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IAL 2024 Jan S2 Q2

A Level / Edexcel / S2

IAL 2024 Jan Paper · Question 2

题目

Problem

The length of pregnancy for a randomly selected pregnant sheep is D days where

D ~ N(112.4, σ²)

Given that 5% of pregnant sheep have a length of pregnancy of less than 108 days,

(a) find the value of σ

(3)

Qiang selects 25 pregnant sheep at random from a large flock.

(b) Find the probability that more than 3 of these pregnant sheep have a length of pregnancy of less than 108 days.

(2)

Charlie takes 200 random samples of 25 pregnant sheep.

(c) Use a Poisson approximation to estimate the probability that at least 2 of the samples have more than 3 pregnant sheep with a length of pregnancy of less than 108 days.

(3)

(Total for Question 2 is 8 marks)

题目中文翻译

随机选择的一只怀孕母羊的妊娠天数为 D 天，其中

D ~ N(112.4, σ²)

已知 5% 的怀孕母羊的妊娠天数少于 108 天，

(a) 求 σ 的值

强从一大群怀孕母羊中随机选择 25 只。

(b) 求超过 3 只怀孕母羊的妊娠天数少于 108 天的概率。

查理随机抽取 200 个 25 只怀孕母羊的样本。

(c) 使用泊松近似来估计至少有 2 个样本中超过 3 只怀孕母羊的妊娠天数少于 108 天的概率。

（第 2 题共 8 分）

解答

(a)

解法一

思路

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把 D<108D<108 标准化。已知该概率为 0.050.05，对应标准正态分布的第 5 百分位数 −1.6449-1.6449；建立关于 σ\sigma 的方程并求解。

答题过程

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Since D∼N⁡(112.4,σ2)D\sim\operatorname{N}(112.4,\sigma^2),

P(D<108)=P(Z<108−112.4σ)=0.05.P(D<108) =P\left(Z<\frac{108-112.4}{\sigma}\right) =0.05.

The lower 5%5\% point of the standard normal distribution is −1.6449-1.6449. Therefore,

108−112.4σ= −1.6449,σ= 4.41.6449= 2.675…\begin{align*} \frac{108-112.4}{\sigma}=&\,-1.6449,\\ \sigma=&\,\frac{4.4}{1.6449}\\ =&\,2.675\ldots \end{align*}

Hence

σ=2.68 days(3 s.f.).\boxed{\sigma=2.68\text{ days}\quad\text{(3 s.f.)}}.

(b)

解法一

思路

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每只母羊妊娠期少于 108 天的概率为 0.050.05。25 只来自大羊群的随机样本可建模为独立二项试验；“超过 3 只”即至少 4 只，用补事件求概率。

答题过程

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Let JJ be the number of sheep whose pregnancy lasts less than 108 days. Then

J∼B⁡(25,0.05).J\sim\operatorname{B}(25,0.05).

Therefore,

P(J>3)= 1−P(J⩽3)= 0.034090…= 0.0341(3 s.f.).\begin{align*} P(J>3)=&\,1-P(J\leqslant3)\\ =&\,0.034090\ldots\\ =&\,\boxed{0.0341}\quad\text{(3 s.f.)}. \end{align*}

(c)

解法一

思路

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把“一个样本中超过 3 只妊娠期少于 108 天”视为一次成功，其概率由 (b) 得到。200 个样本中的成功数原本服从二项分布，按题意用均值 200p200p 的泊松分布近似；“至少 2 个”用补事件减去 0 个和 1 个。

答题过程

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Let TT be the number of samples containing more than 3 such sheep. Using the result from part (b),

λ=200(0.034090…)=6.8181…\lambda=200(0.034090\ldots)=6.8181\ldots

Using a Poisson approximation,

T∼Po⁡(6.8181…).T\sim\operatorname{Po}(6.8181\ldots).

Hence

P(T⩾2)= 1−P(T=0)−P(T=1)= 1−e−λ−λe−λ= 0.99144…= 0.991(3 s.f.).\begin{align*} P(T\geqslant2)=&\,1-P(T=0)-P(T=1)\\ =&\,1-\mathrm{e}^{-\lambda} -\lambda\mathrm{e}^{-\lambda}\\ =&\,0.99144\ldots\\ =&\,\boxed{0.991}\quad\text{(3 s.f.)}. \end{align*}