Cambridge Y13 Statistics: Mock Unit Test with Worked Solutions | 剑桥Y13统计：单元模拟测试详解

📚 Cambridge Y13 Statistics: Mock Unit Test with Worked Solutions | 剑桥Y13统计：单元模拟测试详解

This article presents a complete mock unit test covering core topics from the Cambridge Y13 Statistics syllabus, including Poisson distribution, normal approximations, hypothesis tests for proportions and Poisson means, continuous random variables, sampling distributions, confidence intervals, and linear combinations of normal variables. Each question is followed by detailed step-by-step solutions with bilingual explanations to reinforce key concepts and exam technique.

本文提供一套完整的模拟单元测试，涵盖剑桥Y13统计课程的核心主题，包括泊松分布、正态近似、比例与泊松均值的假设检验、连续随机变量、抽样分布、置信区间以及正态变量的线性组合。每道题后附有详尽的逐步解答，并用中英双语解释，以巩固关键概念和应试技巧。

1. Poisson Distribution | 泊松分布

Question: The number of calls received by a call centre follows a Poisson distribution with mean 4 per 5‑minute interval. Find the probability that in a 10‑minute interval: (a) there are exactly 6 calls; (b) there are more than 10 calls; (c) determine the most likely number of calls.

问题： 某呼叫中心接到的电话数服从泊松分布，每5分钟平均4通。求在一个10分钟间隔内：(a) 恰好有6通电话的概率；(b) 超过10通电话的概率；(c) 最可能出现的电话数。

Solution:

First, calculate the mean for 10 minutes. Since the rate is 4 per 5 minutes, for 10 minutes λ = 2 × 4 = 8.

首先，计算10分钟的平均值。因为每5分钟4通，所以10分钟的λ = 2 × 4 = 8。

(a) P(X = 6) = e⁻⁸ · 8⁶ / 6! ≈ 0.1221 (4 d.p.). The calculation uses the Poisson probability mass function.

(a) P(X = 6) = e⁻⁸ · 8⁶ / 6! ≈ 0.1221（保留四位小数）。计算使用了泊松概率质量函数。

(b) P(X > 10) = 1 − P(X ≤ 10). Using cumulative Poisson probabilities for λ = 8, P(X ≤ 10) ≈ 0.8159, so P(X > 10) ≈ 0.1841.

(b) P(X > 10) = 1 − P(X ≤ 10)。查λ = 8的累积泊松概率，P(X ≤ 10) ≈ 0.8159，因此 P(X > 10) ≈ 0.1841。

(c) For a Poisson distribution with λ = 8, the mode occurs at the integer(s) where P(X = k) is maximised. Here P(X=7) ≈ P(X=8) ≈ 0.1396, so the most likely number of calls is 7 and 8 (bimodal).

(c) 对于λ = 8的泊松分布，众数出现在P(X = k)最大的整数处。此处P(X=7) ≈ P(X=8) ≈ 0.1396，因此最可能的呼叫次数是7和8（双众数）。

2. Normal Approximation to Poisson | 泊松分布的正态近似

Question: Emergency admissions to a hospital follow a Poisson process with a mean of 14 per day. Use a normal approximation to estimate the probability that on a given day there are more than 18 admissions. Justify the use of the approximation.

问题： 某医院急诊入院人数服从泊松过程，日均14人。利用正态近似估计在给定一天超过18人入院的概率，并说明使用近似的依据。

Solution:

The Poisson mean is λ = 14. Since λ > 10, the normal approximation N(λ, λ) is appropriate. Thus we use Y ~ N(14, 14). To improve accuracy, a continuity correction is applied. For P(X > 18) we consider P(Y > 18.5).

泊松均值为λ = 14。由于λ > 10，适合使用正态近似N(λ, λ)。因此我们采用Y ~ N(14, 14)。为提高精度，应用连续性校正。对于P(X > 18)，我们考虑P(Y > 18.5)。

Standardize: Z = (18.5 − 14) / √14 ≈ 4.5 / 3.7417 ≈ 1.202. Then P(Z > 1.202) ≈ 1 − 0.8853 = 0.1147 (using standard normal table). So the approximate probability is 0.1147.

标准化：Z = (18.5 − 14) / √14 ≈ 4.5 / 3.7417 ≈ 1.202

Published by TutorHao | Year 13 统计 Revision Series | aleveler.com

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