题目

Problem

A firm sells cheese in packets. The mass of cheese in the packets is normally distributed with mean $1.12\,\mathrm{kg}$ and standard deviation $0.15\,\mathrm{kg}$ . The firm makes a profit of $0.32 per kilogram of cheese sold.

Find the probability that the total profit made on $50$ randomly chosen packets of cheese is less than $18.00.

(6)

题目中文翻译

一家公司按袋销售奶酪。每袋奶酪的质量服从正态分布，均值为 $1.12\,\mathrm{kg}$ ，标准差为 $0.15\,\mathrm{kg}$ 。公司每售出一千克奶酪可获利 $0.32$ 美元。

求随机选取的 $50$ 袋奶酪带来的总利润低于 $18.00$ 美元的概率。

(6)

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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先求五十包总质量的分布，再乘每千克利润率。总和使方差乘以五十；金额缩放使方差乘以利润率的平方。

答题过程

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Let $S$ be the total mass, in kilograms, of the $50$ independently chosen packets. Then

\begin{align*} S\sim&\,\mathrm N\big(50(1.12),\,50(0.15^2)\big)\\[4mm] =&\,\mathrm N(56,1.125). \end{align*}

The total profit is $P=0.32S$ , so

\begin{align*} \mathrm E(P)=&\,0.32(56)=17.92,\\[4mm] \operatorname{Var}(P)=&\,0.32^2(1.125)=0.1152. \end{align*}

Thus $P\sim\mathrm N(17.92,0.1152)$ , and

\begin{align*} \mathrm P(P<18) =&\,\Phi\bigg(\frac{18-17.92}{\sqrt{0.1152}}\bigg)\\[4mm] =&\,\Phi(0.235702\ldots)\\[4mm] \approx&\,\boxed{0.593}. \end{align*}

CIE 9709 2026 March Paper 62 Q4

题目

解答

解法一

思路

答题过程