📚 A-Level Edexcel Statistics 1: Mock Unit Test Paper Solutions | A-Level Edexcel S1: 单元测试模拟卷解析
This article provides a step-by-step walkthrough of a mock unit test for the Edexcel A-Level Statistics 1 module. We cover typical exam-style questions, including box plots, probability, discrete random variables, normal distribution, regression, and more. Each solution is accompanied by clear explanations to reinforce key concepts and techniques needed for the actual examination.
本文为 Edexcel A-Level 统计学 S1 单元测试模拟卷提供逐步解析。涵盖箱线图、概率、离散随机变量、正态分布、回归分析等典型考题。每道题的解答都配有清晰讲解，以巩固考试所需的核心概念与解题技巧。
1. Stem-and-Leaf Plot and Box Plot: Outlier Analysis | 茎叶图与箱线图：异常值分析
A stem-and-leaf diagram displays the following data (in mm): 42, 45, 48, 51, 53, 53, 55, 56, 58, 60, 62, 64, 65, 68, 70, 72, 75, 78, 80, 85, 90, 105. Construct a box plot and identify any outliers.
茎叶图展示了以下数据（单位：mm）：42, 45, 48, 51, 53, 53, 55, 56, 58, 60, 62, 64, 65, 68, 70, 72, 75, 78, 80, 85, 90, 105。请绘制箱线图并识别异常值。
First, locate the median. With n=22, the median is the average of the 11th and 12th values: (62+64)/2 = 63. The lower quartile Q1 is the median of the first 11 values: the 6th value is 53. The upper quartile Q3 is the median of the upper 11 values: the 17th value is 75. IQR = Q3 – Q1 = 75 – 53 = 22.
首先确定中位数。n=22，中位数为第11和12个数据的平均值：(62+64)/2 = 63。下四分位数 Q1 为前11个数据的中位数：第6个数据是53。上四分位数 Q3 为后11个数据的中位数：第17个数据是75。四分位距 IQR = Q3 – Q1 = 22。
Outlier boundaries are Q1 – 1.5×IQR = 53 – 33 = 20 and Q3 + 1.5×IQR = 75 + 33 = 108. Any value below 20 or above 108 is an outlier. The value 105 is inside the upper fence, so no outliers are present. The box plot extends whiskers to the minimum 42 and maximum 105.
异常值界限为 Q1 – 1.5×IQR = 53 – 33 = 20 以及 Q3 + 1.5×IQR = 75 + 33 = 108。低于20或高于108的值视为异常值。数据105位于上界限之内，因此没有异常值。箱线图的须线延伸至最小值42和最大值105。
2. Venn Diagrams and Combined Probability | 维恩图与组合概率
Events A and B are such that P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3. Find P(A ∪ B), P(A’ ∩ B) and determine whether A and B are independent.
事件 A 和 B 满足 P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3。求 P(A ∪ B)，P(A’ ∩ B) 并判断 A 与 B 是否独立。
Using the addition formula: P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.6 + 0.5 – 0.3 = 0.8. For P(A’ ∩ B), this represents the part of B not in A, so P(A’ ∩ B) = P(B) – P(A ∩ B) = 0.5 – 0.3 = 0.2.
利用加法公式：P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.6 + 0.5 – 0.3 = 0.8。P(A’ ∩ B) 表示 B 中不属于 A 的部分，因此 P(A’ ∩ B) = P(B) – P(A ∩ B) = 0.5 – 0.3 = 0.2。
To test independence, check if P(A ∩ B) = P(A)×P(B). Here 0.3 ≠ 0.6×0.5 = 0.3; they are equal, so the events are independent. A Venn diagram would show the intersection exactly equal to the product of the individual probabilities.
检验独立性，需验证 P(A ∩ B) 是否等于 P(A)×P(B)。此处 0.3 = 0.6×0.5，成立，因此事件独立。维恩图中交集部分恰好等于各自概率的乘积。
3. Discrete Random Variable: Expectation and Variance | 离散随机变量：期望与方差
The probability distribution of a discrete random variable X is given in the table:
离散随机变量 X 的概率分布如下表：
| x | 1 | 2 | 3 | 4 |
|---|---|---|---|---|
| P(X=x) | 0.2 | 0.3 | 0.1 | 0.4 |
Calculate E(X), Var(X), and hence find E(3X – 2) and Var(3X – 2).
