Pre-U CIE Statistics: Mock Exam Paper Walkthrough | Pre-U CIE 统计：模拟试卷解析

📚 Pre-U CIE Statistics: Mock Exam Paper Walkthrough | Pre-U CIE 统计：模拟试卷解析

A well-designed mock examination is one of the most effective tools for consolidating statistical understanding before the Pre-U CIE assessment. This walkthrough dissects a typical paper covering probability, discrete and continuous distributions, sampling theory, inference, and regression. Every solution is broken into clear steps, highlighting common pitfalls and examiner expectations so that you can approach your real exam with confidence.

精心设计的模拟试卷是巩固 Pre-U CIE 统计知识的有效工具之一。本解析拆解了一份涵盖概率、离散与连续分布、抽样理论、推断和回归的典型试卷。每个题的解答都分解为清晰的步骤，突出常见错误和考官期望，帮助您自信应对正式考试。

1. Probability and Counting Principles | 概率与计数原理

Question: A bag contains 5 red, 3 blue and 2 green marbles. Three marbles are drawn at random without replacement. Find the probability that exactly two are red.

问题：一个袋子里有5颗红色、3颗蓝色和2颗绿色弹珠。随机抽取三颗且不放回。求恰好有两颗红色的概率。

Step 1: Determine the total number of equally likely outcomes. Three marbles are chosen from 10, giving 10C3 = 120 possible combinations.

步骤1：确定等可能结果的总数。从10颗中选3颗，共有10选3的组合数，即120种可能。

Step 2: For exactly two reds, we must choose 2 reds from the 5 available and 1 non-red from the remaining 5 (3 blue + 2 green). The number of favourable combinations is 5C2 × 5C1 = 10 × 5 = 50.

步骤2：要恰好有两颗红色，必须从5颗红色中选2颗，并从其余5颗非红色（3蓝+2绿）中选1颗。有利组合数为5选2乘以5选1，即10×5=50。

Step 3: The required probability is 50 / 120 = 5/12 ≈ 0.4167.

步骤3：所求概率为50/120=5/12≈0.4167。

A common mistake is to treat the draws as ordered or to forget that the non-reds must be counted together. Always define the sample space consistently as combinations when order does not matter.

常见错误是把抽取视为有序或忘记把非红色合并计数。当顺序无关时，始终将样本空间统一视为组合。

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2. Discrete Random Variables and Expectation | 离散随机变量与期望

Question: A discrete random variable X takes values 0, 1, 2 and 3 with probabilities given by P(X = x) = k(x + 1). Find the value of k, E(X) and Var(X).

问题：离散随机变量X取值为0,1,2,3，概率由P(X=x)=k(x+1)给出。求k的值、E(X)和Var(X)。

Step 1: Sum the probabilities over all possible values of x and set equal to 1. k[(0+1)+(1+1)+(2+1)+(3+1)] = k(1+2+3+4) = 10k = 1, so k = 0.1.

步骤1：对所有可能的x值概率求和并令其等于1。k[(0+1)+(1+1)+(2+1)+(3+1)]=k(1+2+3+4)=10k=1，所以k=0.1。

Step 2: Compute the probability distribution.

x	0	1	2	3
P(X=x)	0.1	0.2	0.3	0.4

步骤2：计算概率分布如上表。

Step 3: E(X) = Σ x · P(X=x) = (0)(0.1)+(1)(0.2)+(2)(0.3)+(3)(0.4) = 0 + 0.2 + 0.6 + 1.2 = 2.0.

步骤3：E(X)=Σ x·P(X=x)=0+0.2+0.6+1.2=2.0。

Step 4: E(X²) = Σ x²·P(X=x) = (0²)(0.1)+(1²)(0.2)+(2²)(0.3)+(3²)(0.4)=0 + 0.2 + 1.2 + 3.6 = 5.0.

步骤4：E(X²)=0+0.2+1.2+3.6=5.0。

Step 5: Var(X) = E(X²) – [E(X)]² = 5.0 – (2.0)² = 1.0.

步骤5：Var(X)=5.0–4.0=1.0。

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3. Binomial Distribution | 二项分布

Question: X ~ B(10, 0.3). Find P(X ≥ 4).

问题：X 服从二项分布 B(10,0.3)。求 P(X ≥ 4)。

Step 1: Recognise that P(X ≥ 4) = 1 – P(X ≤ 3) is far easier to compute using cumulative binomial tables or formula.

