Year 11 WJEC Statistics: Unit Test Mock Exam Walkthrough | WJEC统计单元测试模拟卷解析

📚 Year 11 WJEC Statistics: Unit Test Mock Exam Walkthrough | WJEC统计单元测试模拟卷解析

This article provides a complete walkthrough of a mock Unit 1 test for the WJEC GCSE Statistics specification. Each question is followed by a step‑by‑step solution in both English and Chinese, covering the full range of topics you need to master: data types, sampling, frequency analysis, cumulative frequency diagrams, box plots, probability, expectation, correlation, time series, index numbers and quality assurance. Use this resource to consolidate your revision and identify any gaps before your real exam.

本文为WJEC GCSE统计学科目第一单元模拟卷提供完整的逐题解析。每道题目均配有中英双语的详细解题步骤，涵盖数据类型、抽样方法、频数分析、累积频率图、箱线图、概率、期望值、相关性、时间序列、指数以及质量控制等全部核心考点。请利用这份材料巩固复习，在正式考试前找出自己的薄弱环节。

1. Data Types and Sampling Methods | 数据类型与抽样方法

A researcher records the following three variables: shoe size (UK), height (cm), and number of siblings. Classify each variable as qualitative, discrete quantitative or continuous quantitative. The researcher then wishes to select a stratified sample of 60 students from a school with 240 Year 10, 180 Year 11 and 100 Year 12 students. Determine how many pupils should be chosen from each year group.

一位研究人员记录了三个变量：鞋码（英制）、身高（厘米）和兄弟姐妹人数。请将每个变量分类为定性数据、离散定量数据或连续定量数据。随后，该研究人员计划从一所拥有 240 名 10 年级、180 名 11 年级和 100 名 12 年级学生的学校中，采用分层抽样方法抽取 60 名学生。请计算每个年级应抽取的人数。

Shoe size takes whole and half values but is measured on a scale — it is discrete quantitative because you can list all possible sizes. Height is measured and can take any value within a range, so it is continuous quantitative. Number of siblings is a count, hence discrete quantitative. For the stratified sample, the total population is 240 + 180 + 100 = 520. The sampling fraction is 60 / 520. Year 10: (240 ÷ 520) × 60 ≈ 27.69 → 28 students. Year 11: (180 ÷ 520) × 60 ≈ 20.77 → 21 students. Year 12: (100 ÷ 520) × 60 ≈ 11.54 → 11 students. Check: 28 + 21 + 11 = 60.

鞋码有整数和半码值，但仍是可列出的测量尺度，属于离散定量数据。身高可测量并在某区间内取任意值，因此为连续定量数据。兄弟姐妹人数是计数结果，属于离散定量数据。分层抽样中，总人数为 240 + 180 + 100 = 520。抽样比为 60/520。10 年级：(240÷520)×60 ≈ 27.69 → 28 人；11 年级：(180÷520)×60 ≈ 20.77 → 21 人；12 年级：(100÷520)×60 ≈ 11.54 → 11 人。核对：28+21+11 = 60。

2. Frequency Distributions and the Mean | 频数分布与均值

The table shows the number of books read by 40 students in a month. Use it to estimate the mean number of books read.

Books	0‑2	3‑5	6‑8	9‑11
Frequency	8	14	12	6

上表显示了 40 名学生在某月阅读的书籍数量。请据此估计平均阅读数量。

Find the midpoint of each class: 1, 4, 7, 10. Multiply each midpoint by its frequency: 1×8 = 8, 4×14 = 56, 7×12 = 84, 10×6 = 60. Sum = 208. The estimated mean is 208 ÷ 40 = 5.2 books. Alternatively, using an assumed mean of 4 gives deviations -3, 0, 3, 6, then Σfd = 8×(-3)+14×0+12×3+6×6 = -24+0+36+36 = 48, mean = 4 + 48/40 = 4 + 1.2 = 5.2. This confirms the direct calculation.

计算每组中点：1, 4, 7, 10。各中点乘以对应频数：1×8=8，4×14=56，7×12=84，10×6=60。总和为 208。估计均值 = 208÷40 = 5.2 本。或者使用假定均值 4，偏差为 -3, 0, 3, 6，Σfd = 8×(-3)+14×0+12×3+6×6 = -24+0+36+36 = 48，均值 = 4 + 48/40 = 5.2。两种方法结果一致。

3. Cumulative Frequency and Median | 累积频率与中位数

The grouped frequency distribution below shows the daily screen time (hours) of 80 teenagers. Construct a cumulative frequency table, draw the cumulative frequency curve and estimate the median and interquartile range.

