📚 Year 11 Eduqas Statistics: Unit Test Mock Paper Walkthrough | 英国 Year 11 Eduqas 统计单元测试模拟卷解析
This walkthrough is designed to guide Year 11 students through a complete Eduqas Statistics unit test mock paper. Each section breaks down a representative exam-style question, unpicks the key concepts, and models the step-by-step working required to secure full marks. From sampling methods to normal distribution, every major topic is covered with clear explanations and examiner-friendly presentation.
本模拟卷解析旨在为 Year 11 学生提供一份完整的 Eduqas 统计单测模拟卷逐题精讲。每一节都拆解一道典型考试风格题目，剖析核心概念，并演示解题步骤，帮助同学们拿到满分。从抽样方法到正态分布，各个核心主题都配有清晰的解释和易于得分的解答格式。
1. Mock Paper Overview and Key Topics | 模拟卷概述与核心考点
The mock paper mirrors the structure of an Eduqas GCSE Statistics unit test, with a mix of short calculation questions, graph interpretation, and longer problem-solving tasks. Topics tested include data collection, box plots, measures of average and spread, cumulative frequency, probability trees, binomial distribution, normal distribution, scatter graphs, and time series. A solid understanding of these areas and the ability to apply formulas will be essential.
这份模拟卷仿照 Eduqas GCSE 统计单元测的题型设置，包含计算题、图表解读题和较长的应用题。涉及的数据收集、箱线图、集中趋势与离散程度、累积频率、树状概率、二项分布、正态分布、散点图以及时间序列是考试重点。牢固掌握这些领域并能灵活运用公式是得分的关键。
2. Stratified Sampling Calculation | 分层抽样比例计算
Question: A school has 500 students in Year 7, 450 in Year 8, 400 in Year 9, 350 in Year 10 and 300 in Year 11. A stratified sample of 80 students is to be selected. Determine how many students should be taken from each year group.
题目：某校七年级有 500 名学生，八年级 450 名，九年级 400 名，十年级 350 名，十一年级 300 名。需要抽取一个容量为 80 的分层样本。计算每个年级应抽取的人数。
The total number of students is 500 + 450 + 400 + 350 + 300 = 2000. For a stratified sample, the number from each stratum is proportional to its size. Multiply the total stratum size by the sampling fraction 80/2000 = 0.04. Year 7: 500 × 0.04 = 20; Year 8: 450 × 0.04 = 18; Year 9: 400 × 0.04 = 16; Year 10: 350 × 0.04 = 14; Year 11: 300 × 0.04 = 12.
总学生人数为 500 + 450 + 400 + 350 + 300 = 2000。在分层抽样中，每层抽取的人数与该层总数成正比。用各层人数乘以抽样比例 80/2000 = 0.04。七年级：500 × 0.04 = 20；八年级：18；九年级：16；十年级：14；十一年级：12。
3. Box Plot Interpretation and Comparison | 箱线图解读与分布比较
Question: The box plots below summarise test scores for Class A and Class B. Compare the two distributions, referring to median, interquartile range, and overall spread.
题目：下面的箱线图汇总了 A 班和 B 班的测试成绩。请比较两个班的分布，提及中位数、四分位距和整体分散程度。
Assume the five-number summaries are: Class A – min 32, Q1 45, median 56, Q3 68, max 85; Class B – min 40, Q1 50, median 58, Q3 72, max 90. The median of Class B (58) is slightly higher than that of Class A (56), suggesting a small difference in central tendency. Class A shows a larger interquartile range (IQR = 68 – 45 = 23) compared with Class B (IQR = 72 – 50 = 22), so the middle 50% is similarly spread. The overall range for Class A is 85 – 32 = 53, while Class B has a range of 90 – 40 = 50. Additionally, Class A’s lower whisker extends further down, indicating a possible low outlier or greater lower tail. Both distributions appear roughly symmetric, with medians near the centre of the box.
假设五数综合为：A 班 — 最小值 32，下四分位数 45，中位数 56，上四分位数 68，最大值 85；B 班 — 最小值 40，下四分位数 50，中位数 58，上四分位数 72，最大值 90。B 班中位数 (58) 略高于 A 班 (56)，中心位置差别不大。A 班四分位距 IQR = 68 − 45 = 23，B 班 IQR = 72 − 50 = 22，中间 50% 的分布相近。A 班全距为 85 − 32 = 53，B 班全距为 50。另外 A 班下须线延伸得更低，提示可能存在低异常值或下尾更重。两个分布大致对称，中位数均靠近箱体中央。
4. Mean and Standard Deviation | 平均数与标准差
Question: Calculate the mean and standard deviation of the following data set: 12, 15, 18, 21, 24.
题目：计算下列数据集的均值和标准差：12, 15, 18, 21, 24。
First find the mean: (12 + 15 + 18 + 21 + 24) / 5 = 90 / 5 = 18. To obtain the standard deviation, use the formula for population standard deviation σ or sample standard deviation s. Examiners often expect the standard deviation with divisor n for a defined dataset. Compute deviations from the mean: 12-18 = -6, 15-18 = -3, 18-18 = 0, 21-18 = 3, 24-18 = 6. Square each: 36, 9, 0, 9, 36. Sum of squared deviations = 90. Divide by n = 5 gives variance = 18. Standard deviation σ = √18 ≈ 4.24 (to 3 s.f.).
