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IAL 2020 Oct S3 Q7

A Level / Edexcel / S3

IAL 2020 Oct Paper · Question 7

题目

Problem

A company makes cricket balls and tennis balls.

The weights of cricket balls, CC grams, follow a normal distribution

C∼N(160,1.252)C \sim N(160, 1.25^2)

Three cricket balls are selected at random.

(a) Find the probability that their total weight is more than 475.8 grams.

(4)

The weights of tennis balls, TT grams, follow a normal distribution

T∼N(60,22)T \sim N(60, 2^2)

Five tennis balls and two cricket balls are selected at random.

(b) Find the probability that the total weight of the five tennis balls and the two cricket balls is more than 625 grams.

(4)

A random sample of nn tennis balls T1,T2,T3,…,TnT_1, T_2, T_3, \dots, T_n is taken.

Y=(n−1)T1−∑r=2nTrY = (n - 1)T_1 - \sum_{r=2}^{n} T_r

Given that P(Y>40)=0.0838P(Y > 40) = 0.0838, correct to 4 decimal places,

(c) find nn.

(8)
(Total 16 marks)
题目中文翻译

一家公司生产板球和网球。

板球的重量 CC（克）服从正态分布

C∼N(160,1.252)C \sim N(160, 1.25^2)

随机抽取 3 个板球。

(a) 求它们总重量超过 475.8 克的概率。

网球的重量 TT（克）服从正态分布

T∼N(60,22)T \sim N(60, 2^2)

随机抽取 5 个网球和 2 个板球。

(b) 求 5 个网球和 2 个板球的总重量超过 625 克的概率。

随机抽取一个由 nn 个网球 T1,T2,T3,…,TnT_1, T_2, T_3, \dots, T_n 组成的样本。

Y=(n−1)T1−∑r=2nTrY = (n - 1)T_1 - \sum_{r=2}^{n} T_r

已知 P(Y>40)=0.0838P(Y > 40) = 0.0838，保留到小数点后 4 位。

(c) 求 nn。

解答

(a)

解法一

思路

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三个板球重量相互独立且都服从同一正态分布，因此总重量仍服从正态分布。总和的均值是三个均值之和，方差是三个方差之和；建立总重量分布后求超过 475.8475.8 克的右尾概率。

答题过程

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Let

S=C1+C2+C3.S=C_1+C_2+C_3.

Since the cricket ball weights are independent,

E(S)= 3(160)=480,Var⁡(S)= 3(1.252)=4.6875.\begin{align*} E(S)=&\,3(160)=480,\\ \operatorname{Var}(S)=&\,3(1.25^2)=4.6875. \end{align*}

Therefore,

S∼N(480,4.6875).S\sim N(480,4.6875).

Hence

P(S>475.8)= P(Z>475.8−4804.6875)= P(Z>−1.9399…)= 0.9738…\begin{align*} P(S>475.8)=&\,P\bigg(Z>\frac{475.8-480}{\sqrt{4.6875}}\bigg)\\ =&\,P(Z>-1.9399\ldots)\\ =&\,0.9738\ldots \end{align*}

Thus, to 3 significant figures,

P(S>475.8)=0.974.\boxed{P(S>475.8)=0.974}.

(b)

解法一

思路

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把五个网球与两个板球的重量全部相加。由于七个球的重量相互独立，总重量仍服从正态分布；均值直接相加，方差也直接相加，然后求总重量超过 625625 克的右尾概率。

答题过程

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Let

W=T1+T2+T3+T4+T5+C1+C2.W=T_1+T_2+T_3+T_4+T_5+C_1+C_2.

Then

E(W)= 5(60)+2(160)= 620,\begin{align*} E(W)=&\,5(60)+2(160)\\ =&\,620, \end{align*}

and

Var⁡(W)= 5(22)+2(1.252)= 23.125.\begin{align*} \operatorname{Var}(W)=&\,5(2^2)+2(1.25^2)\\ =&\,23.125. \end{align*}

Therefore,

W∼N(620,23.125).W\sim N(620,23.125).

The required probability is

P(W>625)= P(Z>625−62023.125)= P(Z>1.0398…)= 0.1492…\begin{align*} P(W>625)=&\,P\bigg(Z>\frac{625-620}{\sqrt{23.125}}\bigg)\\ =&\,P(Z>1.0398\ldots)\\ =&\,0.1492\ldots \end{align*}

Hence, to 3 significant figures,

P(W>625)=0.149.\boxed{P(W>625)=0.149}.

(c)

解法一

思路

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YY 是独立正态变量的线性组合，所以仍服从正态分布。T1T_1 的系数是 n−1n-1，其余 n−1n-1 个变量的系数都是 −1-1；计算时均值要保留负号，而方差中的系数要平方。由给定右尾概率查得标准正态值约为 1.381.38，再解关于 nn 的方程，并舍去不可能的负根。

答题过程

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The mean of YY is

E(Y)= (n−1)E(T1)−∑r=2nE(Tr)= 60(n−1)−60(n−1)= 0.\begin{align*} E(Y)=&\,(n-1)E(T_1)-\sum_{r=2}^{n}E(T_r)\\ =&\,60(n-1)-60(n-1)\\ =&\,0. \end{align*}

Since all the tennis ball weights are independent,

Var⁡(Y)= (n−1)2(22)+(n−1)(22)= 4(n−1)2+4(n−1)= 4n2−4n.\begin{align*} \operatorname{Var}(Y)=&\,(n-1)^2(2^2)+(n-1)(2^2)\\ =&\,4(n-1)^2+4(n-1)\\ =&\,4n^2-4n. \end{align*}

Therefore,

Y∼N(0,4n2−4n).Y\sim N(0,4n^2-4n).

Given that

P(Y>40)=0.0838,P(Y>40)=0.0838,

the corresponding upper-tail standard normal value is

P(Z>1.38)=0.0838P(Z>1.38)=0.0838

to 4 decimal places. Hence

404n2−4n=1.38.\frac{40}{\sqrt{4n^2-4n}}=1.38.

Squaring and rearranging gives

4n2−4n= (401.38)2= 840.16…,4n2−4n−840.16…= 0.\begin{align*} 4n^2-4n=&\,\bigg(\frac{40}{1.38}\bigg)^2\\ =&\,840.16\ldots,\\ 4n^2-4n-840.16\ldots=&\,0. \end{align*}

Solving this quadratic gives

n≈15.0orn≈−14.0.n\approx15.0\quad\text{or}\quad n\approx-14.0.

Since nn is a positive integer sample size, the negative root is rejected. Thus

n=15.\boxed{n=15}.

Indeed, when n=15n=15,

P(Z>40840)=0.0838P\bigg(Z>\frac{40}{\sqrt{840}}\bigg)=0.0838

to 4 decimal places, as required.