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CIE 9709 2022 June Paper 62 Q7

A Level / CIE / S2

CIE 9709 2022 June Paper 62 Paper · Question 7

题目

Problem

XX is a random variable with distribution Po(2.90)\mathrm{Po}(2.90). A random sample of 100100 values of XX is taken.

Find the probability that the sample mean is less than 2.882.88.

(5)
题目中文翻译

随机变量 XX 服从 Po(2.90)\mathrm{Po}(2.90) 分布。从 XX 中随机抽取 100100 个取值作为样本。

求样本均值小于 2.882.88 的概率。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

解法一

思路

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100个独立泊松值的总和服从Po290。样本均值小于2.88等价于整数总和不超过287，再用正态近似及连续性修正。

答题过程

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Let SS be the sum of the 100 sampled values. Then

S∼ Po(100(2.90))=Po(290)≈ N(290,290).\begin{align*} S\sim&\,\mathrm{Po}(100(2.90))=\mathrm{Po}(290)\\[4mm] \approx&\,\mathrm{N}(290,290). \end{align*}

The event for the sample mean is

X‾<2.88⟺ S<288⟺ S≤287.\begin{align*} \overline X<2.88\quad\Longleftrightarrow&\,S<288\\[4mm] \Longleftrightarrow&\,S\leq287. \end{align*}

Using a continuity correction,

P(X‾<2.88)≈ Φ(287.5−290290)= 0.441643…= 0.442(3 s.f.).\begin{align*} \mathrm{P}(\overline X<2.88) \approx&\,\Phi\bigg(\frac{287.5-290}{\sqrt{290}}\bigg)\\[4mm] =&\,0.441643\ldots\\[4mm] =&\,\boxed{0.442}\quad\text{(3 s.f.).} \end{align*}