先求均值：(12 + 15 + 18 + 21 + 24) ÷ 5 = 90 ÷ 5 = 18。标准差常使用除以 n 的总体标准差。计算各值与均值的离差：12−18 = −6，15−18 = −3，18−18 = 0，21−18 = 3，24−18 = 6。平方得 36、9、0、9、36。离差平方和为 90。除以 n=5 得方差 18。标准差 σ = √18 ≈ 4.24（保留三位有效数字）。
σ = √( Σ(x – x̄)² / n ) = √18 ≈ 4.24
5. Cumulative Frequency and Percentiles | 累积频率与百分位数
Question: The cumulative frequency table shows marks of 60 students. Draw a cumulative frequency curve and estimate the median and the interquartile range.
题目：累积频率表给出了 60 名学生的分数。绘制累积频率曲线图，并估计中位数和四分位距。
Suppose the grouped data are: 0≤ m <10 (cf 5), 10≤ m <20 (cf 18), 20≤ m <30 (cf 35), 30≤ m <40 (cf 50), 40≤ m <50 (cf 58), 50≤ m <60 (cf 60). Plot upper class boundaries against cumulative frequency and join with a smooth curve. The median is the mark at ½ × 60 = 30th value: reading from the curve gives roughly 26 marks. The lower quartile (15th value) is about 18, and the upper quartile (45th value) is about 35, so IQR ≈ 35 - 18 = 17 marks. Always show construction lines on the graph to secure method marks.
假设分组数据为：0≤ m <10（累积频数 5），10≤ m <20（18），20≤ m <30（35），30≤ m <40（50），40≤ m <50（58），50≤ m <60（60）。以上组界对累积频数描点，并用平滑曲线连接。中位数位置为 ½ × 60 = 30，从曲线上读取约 26 分。下四分位数（第 15 位）约为 18，上四分位数（第 45 位）约为 35，因此 IQR ≈ 35 − 18 = 17 分。答题时务必在图上画出辅助线，以获得方法分。
6. Probability without Replacement and Tree Diagrams | 不放回概率与树状图
Question: A bag contains 3 red balls and 5 green balls. Two balls are drawn at random without replacement. Find the probability that both are red, and the probability that at least one is green.
题目：一个袋子里有 3 个红球和 5 个绿球。随机抽取两个球且不放回。求两个球都是红色的概率，以及至少有一个绿球的概率。
Draw a probability tree. First draw: P(Red) = 3/8, P(Green) = 5/8. Second draw depends on first outcome. P(both red) = 3/8 × 2/7 = 6/56 = 3/28. For ‘at least one green’, it is easier to use the complement: P(at least one green) = 1 – P(both red) = 1 – 3/28 = 25/28. Alternatively, add probabilities of (Red, Green) and (Green, Red) and (Green, Green). The tree makes these pathways clear.
画出概率树状图。第一次抽取：P(红) = 3/8，P(绿) = 5/8。第二次抽取概率与第一次结果有关。P(两红) = 3/8 × 2/7 = 6/56 = 3/28。求“至少一个绿球”时，用补集更为简便：P(至少一绿) = 1 − P(两红) = 1 − 3/28 = 25/28。也可将 (红,绿)、(绿,红)、(绿,绿) 的概率相加，树状图能清晰显示各条路径。
P(both red) = 3/8 × 2/7 = 3/28
7. Binomial Distribution Calculation | 二项分布概率计算
Question: The probability that a light bulb is defective is 0.1. A random sample of 6 bulbs is taken. Find the probability that exactly 2 bulbs are defective.
题目：一个灯泡为次品的概率是 0.1。随机抽取 6 个灯泡组成样本，求恰好有 2 个次品的概率。
Let X be the number of defective bulbs. Then X ~ B(6, 0.1). The probability of exactly 2 defectives is given by the binomial formula: P(X=2) = C(6,2) × 0.1² × 0.9⁴. C(6,2) = 15, 0.1² = 0.01, 0.9⁴ = 0.6561. Multiplying gives 15 × 0.01 × 0.6561 = 0.098415 ≈ 0.0984 (to 3 s.f.). Always state the distribution clearly and show the substitution into the formula.
设 X 为次品灯泡数，则 X ~ B(6, 0.1)。恰好 2 个次品的概率由二项概率公式给出：P(X=2) = C(6,2) × 0.1² × 0.9⁴。C(6,2) = 15，0.1² = 0.01，0.9⁴ = 0.6561。相乘得 15 × 0.01 × 0.6561 = 0.098415 ≈ 0.0984（三位有效数字）。作答时应明确写出分布并代入数值。
P(X = 2) = ⁶C₂ × (0.1)² × (0.9)⁴ = 0.0984
8. Normal Distribution and the Empirical Rule | 正态分布与经验法则
Question: The weights of apples are normally distributed with a mean of 150 g and a standard deviation of 12 g. Use the empirical rule to estimate the percentage of apples weighing between 126 g and 174 g.
