题目

Problem

The weights of large bags of pasta produced by a company are normally distributed with mean $1.5\,\mathrm{kg}$ and standard deviation $0.05\,\mathrm{kg}$ .

(a) Find the probability that a randomly chosen large bag of pasta weighs between $1.42\,\mathrm{kg}$ and $1.52\,\mathrm{kg}$ .

(3)

The weights of small bags of pasta produced by the company are normally distributed with mean $0.75\,\mathrm{kg}$ and standard deviation $\sigma\,\mathrm{kg}$ . It is found that 68% of these small bags have weight less than $0.9\,\mathrm{kg}$ .

(b) Find the value of $\sigma$ .

(3)

题目中文翻译

某公司生产的大袋意大利面的重量服从正态分布，均值为 $1.5\,\mathrm{kg}$ ，标准差为 $0.05\,\mathrm{kg}$ 。

(a) 随机选择一大袋意大利面，求其重量在 $1.42\,\mathrm{kg}$ 与 $1.52\,\mathrm{kg}$ 之间的概率。

(3)

该公司生产的小袋意大利面的重量服从正态分布，均值为 $0.75\,\mathrm{kg}$ ，标准差为 $\sigma\,\mathrm{kg}$ 。发现这些小袋意大利面中有 68% 的重量小于 $0.9\,\mathrm{kg}$ 。

(b) 求 $\sigma$ 的值。

(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

标准化两个重量边界后取累计概率之差。

答题过程

展开

Let $W\sim\mathrm{N}(1.5,0.05^2)$ . Then

\begin{align*} \mathrm{P}(1.42<W<1.52) =&\,\Phi\biggl(\frac{1.52-1.5}{0.05}\biggr) -\Phi\biggl(\frac{1.42-1.5}{0.05}\biggr)\\[4mm] =&\,\Phi(0.4)-\Phi(-1.6)\\[4mm] =&\,0.600622449\ldots\approx\boxed{0.601}. \end{align*}

(b)

解法一

思路

展开

左尾0.68对应正分位点，求标准差。

答题过程

展开

\frac{0.9-0.75}{\sigma}=\Phi^{-1}(0.68)=0.467698799\ldots.

Hence

\sigma=\frac{0.15}{0.467698799\ldots} =0.320719232\ldots\approx\boxed{0.321}.

CIE 9709 2023 November Paper 53 Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程