步骤1：认识到 P(X ≥ 4)=1–P(X ≤ 3) 利用累积二项分布表或公式计算更为简便。

Step 2: Using the binomial formula, P(X = r) = 10Cr (0.3)^r (0.7)^(10–r).

步骤2：利用二项分布公式 P(X=r)=₁₀Cᵣ (0.3)^r (0.7)^(10–r)。

Step 3: Compute P(X=0) = 1×(0.3)⁰×(0.7)¹⁰ ≈ 0.0282, P(X=1) = 10×0.3×0.7⁹ ≈ 0.1211, P(X=2) = 45×0.3²×0.7⁸ ≈ 0.2335, P(X=3) = 120×0.3³×0.7⁷ ≈ 0.2668.

步骤3：计算得 P(X=0)≈0.0282, P(X=1)≈0.1211, P(X=2)≈0.2335, P(X=3)≈0.2668。

Step 4: Summing gives P(X ≤ 3) ≈ 0.0282+0.1211+0.2335+0.2668 = 0.6496. Thus P(X ≥ 4) = 1 – 0.6496 = 0.3504.

步骤4：求和得 P(X ≤ 3)≈0.6496。因此 P(X ≥ 4)=0.3504。

Always verify whether the question expects an exact calculation or use of tables; here the working is shown to a reasonable degree of accuracy.

务必确认题目要求精确计算还是查表；此处展示适当精度的计算过程。

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4. Poisson Distribution | 泊松分布

Question: Calls arrive at a switchboard at a constant average rate of 2 per minute. Find the probability of receiving exactly 5 calls in a 3-minute period.

问题：电话总机平均每分钟接到2次呼叫。求在3分钟时段内恰好接到5次呼叫的概率。

Step 1: For a 3‑minute interval, the mean λ = 2 × 3 = 6 calls. The number of calls, Y, follows Po(6).

步骤1：对于3分钟间隔，均值 λ = 2×3 = 6次。呼叫次数 Y 服从泊松分布 Po(6)。

Step 2: The Poisson probability mass function is P(Y = y) = (e^(–λ) λ^y) / y!. For y = 5 and λ = 6, P(Y = 5) = (e^(–6) × 6⁵) / 5!.

步骤2：泊松概率质量函数为 P(Y=y)=(e^(–λ)λ^y)/y!。代入 y=5,λ=6，P(Y=5)=(e^(–6)×6⁵)/5!。

Step 3: 6⁵ = 7776, 5! = 120, so the expression becomes (e^(–6) × 7776) / 120 = 64.8 × e^(–6). Using e^(–6) ≈ 0.00247875, probability ≈ 64.8 × 0.00247875 ≈ 0.1606.

步骤3：6⁵=7776,5!=120，表达式为(e^(–6)×7776)/120=64.8×e^(–6)。利用e^(–6)≈0.00247875，概率≈64.8×0.00247875≈0.1606。

Thus there is about a 16.1% chance of exactly 5 calls during three minutes.

因此在三分钟内恰好接到5次呼叫的概率约为16.1%。

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5. Normal Distribution | 正态分布

Question: The heights of a population are normally distributed with mean μ = 170 cm and standard deviation σ = 6 cm. Find the proportion of individuals with heights between 165 cm and 180 cm.

问题：某人群身高服从正态分布，均值μ=170 cm，标准差σ=6 cm。求身高在165 cm至180 cm之间的人数比例。

Step 1: Standardise the bounds: Z₁ = (165 – 170)/6 = –0.8333, Z₂ = (180 – 170)/6 = 1.6667.

步骤1：标准化界限：Z₁=(165–170)/6=–0.8333，Z₂=(180–170)/6=1.6667。

Step 2: The required proportion is Φ(1.6667) – Φ(–0.8333). By symmetry, Φ(–0.8333) = 1 – Φ(0.8333).

步骤2：所求比例为 Φ(1.6667)–Φ(–0.8333)。由对称性，Φ(–0.8333)=1–Φ(0.8333)。

Step 3: From standard normal tables, Φ(1.6667) ≈ 0.9522, Φ(0.8333) ≈ 0.7976. Thus the required proportion = 0.9522 – (1 – 0.7976) = 0.9522 – 0.2024 = 0.7498.

步骤3：查标准正态表，Φ(1.6667)≈0.9522，Φ(0.8333)≈0.7976。因此比例=0.9522–(1–0.7976)=0.9522–0.2024=0.7498。

About 74.98% of the population have heights in that range. Always sketch the curve to check that the difference of cumulative areas makes sense.

