AS Cambridge Statistics: Unit Test Mock Paper Walkthrough | AS剑桥统计：单元测试模拟卷解析

📚 AS Cambridge Statistics: Unit Test Mock Paper Walkthrough | AS剑桥统计：单元测试模拟卷解析

This detailed walkthrough breaks down a carefully designed AS-level Statistics mock paper covering stem-and-leaf diagrams, measures of spread, probability, discrete random variables, the binomial and normal distributions. Each question is solved step by step, explaining both the reasoning and the correct use of standard notation to help you build confidence for your unit test.

这份详尽的解析拆解了一套精心设计的AS统计模拟试卷，涵盖茎叶图、离散程度、概率、离散随机变量、二项分布与正态分布。每道题都逐步求解，解释推理过程及标准符号的正确用法，帮助你建立对单元测试的信心。

1. Stem-and-Leaf Diagram and Box Plot | 茎叶图与箱形图

A stem-and-leaf diagram shows the marks of 20 students in a test. Key: 6|2 means 62. Stem: 4 | 5 8 ; 5 | 0 1 2 3 5 7 9 ; 6 | 2 4 4 6 8 8 ; 7 | 1 3 5 ; 8 | 0. We need to find the median, quartiles, draw a box plot and check for outliers using the IQR method.

一张茎叶图显示了20名学生的考试成绩。图例：6|2表示62分。茎：4|5 8；5|0 1 2 3 5 7 9；6|2 4 4 6 8 8；7|1 3 5；8|0。要求找出中位数、四分位数、绘制箱形图并用IQR法检测异常值。

The ordered data set is: 45, 48, 50, 51, 52, 53, 55, 57, 59, 62, 64, 64, 66, 68, 68, 71, 73, 75, 80. With n = 20, the position of the median is (n+1)/2 = 10.5, so the median lies between the 10th and 11th values.

按顺序排列的数据为：45, 48, 50, 51, 52, 53, 55, 57, 59, 62, 64, 64, 66, 68, 68, 71, 73, 75, 80。n=20，中位数位于第(n+1)/2=10.5位，即第10和第11个值之间。

Median = (59 + 62) / 2 = 60.5

The lower quartile Q₁ is at position (n+1)/4 = 5.25, interpolating between the 5th (52) and 6th (53) values: Q₁ = 52 + 0.25 × (53 − 52) = 52.25. The upper quartile Q₃ is at position 3(n+1)/4 = 15.75, between the 15th and 16th values (both are 68), giving Q₃ = 68.

下四分位数Q₁位于第(n+1)/4=5.25位，在第5个（52）和第6个（53）之间插值：Q₁ = 52 + 0.25×(53−52) = 52.25。上四分位数Q₃位于第3(n+1)/4=15.75位，在第15个和第16个（均为68）之间，得到Q₃ = 68。

Interquartile range IQR = Q₃ − Q₁ = 68 − 52.25 = 15.75. Lower fence = Q₁ − 1.5 × IQR = 28.625; upper fence = Q₃ + 1.5 × IQR = 91.625. No data points fall outside these fences, so there are no outliers. Draw the box plot with whiskers from 45 to 80 and a box from 52.25 to 68, with a median line at 60.5.

四分位距IQR = Q₃−Q₁ = 68 − 52.25 = 15.75。下界 = Q₁−1.5×IQR = 28.625；上界 = Q₃+1.5×IQR = 91.625。没有数据点超出边界，因此无异常值。绘制箱形图时，触须从45到80，盒体从52.25到68，中位线在60.5。

2. Mean, Variance and Standard Deviation | 均值、方差与标准差

Given the data set: 10, 12, 15, 9, 14, 8, 16, 11. Calculate the mean and standard deviation, then use the coding y = x − 10 to show how coding affects these measures.