Hours	0‑<2	2‑<4	4‑<6	6‑<8	8‑10
Freq	6	18	32	16	8

以下分组频数表显示了 80 名青少年每日屏幕使用时间（小时）。请构建累积频数表，绘制累积频率曲线，并估算中位数和四分位距。

Cumulative frequencies: upper boundary 2 → 6, 4 → 24, 6 → 56, 8 → 72, 10 → 80. Plot points at (2,6), (4,24), (6,56), (8,72), (10,80) and join with a smooth curve. The median is the 40th value: from the graph, draw from 40 on the cumulative axis across to the curve and down to the x‑axis, giving about 5.2 hours. The lower quartile (20th value) ≈ 3.7 hours, upper quartile (60th value) ≈ 6.7 hours. IQR = 6.7 − 3.7 = 3.0 hours. The curve helps to read off any percentile accurately.

累积频数：上限 2 → 6，4 → 24，6 → 56，8 → 72，10 → 80。在坐标纸上描点 (2,6), (4,24), (6,56), (8,72), (10,80)，用光滑曲线连接。中位数对应第 40 个值：从累积轴 40 处画水平线交曲线，再垂直向下读取，约为 5.2 小时。下四分位数 (第 20 个值) ≈ 3.7 小时，上四分位数 (第 60 个值) ≈ 6.7 小时。四分位距 = 6.7 − 3.7 = 3.0 小时。累积频率曲线可精确读取任何百分位数。

4. Box Plots and Outliers | 箱线图与异常值

A data set gives a five‑number summary: minimum = 8, Q₁ = 14, median = 20, Q₃ = 26, maximum = 45. Construct a box plot and determine whether the maximum is an outlier, using the 1.5 × IQR rule.

某数据集的五数概要为：最小值 = 8，Q₁ = 14，中位数 = 20，Q₃ = 26，最大值 = 45。请绘制箱线图，并用 1.5 × IQR 规则判断最大值是否为异常值。

Draw a scale. Box from 14 to 26 with a line at 20. Whiskers from 8 to 14 and from 26 to 45. Compute IQR = 26 − 14 = 12. Upper fence = Q₃ + 1.5×IQR = 26 + 18 = 44. Since the maximum 45 > 44, it is an outlier. Mark it with a separate point and extend the upper whisker to the largest non‑outlier, which could be 44 if present, but here the next value is 45 itself; without further data, we indicate 45 as an outlier. In practice, the box plot would show the whisker stopping at the largest value inside the fence.

先绘制数轴。箱体从 14 到 26，中位数位置画线于 20。触须线从 8 到 14，以及从 26 到 45。计算 IQR = 26 − 14 = 12。上界限 = Q₃ + 1.5×IQR = 26 + 18 = 44。因为最大值 45 > 44，所以 45 属于异常值，应单独用点标出，上触须仅延伸至非异常值的最大值。若 44 不存在，则上触须终止于 44 以内。本例中 45 被标记为异常值。

5. Probability Tree Diagrams | 概率树图

A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. Draw a tree diagram and find the probability that both marbles are the same colour.

一个袋子里有 5 个红球和 3 个蓝球。不放回地依次取出两个球。画出树图，并求两球同色的概率。

First draw: P(R) = 5/8, P(B) = 3/8. Second draw: if first is R, remaining: 4 R, 3 B → P(R|R) = 4/7, P(B|R) = 3/7. If first is B, remaining: 5 R, 2 B → P(R|B) = 5/7, P(B|B) = 2/7. Probability both red = 5/8 × 4/7 = 20/56. Both blue = 3/8 × 2/7 = 6/56. Sum = 26/56 = 13/28 ≈ 0.464. The tree diagram visually confirms these conditional probabilities and helps avoid common errors with replacement misconceptions.