计算 E(X)、Var(X)，并由此求 E(3X – 2) 和 Var(3X – 2)。
E(X) = Σ x·P(X=x) = 1×0.2 + 2×0.3 + 3×0.1 + 4×0.4 = 0.2 + 0.6 + 0.3 + 1.6 = 2.7. Next, E(X²) = 1²×0.2 + 2²×0.3 + 3²×0.1 + 4²×0.4 = 0.2 + 1.2 + 0.9 + 6.4 = 8.7. Var(X) = E(X²) – [E(X)]² = 8.7 – 2.7² = 8.7 – 7.29 = 1.41.
期望 E(X) = Σ x·P(X=x) = 1×0.2 + 2×0.3 + 3×0.1 + 4×0.4 = 2.7。再求 E(X²)=1²×0.2+4×0.3+9×0.1+16×0.4=8.7。方差 Var(X)=E(X²)–[E(X)]²=8.7–7.29=1.41。
The linear transformation rules give: E(3X – 2) = 3E(X) – 2 = 3×2.7 – 2 = 6.1. Var(3X – 2) = 3² Var(X) = 9 × 1.41 = 12.69.
线性变换规则：E(3X – 2) = 3×2.7 – 2 = 6.1；Var(3X – 2) = 3²×1.41 = 12.69。
4. Normal Distribution: Standardisation and Inverse | 正态分布：标准化与逆运算
The random variable X follows a normal distribution with mean 50 and standard deviation 16, i.e. X ~ N(50, 16²). Find P(X < 60), P(X > 40), and the value of k such that P(X < k) = 0.85.
随机变量 X 服从均值为50、标准差为16的正态分布，即 X ~ N(50, 16²)。求 P(X < 60)、P(X > 40)，以及满足 P(X < k) = 0.85 的 k 值。
Standardise to Z: Z = (X – 50)/16. For X = 60, Z = (60 – 50)/16 = 0.625. Using the normal table, P(Z < 0.625) ≈ 0.7340 (interpolating between 0.62 and 0.63). So P(X < 60) ≈ 0.734.
化为标准正态：Z = (X – 50)/16。当 X = 60 时，Z = 0.625。查表得 P(Z < 0.625) 约 0.7340（在0.62与0.63之间插值），故 P(X < 60) ≈ 0.734。
For P(X > 40), Z = (40 – 50)/16 = –0.625. By symmetry, P(Z < –0.625) = P(Z > 0.625) = 1 – 0.7340 = 0.2660. Hence P(X > 40) = 1 – 0.2660 = 0.7340 (or directly: 1 – P(Z < –0.625) = 0.7340).
计算 P(X > 40)，Z = –0.625，利用对称性，P(Z > 0.625)=0.2660，那么 P(X > 40) = 1 – 0.2660 = 0.7340。
To find k for a probability of 0.85, look up the Z-value with Φ(z) = 0.85. The table gives approximately z = 1.0364. Then k = μ + zσ = 50 + 1.0364×16 ≈ 66.58.
求满足累积概率0.85的k值：查表得 Φ(z)=0.85 时 z ≈ 1.0364，故 k = 50 + 1.0364×16 ≈ 66.58。
5. Product Moment Correlation and Regression Line | 积矩相关系数与回归线
Five paired observations give: Σx = 30, Σy = 40, Σx² = 220, Σy² = 370, Σxy = 275. Calculate the product moment correlation coefficient r, and find the regression line of y on x in the form y = a + bx.
五对观测数据：Σx=30, Σy=40, Σx²=220, Σy²=370, Σxy=275。计算积矩相关系数 r，并求 y 对 x 的回归线，形式为 y = a + bx。
First compute summary statistics: Sxx = Σx² – (Σx)²/n = 220 – 30²/5 = 220 – 180 = 40. Syy = 370 – 40²/5 = 370 – 320 = 50. Sxy = 275 – (30×40)/5 = 275 – 240 = 35.
首先计算汇总统计量：Sxx = 220 – 900/5 = 40；Syy = 370 – 1600/5 = 50；Sxy = 275 – 1200/5 = 35。
r = Sxy / √(Sxx × Syy) = 35 / √(40 × 50) = 35 / √2000 ≈ 35 / 44.721 = 0.7826
回归系数 b = Sxy / Sxx = 35 / 40 = 0.875. Means: x̄ = 30/5 = 6, ȳ = 40/5 = 8. Intercept a = ȳ – b x̄ = 8 – 0.875×6 = 8 – 5.25 = 2.75.
