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CIE 9709 2023 June Paper 62 Q2

A Level / CIE / S2

CIE 9709 2023 June Paper 62 Paper · Question 2

题目

Problem

(a) The random variable WW has a Poisson distribution.

State the relationship between E(W)\mathrm{E}(W) and Var(W)\mathrm{Var}(W).

(1)

(b) The random variable XX has the distribution B(n,p)\mathrm{B}(n,p). Jyothi wishes to use a Poisson distribution as an approximate distribution for XX.

Use the formulae for E(X)\mathrm{E}(X) and Var(X)\mathrm{Var}(X) to explain why it is necessary for pp to be close to 00 for this to be a reasonable approximation.

(1)

(c) Given that YY has the distribution B(20 000,0.000 07)\mathrm{B}(20\,000,0.000\,07), use a Poisson distribution to calculate an estimate of P(Y>2)\mathrm{P}(Y>2).

(3)
题目中文翻译

(a) 随机变量 WW 服从泊松分布。

写出 E(W)\mathrm{E}(W) 与 Var(W)\mathrm{Var}(W) 的关系。

(1)

(b) 随机变量 XX 服从 B(n,p)\mathrm{B}(n,p) 分布。Jyothi 想用泊松分布来近似 XX 的分布。

利用 E(X)\mathrm{E}(X) 和 Var(X)\mathrm{Var}(X) 的公式,解释为什么 pp 必须接近 00,这种近似才合理。

(1)

(c) 已知 YY 服从 B(20 000,0.000 07)\mathrm{B}(20\,000,0.000\,07) 分布,用泊松分布估算 P(Y>2)\mathrm{P}(Y>2)。

(3)
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

(a)

解法一

思路

展开

泊松分布的均值与方差相等。

答题过程

展开 E(W)=Var(W).\boxed{\mathrm{E}(W)=\mathrm{Var}(W)}.

(b)

解法一

思路

展开

二项均值为np、方差为np乘1减p。要接近均值等于方差的泊松分布,需要1减p接近1。

答题过程

展开

For X∼B(n,p)X\sim\mathrm{B}(n,p),

E(X)=np,Var(X)=np(1−p).\mathrm{E}(X)=np,\qquad\mathrm{Var}(X)=np(1-p).

A Poisson distribution has equal mean and variance. Thus we need np(1−p)≈npnp(1-p)\approx np, requiring 1−p≈11-p\approx1, so pp must be close to 0.

(c)

解法一

思路

展开

近似泊松参数为20000乘0.00007等于1.4,超过2取不超过2的补集。

答题过程

展开

Using Y≈Po(20000(0.00007))=Po(1.4)Y\approx\mathrm{Po}(20000(0.00007))=\mathrm{Po}(1.4),

P(Y>2)≈ 1−e−1.4(1+1.4+1.422!)= 0.166502…= 0.167(3 s.f.).\begin{align*} \mathrm{P}(Y>2)\approx&\,1-e^{-1.4}\bigg(1+1.4+\frac{1.4^2}{2!}\bigg)\\[4mm] =&\,0.166502\ldots\\[4mm] =&\,\boxed{0.167}\quad\text{(3 s.f.).} \end{align*}