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S2 2024 Jan 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 DD days where

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

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

(a) find the value of σ\sigma

(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 DD 天，其中

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

已知 5% 的怀孕母羊的妊娠天数少于 108 天，

(a) 求 σ\sigma 的值

(3)

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。

(2)

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。

(3)

（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 百分位数约为 −1.6448536…-1.6448536\ldots；建立关于 σ\sigma 的方程并求解。

答题过程

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Let Z∼N⁡(0,1)Z\sim\operatorname{N}(0,1). Since D∼N⁡(112.4,σ2)D\sim\operatorname{N}(112.4,\sigma^2),

 P(D<108) =P(Z<108−112.4σ) =0.05.\begin{align*} &\,P(D<108)\\[2mm] &\,=P\Big(Z<\frac{108-112.4}{\sigma}\Big)\\ &\,=0.05. \end{align*}

The lower 5%5\% point of the standard normal distribution is −1.6448536…-1.6448536\ldots. Therefore,

 108−112.4σ =−1.6448536…, σ=4.41.6448536… =2.675010…\begin{align*} &\,\frac{108-112.4}{\sigma}\\[2mm] &\,=-1.6448536\ldots,\\[2mm] &\,\sigma=\frac{4.4}{1.6448536\ldots}\\[2mm] &\,=2.675010\ldots \end{align*}

Hence

σ=2.68 days.\boxed{\sigma=2.68\text{ days}}.

(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.\begin{align*} &\,P(J>3)\\[2mm] &\,=1-P(J\leqslant3)\\ &\,=0.034090\ldots\\ &\,\approx\boxed{0.0341}. \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…\begin{align*} \lambda=&\,200(0.034090\ldots)\\ &\,=6.8181\ldots \end{align*}

The exact model is T∼B⁡(200,p)T\sim\operatorname{B}(200,p), where pp is the probability from part (b). Using a Poisson approximation,

 T≈U, U∼Po⁡(λ).\begin{align*} &\,T\approx U,\\ &\,U\sim\operatorname{Po}(\lambda). \end{align*}

Hence

 P(T⩾2) ≈1−P(U=0) −P(U=1) =1−e−λ−λe−λ =0.99144… ≈0.991.\begin{align*} &\,P(T\geqslant2)\\[2mm] &\,\approx1-P(U=0)\\ &\,\hspace{2pt}-P(U=1)\\ &\,=1-\mathrm{e}^{-\lambda} -\lambda\mathrm{e}^{-\lambda}\\ &\,=0.99144\ldots\\ &\,\approx\boxed{0.991}. \end{align*}