给定数据集：10, 12, 15, 9, 14, 8, 16, 11。计算均值与标准差，然后使用编码 y = x − 10 说明编码对这些测度的影响。

Sum of x = 10+12+15+9+14+8+16+11 = 95. n = 8. Mean x̄ = 95/8 = 11.875. For variance we use the formula s² = (Σx² − n x̄²) / (n−1).

x的总和 = 95，n = 8，均值x̄ = 11.875。计算方差使用公式 s² = (Σx² − n x̄²) / (n−1)。

Σx² = 10² + 12² + 15² + 9² + 14² + 8² + 16² + 11² = 100 + 144 + 225 + 81 + 196 + 64 + 256 + 121 = 1187

Then s² = (1187 − 8 × 11.875²) / 7 = (1187 − 8 × 141.015625) / 7 = (1187 − 1128.125) / 7 = 58.875 / 7 ≈ 8.4107. Standard deviation s ≈ √8.4107 ≈ 2.9001.

于是 s² ≈ 8.4107，标准差 s ≈ 2.9001。

Using y = x − 10 gives the values 0, 2, 5, −1, 4, −2, 6, 1. Clearly Σy = −5+? No, check: 0+2+5−1+4−2+6+1 = 15. Mean ȳ = 15/8 = 1.875. Notice that x̄ = ȳ + 10, which is a direct translation. The variances are identical: s_y² = s_x², because subtracting a constant does not change spread. Indeed, we can verify s_y² = (Σy² − n ȳ²)/(n−1) = (0+4+25+1+16+4+36+1 − 8×1.875²)/7 = (87 − 28.125)/7 = 58.875/7, same as before.

使用 y = x − 10 得到数据 0, 2, 5, −1, 4, −2, 6, 1。总和为15，均值ȳ = 1.875。注意到 x̄ = ȳ + 10，正是平移结果。方差保持不变，因为减去一个常数不改变离散程度。验证可得 s_y² 与 s_x² 相同。

3. Probability and Venn Diagrams | 概率与维恩图

In a group of 50 students, 30 study Mathematics (M), 20 study Physics (P), and 10 study both. Construct a Venn diagram and calculate various probabilities.

一组50名学生中，30人学习数学(M)，20人学习物理(P)，10人两者都学。构建维恩图并计算各种概率。

M only	20
P only	10
Both	10
Neither	10

From the diagram, P(M) = 30/50 = 0.6, P(P) = 20/50 = 0.4. P(M ∪ P) = (20+10+10)/50 = 40/50 = 0.8. For conditional probability P(P | M), we restrict attention to the 30 students studying Mathematics: among them, 10 also study Physics, so P(P | M) = 10/30 = 1/3.

由图可知，P(M)=0.6，P(P)=0.4。P(M ∪ P)=0.8。条件概率 P(P|M) 限定在30名学数学的学生中，其中10人也学物理，所以 P(P|M)=1/3。

Independence check: P(M ∩ P) = 10/50 = 0.2, while P(M) × P(P) = 0.6 × 0.4 = 0.24. Since 0.2 ≠ 0.24, the events M and P are not independent.

独立性检验：P(M∩P)=0.2，而 P(M)×P(P)=0.24。由于不相等，M与P不独立。

4. Conditional Probability and Tree Diagrams | 条件概率与树状图

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, and that at least one is red.

一个袋子有5颗红球和3颗蓝球，不放回地连续抽取两颗。画出树状图，求两颗同色及至少一颗红色的概率。

First draw: P(R₁) = 5/8, P(B₁) = 3/8. Second draw probabilities depend on the first: if the first is red, P(R₂|R₁) = 4/7, P(B₂|R₁) = 3/7; if first is blue, P(R₂|B₁) = 5/7, P(B₂|B₁) = 2/7.

第一次抽取：P(R₁)=5/8, P(B₁)=3/8。第二次概率依赖于第一次结果：若第一次为红，P(R₂|R₁)=4/7, P(B₂|R₁)=3/7；若第一次为蓝，P(R₂|B₁)=5/7, P(B₂|B₁)=2/7。

Both same colour = red both + blue both = (5/8)×(4/7) + (3/8)×(2/7) = 20/56 + 6/56 = 26/56 = 13/28. At least one red = 1 − P(both blue) = 1 − 6/56 = 50/56 = 25/28.