斜率 b = 35/40 = 0.875。均值 x̄=6, ȳ=8，截距 a = 8 – 0.875×6 = 2.75。
Regression equation: y = 2.75 + 0.875x
The correlation r ≈ 0.7826 indicates a moderate positive linear relationship. For each unit increase in x, y is expected to rise by 0.875 units on average.
相关系数 r 约 0.7826 表明存在中等程度的正线性相关。x 每增加一个单位，y 平均增加 0.875 个单位。
6. Conditional Probability and Independence | 条件概率与独立性
Given P(A) = 0.7, P(B) = 0.4 and P(A ∩ B) = 0.28. Find P(B | A) and P(A | B’), and state, with a reason, whether A and B are independent.
已知 P(A)=0.7, P(B)=0.4, P(A∩B)=0.28。求 P(B|A) 和 P(A|B’)，并判断 A 与 B 是否独立，给出理由。
Conditional probability: P(B | A) = P(A ∩ B)/P(A) = 0.28/0.7 = 0.4. Since P(B | A) = P(B) = 0.4, this already suggests independence. To confirm, check P(A∩B) = P(A)×P(B): 0.28 = 0.7×0.4, which holds exactly.
条件概率：P(B|A) = 0.28/0.7 = 0.4。由于 P(B|A)=P(B)=0.4，这已暗示独立。严格验证：P(A∩B)=0.7×0.4=0.28，恰好成立。
Now find P(A | B’). First compute P(B’) = 1 – 0.4 = 0.6. P(A ∩ B’) = P(A) – P(A ∩ B) = 0.7 – 0.28 = 0.42. Then P(A | B’) = 0.42 / 0.6 = 0.7. This equals P(A), further confirming independence.
再求 P(A|B’)：P(B’)=0.6，P(A∩B’)=0.7–0.28=0.42，则 P(A|B’)=0.42/0.6=0.7，与 P(A) 相等，再次印证独立。
Thus events A and B are independent. In general, independence means the occurrence of one event does not affect the probability of the other, confirmed by all conditional probabilities equalling the original probabilities.
因此事件 A 和 B 独立。一般而言，独立性意味着一个事件的发生不影响另一事件的概率，由所有条件概率均等于原概率得到验证。
7. Histograms and Frequency Density | 直方图与频率密度
A grouped frequency table for the time taken (in minutes) by 80 students is shown. Draw a histogram and estimate the median.
下表为 80 名学生所用时间（分钟）的分组频数表。请绘制直方图并估算中位数。
| Time (min) | Frequency |
|---|---|
| 0–10 | 12 |
| 10–15 | 18 |
| 15–25 | 24 |
| 25–35 | 16 |
| 35–50 | 10 |
We first calculate frequency density = frequency / class width. For 0–10: width 10, FD = 12/10 = 1.2. 10–15: width 5, FD = 18/5 = 3.6. 15–25: width 10, FD = 24/10 = 2.4. 25–35: width 10, FD = 16/10 = 1.6. 35–50: width 15, FD = 10/15 ≈ 0.667. The histogram plots FD on the vertical axis against the time intervals.
首先计算频率密度 = 频数 / 组距。0–10：宽度10, FD=1.2；10–15：宽度5, FD=3.6；15–25：宽度10, FD=2.4；25–35：宽度10, FD=1.6；35–50：宽度15, FD≈0.667。直方图以频率密度为纵轴，时间区间为横轴绘制。
To estimate the median, find the interval containing the 40th value. Cumulative frequencies: 12, 30, 54, 70, 80. The median lies in 15–25. Use linear interpolation: lower boundary 15, cumulative before 30, frequency in class 24, class width 10. Median ≈ 15 + ((40 – 30)/24)×10 = 15 + (10/24)×10 = 15 + 4.17 = 19.17 minutes.
估计中位数：累计频数分别为12, 30, 54, 70, 80，第40个值落在15–25组。线性插值：下界15，前累计30，组内频数24，组距10。中位数 ≈ 15 + (10/24)×10 = 19.17 分钟。
8. Expected Value and Fair Game | 期望值与公平游戏
In a game, a fair coin is tossed twice.
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