题目：苹果的重量服从均值为 150 g、标准差为 12 g 的正态分布。利用经验法则估计重量在 126 g 到 174 g 之间的苹果所占的百分比。
The boundaries 126 g and 174 g are exactly two standard deviations below and above the mean: 150 − 2×12 = 126, 150 + 2×12 = 174. The empirical rule (68-95-99.7) states that approximately 95% of data in a normal distribution lie within two standard deviations of the mean. Therefore, about 95% of apples will weigh between 126 g and 174 g. This quick estimation is often accepted; if a question requires more precise values, it will supply a normal table or the relevant probabilities.
界限 126 g 和 174 g 恰好是均值下方和上方两个标准差处：150 − 2×12 = 126，150 + 2×12 = 174。经验法则（68-95-99.7 规则）指出，正态分布中约 95% 的数据落在均值左右两个标准差之内。因此约有 95% 的苹果重量在 126 g 到 174 g 之间。这种快速估计常见于简答题；若需要更精确的值，题目会提供正态分布表或相应概率。
9. Scatter Graphs and Correlation | 散点图与相关性
Question: A scatter graph shows hours of revision against test score. The points rise steadily from left to right, and a line of best fit has been drawn. Describe the correlation and use the line to estimate the score for 5 hours of revision.
题目：散点图展示了复习时间与测试分数之间的关系。各点从左到右稳步上升，并已画出最佳拟合线。描述相关性，并用此线估计复习时间为 5 小时时的分数。
There is a strong positive correlation: as revision hours increase, test scores tend to increase. The line of best fit allows for interpolation. Locate 5 on the horizontal axis, draw a vertical line to meet the line of best fit, then read horizontally to the vertical axis. Suppose the line gives a value around 45 marks. This estimate is reliable as it lies within the range of the original data. Avoid extrapolation beyond the data range, as the relationship may not hold.
图中呈现强正相关：随着复习时间增加，测试分数呈上升趋势。最佳拟合线可用于内插估计。在横轴上找到 5 小时，向上作垂线与拟合线相交，再从交点水平读取纵轴数值。假设该线给出的分数约为 45 分。该估计值可靠，因为落在原始数据范围内。应避免外推至数据范围之外，因为那可能不再遵循原有的关系。
10. Time Series and Moving Averages | 时间序列与移动平均
Question: Quarterly sales (£000s) for a shop were: Q1 22: 20, Q2: 28, Q3: 35, Q4: 25, Q1 23: 22, Q2: 30, Q3: 38, Q4: 27. Calculate a 4-point moving average and use it to find a trend value for Q3 2023.
题目：某商店季度销售额（千英镑）为：2022 Q1: 20，Q2: 28，Q3: 35，Q4: 25，2023 Q1: 22，Q2: 30，Q3: 38，Q4: 27。计算 4 点移动平均，并据此给出 2023 年 Q3 的趋势值。
Four-point moving averages smooth out seasonal fluctuations. Calculate successive averages: (20+28+35+25)/4 = 27, (28+35+25+22)/4 = 27.5, (35+25+22+30)/4 = 28, (25+22+30+38)/4 = 28.75, (22+30+38+27)/4 = 29.25. These values are centred by averaging pairs: (27+27.5)/2 = 27.25 (centred on Q3 2022), (27.5+28)/2 = 27.75 (Q4), (28+28.75)/2 = 28.375 (Q1 2023), (28.75+29.25)/2 = 29.0 (Q2). The centred trend for Q3 2023 would need the average of (30+38+27+? ) but we can use the last centred value or the moving average centred on Q2 2023 to project a smooth trend; in a typical exam, you would plot the centred moving averages and read the trend line. With given data, the centred moving average for Q2 2023 is 29.0, and the trend is rising, so a reasonable estimate for Q3 2023 from the trend line might be around 29.5 to 30.0, depending on the graph.
四期移动平均可用于平滑季节波动。相继计算移动平均值：(20+28+35+25)/4 = 27，(28+35+25+22)/4 = 27.5，(35+25+22+30)/4 = 28，(25+22+30+38)/4 = 28.75，(22+30+38+27)/4 = 29.25。为得到居中趋势值，对相邻两个移动平均再取均值：(27+27.5)/2 = 27.25（对应 2022 Q3），(27.5+28)/2 = 27.75（Q4），(28+28.75)/2 = 28.375（2023 Q1），(28.75+29.25)/2 = 29.0（Q2）。2023 Q3 的趋势值需依赖后续数据，但在考试中可将已得居中移动平均值描点并绘制趋势线。根据现有数据，Q2 2023 的趋势值为 29.0，且呈上升趋势，因此从趋势线上读取 2023 Q3 的合理估计值约为 29.5 至 30.0，视描点而定。
Published by TutorHao | Statistics Revision Series | aleveler.com
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