两颗同色 = 双双红 + 双双蓝 = 13/28。至少一颗红球 = 1 − 全蓝概率 = 25/28。

Another typical question: Given the first drawn is red, find the probability that the second is also red. This is simply P(R₂|R₁) = 4/7, read directly from the tree.

另一典型问题：已知第一次抽到红球，求第二次也是红球的概率。直接从树状图读取即为 P(R₂|R₁)=4/7。

5. Discrete Random Variables | 离散随机变量

A discrete random variable X has the following probability distribution: x = 1, 2, 3, 4 with P(X=x) = 0.2, p, 0.3, q respectively. Given that E(X) = 2.6, find p and q, then calculate Var(X). Also find E(Y) and Var(Y) for Y = 2X + 1.

一个离散随机变量X的概率分布为：x=1,2,3,4，对应概率0.2, p, 0.3, q。已知E(X)=2.6，求p和q，再算Var(X)以及Y=2X+1的期望和方差。

Since probabilities sum to 1, 0.2 + p + 0.3 + q = 1 → p + q = 0.5. E(X) = 1×0.2 + 2p + 3×0.3 + 4q = 0.2 + 2p + 0.9 + 4q = 1.1 + 2p + 4q = 2.6 → 2p + 4q = 1.5 → p + 2q = 0.75. Subtracting p+q=0.5 from p+2q=0.75 gives q = 0.25, hence p = 0.25.

所有概率之和为1，得 p+q=0.5。期望方程为 1.1+2p+4q=2.6，化简得 p+2q=0.75。与 p+q=0.5 联立解得 q=0.25, p=0.25。

E(X²) = 1²×0.2 + 2²×0.25 + 3²×0.3 + 4²×0.25 = 0.2 + 1.0 + 2.7 + 4.0 = 7.9. Thus Var(X) = E(X²) − [E(X)]² = 7.9 − 2.6² = 7.9 − 6.76 = 1.14.

对于Y=2X+1，应用线性变换性质：E(Y) = 2E(X) + 1 = 2×2.6 + 1 = 6.2；Var(Y) = 2² Var(X) = 4 × 1.14 = 4.56。

使用性质：E(Y) = 6.2, Var(Y) = 4.56。

6. Binomial Distribution | 二项分布

Let X ~ B(10, 0.3). Determine P(X=3), P(X ≤ 3) and P(X > 5).

设 X ~ B(10, 0.3)。求 P(X=3), P(X≤3) 和 P(X>5)。

Using the binomial formula: P(X=k) = ¹⁰Cₖ (0.3)ᵏ (0.7)¹⁰⁻ᵏ. For k=3, ¹⁰C₃ = 120, (0.3)³ = 0.027, (0.7)⁷ ≈ 0.0823543. So P(X=3) ≈ 120 × 0.027 × 0.0823543 = 120 × 0.00222356 = 0.2668 (to 4 d.p.).

使用二项分布公式：P(X=3) ≈ 0.2668。

P(X ≤ 3) = P(0) + P(1) + P(2) + P(3). We can calculate each term or use cumulative tables. For demonstration: P(0)=0.7¹⁰≈0.0282, P(1)=10×0.3×0.7⁹≈0.1211, P(2)=45×0.09×0.7⁸≈0.2335, adding to 0.3828 with P(3) gives ≈0.6496. P(X > 5) = 1 − P(X ≤ 5). Computing P(4) and P(5) continues: P(4)≈0.2001, P(5)≈0.1029, sum ≤5≈0.9526, thus P(X > 5)≈0.0474.

P(X≤3) ≈ 0.6496。P(X>5)=1−P(X≤5)≈0.0474。考试中可直接使用二项分布累積表。

7. Normal Distribution | 正态分布

The weight of cereal filled by a machine is normally distributed with

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AS Cambridge Statistics: Unit Test Mock Paper Walkthrough | AS剑桥统计：单元测试模拟卷解析