约74.98%的人身高于此区间。务必画曲线草图，以确认累积区域之差符合直觉。

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6. Sampling Distributions and the Central Limit Theorem | 抽样分布与中心极限定理

Question: A random sample of size n = 50 is drawn from a population with mean 100 and standard deviation 15. Find the approximate probability that the sample mean x̄ exceeds 103.

问题：从均值为100、标准差为15的总体中抽取容量n=50的随机样本。求样本均值x̄超过103的近似概率。

Step 1: By the Central Limit Theorem, for large n, x̄ ~ N(μ, σ²/n) approximately. Here the standard error = σ/√n = 15/√50 ≈ 2.1213.

步骤1：根据中心极限定理，当n较大时x̄近似服从N(μ,σ²/n)。此处标准误=15/√50≈2.1213。

Step 2: Standardise: Z = (103 – 100) / 2.1213 ≈ 1.4142.

步骤2：标准化：Z=(103–100)/2.1213≈1.4142。

Step 3: P(x̄ > 103) = P(Z > 1.4142) = 1 – Φ(1.4142). Using normal tables, Φ(1.41) ≈ 0.9207, Φ(1.42) ≈ 0.9222; linear interpolation gives Φ(1.4142) ≈ 0.9213. Hence the probability ≈ 1 – 0.9213 = 0.0787.

步骤3：P(x̄>103)=P(Z>1.4142)=1–Φ(1.4142)。查表得Φ(1.41)≈0.9207,Φ(1.42)≈0.9222；线性插值约为0.9213，因此概率≈1–0.9213=0.0787。

Even though the population distribution is not specified as normal, the sample size of 50 is sufficient for the CLT approximation to be reliable.

即使总体分布未指明为正态，样本量50足够使中心极限定理的近似可靠。

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7. Confidence Interval for a Population Mean | 总体均值的置信区间

Question: A random sample of 25 observations yields a sample mean of 52 and a sample standard deviation of 5. Assuming the population is normally distributed, construct a 95% confidence interval for the population mean μ.

问题：一个容量25的随机样本给出样本均值52和样本标准差5。假设总体服从正态分布，构造总体均值μ的95%置信区间。

Step 1: Since the population standard deviation is unknown and the sample size is small, use the t-distribution with df = n – 1 = 24. The 95% critical value t* for two tails is 2.064.

步骤1：由于总体标准差未知且样本量较小，使用自由度df=24的t分布。双侧95%临界值t*为2.064。

Step 2: The standard error is s/√n = 5/√25 = 5/5 = 1.

步骤2：标准误为s/√n=5/5=1。

Step 3: The confidence interval is x̄ ± t* × (s/√n) = 52 ± 2.064×1 = (49.936, 54.064).

步骤3：置信区间为 x̄±t*×(s/√n)=52±2.064×1=(49.936, 54.064)。

We are 95% confident that the population mean lies between 49.94 and 54.06 (to two decimal places).

我们有95%的信心认为总体均值介于49.94和54.06之间（保留两位小数）。

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8. Hypothesis Testing (One-sample t-test) | 假设检验（单样本t检验）

Question: A random sample of 20 observations from a normal population gives a sample mean of 48 and a sample standard deviation of 7. Test at the 5% significance level whether the population mean μ differs from 50.

问题：来自正态总体的一个容量20的随机样本给出样本均值48和样本标准差7。在5%显著性水平下检验总体均值μ是否不等于50。

Step 1: State hypotheses. H₀: μ = 50; H₁: μ ≠ 50 (two-tailed test).

步骤1：陈述假设。H₀: μ=50；H₁: μ≠50（双侧检验）。

Step 2: Compute the test statistic t = (x̄ – μ₀) / (s/√n) = (48 – 50) / (7/√20) = –2 / (7/4.4721) ≈ –2 / 1.5652 ≈ –1.2778.

步骤2：计算检验统计量t=(48–50)/(7/√20)=–2/1.5652≈–1.2778。

Step 3: The critical values for df = 19 at 5% two-tailed are ±2.093. Since –1.2778 lies between –2.093 and 2.093, we fail to reject H₀.

步骤3：自由度19、双侧5%的临界值为±2.093。由于–1.2778介于–2.093与2.093之间，不能拒绝H₀。

Step 4: Conclusion in context: There is insufficient evidence at the 5% level to conclude that the population mean differs from 50.

步骤4：实际结论：在5%显著性水平下，没有足够证据表明总体均值不同于50。

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9. Chi-Squared Test for Independence | 卡方独立性检验

Question: A survey classifies 200 people by gender and preference for a new product. Observed frequencies are: Male–Like: 40, Male–Dislike: 60; Female–Like: 70, Female–Dislike: 30. Test at the 1% significance level whether gender and preference are independent.

