Year 11 OCR Statistics: Unit Test Mock Paper Walkthrough | 11年级OCR统计：单元测试模拟卷解析

📚 Year 11 OCR Statistics: Unit Test Mock Paper Walkthrough | 11年级OCR统计：单元测试模拟卷解析

This walkthrough analyses a mock unit test designed for Year 11 OCR Statistics students. It mirrors the style and content of real exam questions, helping you consolidate key concepts such as data types, probability, distributions, correlation and quality control. Work through each solution carefully to identify common pitfalls and sharpen your exam technique before the final assessment.

这份解析针对为11年级OCR统计学生设计的单元测试模拟卷。它反映了真实考题的风格与内容，帮助你巩固数据类型、概率、分布、相关分析和质量控制等关键概念。请仔细推敲每一题的解答过程，发现常见错误，在最终评估之前打磨你的应试技巧。

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

A survey records each student’s favourite lunch option, the number of siblings they have, and their height in centimetres. (a) Classify each variable as categorical or numerical, stating a more precise type. (b) The school has 320 Year 11 students, of whom 55% are male. Describe how to select a stratified sample of 40 students that reflects the gender split.

一项调查记录了每个学生最喜欢的午餐选择、兄弟姐妹的数量以及身高（厘米）。(a) 将每个变量归类为分类数据或数值数据，并说明更精确的类型。(b) 学校有320名11年级学生，其中55%为男生。请描述如何抽取一个反映性别比例的分层样本，样本容量为40人。

For part (a): ‘favourite lunch option’ is categorical (nominal) because it names categories without order; ‘number of siblings’ is numerical and discrete since it can only take whole-number values; ‘height in cm’ is numerical and continuous because it can take any value within a range. For part (b): calculate the stratum sizes — 40 × 0.55 = 22 males, and 40 × 0.45 = 18 females. List all males and all females separately, assign a number to each, and use a random number generator to select 22 from the male list and 18 from the female list. This ensures proportional representation and removes selection bias.

(a) 部分：“最喜欢的午餐选择”是分类数据（名义型），因为它只命名类别无序次；“兄弟姐妹数量”是数值型且为离散数据，因为它只能是整数值；“身高（厘米）”是数值型且为连续数据，因为它可以在一个区间内取任意值。(b) 部分：先计算各层所需人数——40 × 0.55 = 22名男生，40 × 0.45 = 18名女生。将全体男生和女生分别列表并编号，然后用随机数生成器从男生名单中抽取22人，从女生名单中抽取18人。这样既能保证比例代表，又能消除选择偏差。

2. Stem-and-Leaf Diagram, Median and Box Plot | 茎叶图、中位数与箱线图

The stem-and-leaf diagram below shows the test scores of 15 students. Key: 5|2 represents 52.

下面的茎叶图显示了 15 名学生的测试成绩。图例：5|2 表示 52。

Stem | Leaf
4    | 5 9
5    | 2 4 6 6 8
6    | 1 3 5 7 7 9
7    | 0 2

(a) State the median and interquartile range (IQR). (b) Identify any outlier using the 1.5 × IQR rule. (c) Draw a box plot to represent the data.

(a) 写出中位数和四分位距（IQR）。(b) 使用 1.5 × IQR 准则判断是否存在异常值。(c) 绘制箱线图表示数据。

First, list the ordered data: 45, 49, 52, 54, 56, 56, 58, 61, 63, 65, 67, 67, 69, 70, 72. n = 15. Median is the 8th value: 61. Lower quartile Q₁ is the median of the first 7 values: 54. Upper quartile Q₃ is the median of the last 7 values: 67. IQR = Q₃ – Q₁ = 67 – 54 = 13. Lower boundary for outliers: Q₁ – 1.5 × IQR = 54 – 19.5 = 34.5. Upper boundary: Q₃ + 1.5 × IQR = 67 + 19.5 = 86.5. Since all scores lie between 45 and 72, there is no outlier. The box plot will have whiskers from 45 to 70, with the box from 54 to 67 and a median line at 61.

首先，列出排序后的数据：45, 49, 52, 54, 56, 56, 58, 61, 63, 65, 67, 67, 69, 70, 72。n = 15。中位数是第8个数据：61。下四分位数 Q₁ 是前7个值的中位数：54。上四分位数 Q₃ 是后7个值的中位数：67。IQR = Q₃ – Q₁ = 67 – 54 = 13。异常值的下边界：Q₁ – 1.5 × IQR = 54 – 19.5 = 34.5；上边界：Q₃ + 1.5 × IQR = 67 + 19.5 = 86.5。由于所有分数都在 45 到 72 之间，因此没有异常值。箱线图的须线从 45 延伸到 70，箱子从 54 到 67，中位线在 61 处。

In the exam, remember to draw the box plot with a labelled scale, clearly showing the whiskers, box edges and the median line. An outlier, if present, would be marked as a separate cross beyond the whisker.

