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CIE 9709 2022 November Paper 61 Q4

A Level / CIE / S2

CIE 9709 2022 Nov Paper 61 Paper · Question 4

题目

Problem

Each month a company sells XX kg of brown sugar and YY kg of white sugar, where XX and YY have the independent distributions N(2500,1202)\mathrm{N}(2500,120^2) and N(3700,1302)\mathrm{N}(3700,130^2) respectively.

(a) Find the mean and standard deviation of the total amount of sugar that the company sells in 33 randomly chosen months.

(3)

The company makes a profit of $1.50 per kilogram of brown sugar sold and makes a loss of $0.20 per kilogram of white sugar sold.

(b) Find the probability that, in a randomly chosen month, the total profit is less than $3000.

(5)
题目中文翻译

某公司每月售出 XX 千克红糖和 YY 千克白糖，其中 XX 和 YY 相互独立，分别服从 N(2500,1202)\mathrm{N}(2500,120^2) 和 N(3700,1302)\mathrm{N}(3700,130^2) 分布。

(a) 随机选取 33 个月，求该公司在这 33 个月内售出的糖的总质量的均值和标准差。

(3)

该公司每售出一千克红糖获利 1.501.50 美元，每售出一千克白糖亏损 0.200.20 美元。

(b) 随机选取一个月，求总利润少于 30003000 美元的概率。

(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

(a)

解法一

思路

展开

三个月的独立销售量总和，均值与方差分别相加，最后对总方差开平方。

答题过程

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Let SS be the total amount sold in the three randomly chosen months. Then

E(S)=3(2500+3700)=18600 kg,\mathrm{E}(S)=3(2500+3700)=\boxed{18600\text{ kg}},

and

Var(S)=3(1202+1302)=93900 kg2.\mathrm{Var}(S)=3(120^2+130^2)=93900\text{ kg}^2.

Hence the standard deviation is

93900=306.431…=306 kg(3 s.f.).\sqrt{93900}=306.431\ldots =\boxed{306\text{ kg}}\quad\text{(3 s.f.).}

(b)

解法一

思路

展开

利润为1.5X减0.2Y，损失项的负号影响均值，但其独立方差仍为正系数平方。

答题过程

展开

Let P=1.50X−0.20YP=1.50X-0.20Y be the monthly profit, in dollars. By independence,

E(P)=1.50(2500)−0.20(3700)=3010,\mathrm{E}(P)=1.50(2500)-0.20(3700)=3010,

and

Var(P)= (1.50)2(120)2+(0.20)2(130)2= 33076.\begin{align*} \mathrm{Var}(P)=&\,(1.50)^2(120)^2+(0.20)^2(130)^2\\[4mm] =&\,33076. \end{align*}

Thus P∼N(3010,33076)P\sim\mathrm{N}(3010,33076) and

P(P<3000)= Φ(3000−301033076)≈ 0.478(3 s.f.).\begin{align*} \mathrm{P}(P<3000)=&\,\Phi\bigg(\frac{3000-3010}{\sqrt{33076}}\bigg)\\[4mm] \approx&\,\boxed{0.478}\quad\text{(3 s.f.).} \end{align*}