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CIE 9709 2023 November Paper 52 Q5

A Level / CIE / S1

CIE 9709 2023 Nov Paper 52 Paper · Question 5

题目

Problem

(a) The heights of the members of a club are normally distributed with mean 166 cm166\,\mathrm{cm} and standard deviation 10 cm10\,\mathrm{cm}.

(i) Find the probability that a randomly chosen member of the club has height less than 170 cm170\,\mathrm{cm}.

(2)

(ii) Given that 40% of the members have heights greater than h cmh\,\mathrm{cm}, find the value of hh correct to 2 decimal places.

(3)

(b) The random variable XX is normally distributed with mean μ\mu and standard deviation σ\sigma.

Given that σ=23μ\sigma=\frac{2}{3}\mu, find the probability that a randomly chosen value of XX is positive.

(3)

题目中文翻译

(a) 某俱乐部成员的身高服从正态分布，均值为 166 cm166\,\mathrm{cm}，标准差为 10 cm10\,\mathrm{cm}。

(i) 随机选择该俱乐部的一名成员，求其身高小于 170 cm170\,\mathrm{cm} 的概率。

(2)

(ii) 已知 40% 的成员身高大于 h cmh\,\mathrm{cm}，求 hh 的值，精确到小数点后 2 位。

(3)

(b) 随机变量 XX 服从均值为 μ\mu、标准差为 σ\sigma 的正态分布。

已知 σ=23μ\sigma=\frac{2}{3}\mu，求随机选取的 XX 的值为正数的概率。

(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)(i)

解法一

思路

展开

标准化170，求左尾概率。

答题过程

展开

For H∼N(166,102)H\sim\mathrm{N}(166,10^2),

P(H<170)=Φ(170−16610)=Φ(0.4)=0.655421741…≈0.655.\mathrm{P}(H<170)=\Phi\biggl(\frac{170-166}{10}\biggr) =\Phi(0.4)=0.655421741\ldots\approx\boxed{0.655}.

(a)(ii)

解法一

思路

展开

右尾0.40换为左尾0.60，按题目要求留两位小数。

答题过程

展开 h−16610=Φ−1(0.60)=0.253347….\frac{h-166}{10}=\Phi^{-1}(0.60)=0.253347\ldots.

Therefore

h=166+10(0.253347…)=168.533471…≈168.53 cm.h=166+10(0.253347\ldots)=168.533471\ldots \approx\boxed{168.53\text{ cm}}.

(b)

解法一

思路

展开

非退化正态分布标准差为正，已知比例意味着均值也为正；标准化0后均值约去。

答题过程

展开

Since σ>0\sigma>0 and σ=23μ\sigma=\frac23\mu, we have μ>0\mu>0. Thus

P(X>0)= P(Z>0−μ(2/3)μ)= P(Z>−1.5)=Φ(1.5)= 0.933192798…≈ 0.933.\begin{align*} \mathrm{P}(X>0) =&\,\mathrm{P}\biggl(Z>\frac{0-\mu}{(2/3)\mu}\biggr)\\[4mm] =&\,\mathrm{P}(Z>-1.5)=\Phi(1.5)\\[4mm] =&\,0.933192798\ldots\\[4mm] \approx&\,\boxed{0.933}. \end{align*}