考试时，记得画出的箱线图要带有明确的刻度标签，须线、箱子边界和中位线标示清晰。如果存在异常值，应用叉号在须线之外单独标出。

3. Probability Tree Diagrams and Conditional Probability | 概率树图与条件概率

A bag contains 5 red sweets and 3 blue sweets. Two sweets are drawn at random without replacement. (a) Draw a tree diagram showing all probabilities. (b) Find the probability that both sweets are the same colour. (c) Given that the first sweet is red, find the probability the second sweet is blue.

一个袋子装有 5 颗红色糖果和 3 颗蓝色糖果。随机抽取两颗，不放回。(a) 画出树图，标出所有概率。(b) 求两颗糖果颜色相同的概率。(c) 已知第一颗是红色，求第二颗是蓝色的概率。

First draw: P(Red) = 5/8, P(Blue) = 3/8. After drawing a red, 4 red and 3 blue remain; second draw: P(Red|1st Red) = 4/7, P(Blue|1st Red) = 3/7. After drawing a blue first, 5 red and 2 blue remain; second draw: P(Red|1st Blue) = 5/7, P(Blue|1st Blue) = 2/7. Same colour means RR or BB. P(RR) = (5/8) × (4/7) = 20/56; P(BB) = (3/8) × (2/7) = 6/56. Total = 26/56 = 13/28. For part (c), the condition is that the first sweet is red, so we only consider that branch: the probability the second is blue is directly 3/7; this is P(Blue | 1st Red).

第一次抽取：P(红) = 5/8，P(蓝) = 3/8。先抽到红后，剩余 4 红 3 蓝；第二次：P(红|首红) = 4/7，P(蓝|首红) = 3/7。先抽到蓝后，剩余 5 红 2 蓝；第二次：P(红|首蓝) = 5/7，P(蓝|首蓝) = 2/7。颜色相同即 RR 或 BB。P(RR) = (5/8) × (4/7) = 20/56；P(BB) = (3/8) × (2/7) = 6/56，总和为 26/56 = 13/28。对于 (c) 部分，条件是第一颗为红色，因此我们只关注该分支：第二颗是蓝色的概率就是 3/7；这也就是 P(蓝|首红)。

A common mistake is forgetting that probabilities change after a ‘without replacement’ draw. Always update the denominators on each branch and check your tree diagram for completeness.

常见错误是在“不放回”抽取后忘记了概率会发生变化。一定要在每一分支更新分母，并检查树图是否完整。

4. Binomial Distribution: Calculating Exact Probabilities | 二项分布：计算精确概率

A biased coin lands heads with probability 0.3. It is tossed 10 times. (a) Define the random variable X and state its distribution. (b) Calculate P(X = 3). (c) Find the probability of getting at least one head.

一枚不均匀硬币出现正面的概率为 0.3，抛掷 10 次。(a) 定义随机变量 X 并说明其分布。(b) 计算 P(X = 3)。(c) 求至少出现一次正面的概率。

Let X = number of heads in 10 tosses. X ~ B(10, 0.3). The probability mass function is P(X = r) = C(10, r) × (0.3)ʳ × (0.7)¹⁰⁻ʳ. For r = 3: C(10, 3) = 120. So P(X = 3) = 120 × (0.3)³ × (0.7)⁷ = 120 × 0.027 × 0.0823543… ≈ 0.2668 (4 d.p.). For at least one head, use the complement: P(X ≥ 1) = 1 – P(X = 0). P(X = 0) = (0.7)¹⁰ ≈ 0.02825. Therefore P(X ≥ 1) ≈ 1 – 0.02825 = 0.97175.

设 X = 10 次抛掷中出现正面的次数。X ~ B(10, 0.3)。概率质量函数为 P(X = r) = C(10, r) × (0.3)ʳ × (0.7)¹⁰⁻ʳ。当 r = 3 时：C(10, 3) = 120。因此 P(X = 3) = 120 × (0.3)³ × (0.7)⁷ = 120 × 0.027 × 0.0823543… ≈ 0.2668（保留四位小数）。对于至少一次正面，利用补集：P(X ≥ 1) = 1 – P(X = 0)。P(X = 0) = (0.7)¹⁰ ≈ 0.02825。所以 P(X ≥ 1) ≈ 1 – 0.02825 = 0.97175。

When using the binomial formula, always ensure you correctly identify n, p and the number of successes r. The phrase ‘at least one’ almost always implies the complement approach is faster.