第一次抽取：P(红) = 5/8，P(蓝) = 3/8。第二次：若第一次为红，剩余 4 红 3 蓝，故 P(红|红) = 4/7，P(蓝|红) = 3/7；若第一次为蓝，剩余 5 红 2 蓝，P(红|蓝) = 5/7，P(蓝|蓝) = 2/7。两球全红的概率 = 5/8 × 4/7 = 20/56。两球全蓝 = 3/8 × 2/7 = 6/56。相加得 26/56 = 13/28 ≈ 0.464。概率树图直观展示了这些条件概率，可有效避免与放回情形混淆。

6. Expectation of a Discrete Random Variable | 离散随机变量的期望

The probability distribution of a discrete random variable X is given below. Calculate E(X) and E(X²), and hence find Var(X).

x	1	2	3	4
P(X=x)	0.2	0.3	0.4	0.1

已知离散随机变量 X 的概率分布如上表。计算 E(X) 和 E(X²)，并由此求 Var(X)。

E(X) = Σ x·p = 1×0.2 + 2×0.3 + 3×0.4 + 4×0.1 = 0.2 + 0.6 + 1.2 + 0.4 = 2.4. E(X²) = 1²×0.2 + 2²×0.3 + 3²×0.4 + 4²×0.1 = 0.2 + 1.2 + 3.6 + 1.6 = 6.6. Var(X) = E(X²) − [E(X)]² = 6.6 − (2.4)² = 6.6 − 5.76 = 0.84. The standard deviation is √0.84 ≈ 0.9165. These calculations use the standard formula for variance of a discrete distribution and are essential for understanding spread.

E(X) = Σ x·p = 1×0.2 + 2×0.3 + 3×0.4 + 4×0.1 = 0.2 + 0.6 + 1.2 + 0.4 = 2.4。E(X²) = 1²×0.2 + 2²×0.3 + 3²×0.4 + 4²×0.1 = 0.2 + 1.2 + 3.6 + 1.6 = 6.6。方差 Var(X) = E(X²) − [E(X)]² = 6.6 − (2.4)² = 6.6 − 5.76 = 0.84。标准差为 √0.84 ≈ 0.9165。计算采用了离散分布方差的常规公式，是理解数据离散程度的关键步骤。

7. Scatter Graphs and Spearman’s Rank Correlation | 散点图与斯皮尔曼等级相关

Six students take tests in Mathematics and Statistics. Their scores are: A(45, 52), B(70, 68), C(55, 60), D(80, 78), E(65, 72), F(50, 48). Calculate Spearman’s rank correlation coefficient and comment on the relationship.

六名学生分别参加了数学与统计测验，得分如下：A(45,52), B(70,68), C(55,60), D(80,78), E(65,72), F(50,48)。计算斯皮尔曼等级相关系数，并评价两者关系。

Rank the Maths scores: 45(6), 50(5), 55(4), 65(3), 70(2), 80(1). Rank the Stats scores: 48(6), 52(5), 60(4), 68(3), 72(2), 78(1). Ranks perfectly match: each student has the same rank in both subjects. Differences d = 0 for all pairs. Σd² = 0. Using rₛ = 1 − (6Σd²)/(n(n²−1)) = 1 − 0 = 1. This indicates a perfect positive monotonic correlation — higher Maths scores are always associated with higher Statistics scores. Note that tied ranks would require adjustment, but here no ties occur.

将数学分数排序：45(6), 50(5), 55(4), 65(3), 70(2), 80(1)。统计分数排序：48(6), 52(5), 60(4), 68(3), 72(2), 78(1)。等级完全一致，每位学生在两科中的排名相同。差值 d 均为 0，Σd² = 0。斯皮尔曼等级相关系数 rₛ = 1 − (6Σd²)/(n(n²−1)) = 1 − 0 = 1。这表明两者存在完全正单调相关——数学成绩较高的学生，统计成绩也较高。注意，如果有相同等级需进行修正，但本例无并列情况。

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

Quarterly sales figures (£ thousands) for a shop are: Q1: 22, Q2: 34, Q3: 18, Q4: 40, Q1: 28, Q2: 44, Q3: 24, Q4: 50. Calculate four‑point moving averages and centre them. Plot the original data and the trend line.

