题目

Problem

The masses of the bags of rice made by a company are normally distributed with mean $\mu\,\mathrm{kg}$ and standard deviation $0.14\,\mathrm{kg}$ . The probability that the mass of a randomly chosen bag of this rice is less than $1.48\,\mathrm{kg}$ is $0.22$ .

Find the value of $\mu$ .

(3)

题目中文翻译

某公司生产的袋装大米的质量服从正态分布，均值为 $\mu\,\mathrm{kg}$ ，标准差为 $0.14\,\mathrm{kg}$ 。随机选取一袋这种大米，其质量小于 $1.48\,\mathrm{kg}$ 的概率为 $0.22$ 。

求 $\mu$ 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

解法一

思路

展开

左尾概率0.22对应负标准分数，说明1.48低于均值。标准化后解均值，注意符号。

答题过程

展开

Let $M\sim\mathrm{N}(\mu,0.14^2)$ denote mass in kilograms. Since $\mathrm{P}(M<1.48)=0.22$ ,

\frac{1.48-\mu}{0.14}=\Phi^{-1}(0.22) =-0.772193\ldots.

Therefore,

\begin{align*} \mu=&\,1.48+0.14(0.772193\ldots)\\[4mm] =&\,1.588107\ldots\\[4mm] \approx&\,\boxed{1.59}. \end{align*}

CIE 9709 2025 June Paper 51 Q1

题目

解答

解法一

思路

答题过程