问题：一项调查根据性别和对新产品的偏好将200人分类。观察频数为：男–喜欢40，男–不喜欢60；女–喜欢70，女–不喜欢30。在1%显著性水平下检验性别与偏好是否独立。

Step 1: Compute expected frequencies under independence. Row totals: Male 100, Female 100; Column totals: Like 110, Dislike 90. Expected for Male–Like = (100×110)/200 = 55. Similarly, Male–Dislike = 45, Female–Like = 55, Female–Dislike = 45.

步骤1：计算独立性假设下的期望频数。行合计：男100，女100；列合计：喜欢110，不喜欢90。男–喜欢期望=(100×110)/200=55。类似地，男–不喜欢=45，女–喜欢=55，女–不喜欢=45。

Step 2: Compute the chi-squared statistic χ² = Σ (O – E)² / E. (40–55)²/55 = 4.0909; (60–45)²/45 = 5.0; (70–55)²/55 = 4.0909; (30–45)²/45 = 5.0. Sum = 18.1818.

步骤2：计算卡方统计量χ²=Σ(O–E)²/E。四项分别为4.0909,5.0,4.0909,5.0，总和为18.1818。

Step 3: Degrees of freedom = (rows–1)(columns–1) = 1×1 = 1. The 1% critical value for χ² with df = 1 is 6.635. Since 18.18 > 6.635, we reject H₀.

步骤3：自由度=(2–1)×(2–1)=1。χ²(1)在1%的临界值为6.635。因18.18>6.635，拒绝H₀。

Step 4: There is strong evidence at the 1% level to suggest that gender and product preference are associated.

步骤4：在1%水平下有强证据表明性别与产品偏好之间存在关联。

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10. Correlation and Linear Regression | 相关与线性回归

Question: Given the data: x = 1, 2, 3, 4, 5; y = 2.1, 4.0, 5.8, 8.1, 9.9. Calculate Pearson’s correlation coefficient r and find the equation of the regression line of y on x. Predict y when x = 6.

问题：给出数据x=1,2,3,4,5；y=2.1,4.0,5.8,8.1,9.9。计算皮尔逊相关系数r，并求y对x的回归直线方程。预测x=6时的y值。

Step 1: Compute sums: Σx = 15, Σy = 29.9, Σx² = 55, Σy² = 2.1² + 4.0² + 5.8² + 8.1² + 9.9² = 4.41 + 16.0 + 33.64 + 65.61 + 98.01 = 217.67, Σxy = 1×2.1 + 2×4.0 + 3×5.8 + 4×8.1 + 5×9.9 = 2.1 + 8.0 + 17.4 + 32.4 + 49.5 = 109.4. n = 5.

步骤1：计算各项和：Σx=15, Σy=29.9, Σx²=55, Σy²=217.67, Σxy=109.4, n=5。

Step 2: Use the formula: r = [nΣxy – Σx Σy] / √{ [nΣx² – (Σx)²] [nΣy² – (Σy)²] }. Numerator = 5×109.4 – 15×29.9 = 547 – 448.5 = 98.5. Denominator: nΣx² – (Σx)² = 275 – 225 = 50; nΣy² – (Σy)² = 1088.35 – 894.01 = 194.34; product = 50 × 194.34 = 9717; √9717 ≈ 98.575. Thus r ≈ 98.5 / 98.575 ≈ 0.9992.

步骤2：利用公式计算r。分子=98.5，分母的第一部分=50，第二部分=194.34，乘积为9717，开方≈98.575，r≈0.9992。

Step 3: The regression line of y on x is y = a + bx, where b = (nΣxy – Σx Σy) / (nΣx² – (Σx)²) = 98.5 / 50 = 1.97; a = ȳ – b x̄ = (29.9/5) – 1.97×(15/5) = 5.98 – 1.97×3 = 5.98 – 5.91 = 0.07. Hence the equation is y = 0.07 + 1.97x.

步骤3：y对x的回归线：b=98.5/50=1.97；a=ȳ–b x̄=5.98–5.91=0.07，方程为y=0.07+1.97x。

Step 4: When x = 6, predicted y = 0.07 + 1.97×6 = 0.07 + 11.82 = 11.89.

步骤4：当x=6时，预测y=0.07+11.82=11.89。

The extremely high correlation indicates an almost perfect linear relationship, which justifies the prediction within the range of observed data.

极高的相关系数表明几乎完全线性，这为在观测数据范围内进行预测提供了依据。

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