某店季度销售额（千英镑）为：Q1: 22, Q2: 34, Q3: 18, Q4: 40, Q1: 28, Q2: 44, Q3: 24, Q4: 50。计算四点移动平均并进行中心化处理。绘制原始数据与趋势线。

First moving total: 22+34+18+40 = 114, second: 34+18+40+28 = 120, third: 18+40+28+44 = 130, fourth: 40+28+44+24 = 136, fifth: 28+44+24+50 = 146. Four‑point moving averages (uncentred): 114/4=28.5, 120/4=30, 130/4=32.5, 136/4=34, 146/4=36.5. Centre them by averaging adjacent pairs: (28.5+30)/2 = 29.25, (30+32.5)/2 = 31.25, (32.5+34)/2 = 33.25, (34+36.5)/2 = 35.25. The centred trend is plotted against quarters 2½, 3½, 4½, 5½, showing a steady upward movement. This smooths out seasonal fluctuations.

第一个移动总和：22+34+18+40=114，第二个：34+18+40+28=120，第三个：18+40+28+44=130，第四个：40+28+44+24=136，第五个：28+44+24+50=146。未中心化的四点移动平均分别为：28.5, 30, 32.5, 34, 36.5。将相邻两项平均以实现中心化：(28.5+30)/2 = 29.25，(30+32.5)/2 = 31.25，(32.5+34)/2 = 33.25，(34+36.5)/2 = 35.25。中心化后的趋势值对应第 2.5, 3.5, 4.5, 5.5 季度，呈现出稳步上升的态势，有效消除了季度性波动。

9. Index Numbers | 指数

A basket of goods cost £46 in 2018, £50 in 2019 and £55 in 2020. Using 2018 as the base year, calculate simple index numbers for 2019 and 2020. Then interpret the percentage increase from 2018 to 2020.

一篮子商品在 2018 年花费 46 英镑，2019 年 50 英镑，2020 年 55 英镑。以 2018 年为基年，计算 2019 年和 2020 年的简单指数。解释 2018 年至 2020 年的百分比增幅。

Base year index = 100. For 2019: (50/46) × 100 = 108.7 (to 1 d.p.). For 2020: (55/46) × 100 = 119.6 (to 1 d.p.). The index shows that costs were 8.7% higher in 2019 than in 2018, and 19.6% higher in 2020 compared with 2018. When answering interpretation, you should link the increase to a specific percentage: a rise of around 19.6% over the two‑year period. Always state the base year clearly when reporting index numbers.

基年指数 = 100。2019 年：(50/46)×100 ≈ 108.7（保留一位小数）。2020 年：(55/46)×100 ≈ 119.6。指数表明，2019 年成本较 2018 年上涨了 8.7%，而 2020 年较 2018 年上涨了 19.6%。解释时需明确指出两年间约 19.6% 的增幅。报告指数时务必清楚标明基年。

10. Quality Assurance and Control Charts | 质量控制与控制图

A factory produces bolts whose lengths must have a target mean of 50.0 mm. Regular samples of size 5 are taken and the sample means are plotted. The process standard deviation is known to be 0.2 mm. Set up warning limits at ±2 standard errors and action limits at ±3 standard errors from the target. If the next sample mean is 50.24 mm, what action should be taken?

某工厂生产的螺栓长度目标均值为 50.0 mm。定期抽取容量为 5 的样本并绘制样本均值控制图。已知过程标准差为 0.2 mm。请设定基于目标值 ±2 个标准误的警戒限和 ±3 个标准误的行动限。若下一个样本均值为 50.24 mm，应采取何种措施？

Standard error of the mean = σ/√n = 0.2/√5 = 0.2/2.236 ≈ 0.0894 mm. Warning limits: 50 ± 2×0.0894 → 49.82 mm to 50.18 mm. Action limits: 50 ± 3×0.0894 → 49.73 mm to 50.27 mm. The observed sample mean 50.24 lies outside the warning limit (50.18) but inside the action limit (50.27). According to control chart rules, this is a warning signal: the process may be drifting, so increase sampling frequency or investigate but do not stop production immediately. If it exceeds the action limit, the process should be halted for adjustment.

均值的标准误 = σ/√n = 0.2/√5 ≈ 0.0894 mm。警戒限：50 ± 2×0.0894，即 49.82 mm 至 50.18 mm。行动限：50 ± 3×0.0894，即 49.73 mm 至 50.27 mm。观测到的样本均值 50.24 超出警戒上限 50.18，但仍在行动限 50.27 之内。按控制图规则，这属于警戒信号，表明过程可能发生漂移，应提高抽样频率或展开调查，但无需立即停产。一旦超出行动限，则必须停机调整。

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Year 11 WJEC Statistics: Unit Test Mock Exam Walkthrough | WJEC统计 单元测试模拟卷解析