中位数位于 70-79 组。
Mean x̄ = Σfx ÷ Σf
5. Range and interquartile range | 极差与四分位距
The range measures spread as maximum minus minimum. The interquartile range (IQR) measures the middle 50% of data: IQR = Q₃ – Q₁. It is less affected by extreme values than the range.
极差衡量数据的离散程度，等于最大值减最小值。四分位距衡量中间 50% 数据的离散程度：IQR = Q₃ – Q₁。与极差相比，它受极端值的影响较小。
Mock question: Find the range and interquartile range of: 4, 7, 8, 9, 11, 13, 15, 18, 20.
模拟题：求下列数据的极差和四分位距：4、7、8、9、11、13、15、18、20。
Answer: Range = 20 – 4 = 16. n = 9. Q₁ is the median of the lower half 4, 7, 8, 9, so Q₁ = (7+8)÷2 = 7.5. Q₃ is the median of the upper half 13, 15, 18, 20, so Q₃ = (15+18)÷2 = 16.5. IQR = 16.5 – 7.5 = 9.
答案：极差 = 20 – 4 = 16。n = 9。Q₁ 是下半部分 4、7、8、9 的中位数，所以 Q₁ = (7+8)÷2 = 7.5。Q₃ 是上半部分 13、15、18、20 的中位数，所以 Q₃ = (15+18)÷2 = 16.5。IQR = 16.5 – 7.5 = 9。
6. Box plots and outliers | 箱线图与异常值
A box plot uses five numbers: minimum, Q₁, median, Q₃ and maximum. An outlier is often defined as any value below Q₁ – 1.5×IQR or above Q₃ + 1.5×IQR.
箱线图使用五个数：最小值、Q₁、中位数、Q
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