使用二项公式时，一定要正确识别 n、p 和成功次数 r。“至少一次”这样的表述几乎都意味着用补集方法计算更为快捷。

5. Normal Distribution and Inverse Normal | 正态分布与反向查表

The length of a manufactured bolt is normally distributed with mean 50.0 mm and standard deviation 0.4 mm. (a) Find the proportion of bolts shorter than 49.5 mm. (b) The shortest 5% of bolts are rejected. Find the cut-off length below which a bolt is rejected.

某种螺栓的长度服从均值为 50.0 mm、标准差为 0.4 mm 的正态分布。(a) 求长度短于 49.5 mm 的螺栓所占比例。(b) 最短的 5% 的螺栓将被拒收。求拒收的临界长度。

Let L ~ N(50.0, 0.4²). (a) Standardise: z = (49.5 – 50.0) / 0.4 = -1.25. Using the standard normal table, P(Z < -1.25) = 1 - Φ(1.25) ≈ 1 - 0.8944 = 0.1056. So about 10.6% of bolts are shorter than 49.5 mm. (b) We need the z-score such that P(Z < z) = 0.05. From tables, the z-value is about -1.645. Then unstandardise: length = μ + zσ = 50.0 + (-1.645)×0.4 = 50.0 - 0.658 = 49.342 mm. Bolts shorter than 49.342 mm (approx.) would be rejected.

设 L ~ N(50.0, 0.4²)。(a) 标准化：z = (49.5 – 50.0) / 0.4 = -1.25。查标准正态表，P(Z < -1.25) = 1 - Φ(1.25) ≈ 1 - 0.8944 = 0.1056。因此约 10.6% 的螺栓长度短于 49.5 mm。(b) 需要求出满足 P(Z < z) = 0.05 的 z 值。查表可得 z ≈ -1.645。然后还原为长度：Length = μ + zσ = 50.0 + (-1.645) × 0.4 = 50.0 - 0.658 = 49.342 mm。长度短于约 49.342 mm 的螺栓将被拒收。

Always pay attention to whether the problem asks for a proportion less than, greater than, or between values. For inverse normal calculations, drawing a sketch with the tail area shaded helps avoid sign errors.

始终注意题目问的是“小于”、“大于”还是“介于”某个值的概率。对于反向查表计算，画出带阴影尾部区域的草图有助于避免符号错误。

6. Scatter Graphs, PMCC and Regression Line | 散点图、积差相关系数与回归线

Data on 8 cars show engine size (x litres) and fuel consumption (y km/litre). Summary statistics: Σx = 12.8, Σy = 128, Σx² = 21.8, Σy² = 2196, Σxy = 218.4. (a) Calculate the product moment correlation coefficient (PMCC). (b) Interpret the value. (c) The regression line is y = 22.3 – 3.5x. Predict the fuel consumption for an engine size of 1.6 litres and comment on the reliability.

8 辆汽车的数据显示了发动机排量（x，升）和燃油消耗（y，km/升）。汇总数据：Σx = 12.8, Σy = 128, Σx² = 21.8, Σy² = 2196, Σxy = 218.4。(a) 计算积差相关系数（PMCC）。(b) 解释该数值。(c) 回归线方程为 y = 22.3 – 3.5x。预测发动机排量为 1.6 升时的燃油消耗，并评价其可靠性。

PMCC formula: r = [nΣxy – (Σx)(Σy)] / √[ (nΣx² – (Σx)²)(nΣy² – (Σy)²) ]. n = 8. Numerator: 8 × 218.4 – (12.8 × 128) = 1747.2 – 1638.4 = 108.8. Denominator: √[(8×21.8 – 12.8²) × (8×2196 – 128²)] = √[(174.4 – 163.84) × (17568 – 16384)] = √(10.56 × 1184) = √12503.04 ≈ 111.82. So r ≈ 108.8 / 111.82 ≈ 0.973. This is very close to +1, indicating a strong negative linear correlation — as engine size increases, fuel consumption decreases. Prediction: at x = 1.6, y = 22.3 – 3.5 × 1.6 = 22

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Year 11 OCR Statistics: Unit Test Mock Paper Walkthrough | 11年级OCR统计：单元测试模拟卷解析