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IAL 2022 June S3 Q6

A Level / Edexcel / S3

IAL 2022 June Paper · Question 6

题目

Problem

A particular lift has a maximum load capacity of 700 kg. The weights of men are normally distributed with mean 80 kg and standard deviation 10 kg. The weights of women are normally distributed with mean 69 kg and standard deviation 5 kg. You may assume that weights of people are independent.

(a) Find the probability that when 6 men and 3 women are in the lift, the load exceeds 700 kg.

(4)

A sign in the lift states: “Maximum number of people in the lift is c”

(b) Find the value of c such that the probability of the load exceeding 700 kg is less than 2.5% no matter the gender of the occupants.

(6)
(Total for Question 6 is 10 marks)
题目中文翻译

一部电梯的最大载重为 700 千克。 男性体重服从均值 80 千克、标准差 10 千克的正态分布。 女性体重服从均值 69 千克、标准差 5 千克的正态分布。 你可以假设人的体重相互独立。

(a) 求当电梯中有 6 名男性和 3 名女性时,载重超过 700 千克的概率。

电梯内的标牌写着:“电梯中的最大人数为 c”

(b) 求 c 的值,使得无论乘客性别如何,载重超过 700 千克的概率都小于 2.5%。

解答

(a)

解法一

思路

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独立正态随机变量之和仍服从正态分布。总重量的均值等于各人均值之和,方差等于各人方差之和;注意相加的是方差,而不是标准差。

答题过程

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Let XX be the total weight of the 6 men and 3 women. Since the individual weights are independent,

E(X)=6(80)+3(69)=687,\begin{align*} E(X)=&\,6(80)+3(69)\\ =&\,687, \end{align*}

and

Var(X)=6(102)+3(52)=675.\begin{align*} \operatorname{Var}(X) =&\,6(10^2)+3(5^2)\\ =&\,675. \end{align*}

Therefore,

XN(687,675).X\sim N(687,675).

Hence

P(X>700)=P(Z>700687675)=P(Z>0.5003)=0.30840.308.\begin{align*} P(X>700) =&\,P\bigg( Z>\frac{700-687}{\sqrt{675}} \bigg)\\ =&\,P(Z>0.5003\ldots)\\ =&\,0.3084\ldots\\ \approx&\,\boxed{0.308}. \end{align*}

(b)

解法一

思路

展开

要保证无论乘客性别如何都满足条件,应采用全是男性的最不利情形,因为男性体重的均值和标准差都更大。设人数为 nn,建立总重量的正态分布;右尾概率小于 0.0250.025 等价于对应标准分数大于 1.961.96。解出 nn 的上界后,取符合条件的最大整数作为标牌上的人数。

答题过程

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The greatest risk of exceeding the load occurs when all nn occupants are men. Let YY be their total weight. Then

YN(80n,100n).Y\sim N(80n,100n).

The requirement is

P(Y>700)<0.025.P(Y>700)<0.025.

Since P(Z>1.96)=0.025P(Z>1.96)=0.025, this requires

70080n10n>1.96.\frac{700-80n}{10\sqrt n}>1.96.

Let t=nt=\sqrt n, where t>0t>0. Then

70080t2>19.6t80t2+19.6t700<0.\begin{align*} 700-80t^2>&\,19.6t\\ 80t^2+19.6t-700<&\,0. \end{align*}

The positive root of the corresponding quadratic equation is

t=19.6+19.62+4(80)(700)160=2.8380\begin{align*} t =&\,\frac{-19.6+ \sqrt{19.6^2+4(80)(700)}}{160}\\ =&\,2.8380\ldots \end{align*}

Therefore,

n<2.8380n<8.0546\begin{align*} \sqrt n<&\,2.8380\ldots\\ n<&\,8.0546\ldots \end{align*}

The greatest whole number satisfying this inequality is 88. Hence

c=8.\boxed{c=8}.