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CIE 9709 2024 November Paper 53 Q3

A Level / CIE / S1

CIE 9709 2024 Nov Paper 53 Paper · Question 3

题目

Problem

In Molimba, the heights, in cm\mathrm{cm}, of adult males are normally distributed with mean 176 cm176\,\mathrm{cm} and standard deviation 4.8 cm4.8\,\mathrm{cm}.

(a) Find the probability that a randomly chosen adult male in Molimba has a height greater than 170 cm170\,\mathrm{cm}.

(3)

60%60\% of adult males in Molimba have a height between 170 cm170\,\mathrm{cm} and k cmk\,\mathrm{cm}, where kk is greater than 170170.

(b) Find the value of kk, giving your answer correct to 1 decimal place.

(4)
题目中文翻译

在 Molimba,成年男性的身高(以 cm\mathrm{cm} 为单位)服从均值为 176 cm176\,\mathrm{cm}、标准差为 4.8 cm4.8\,\mathrm{cm} 的正态分布。

(a) 求从 Molimba 随机选出的一名成年男性身高大于 170 cm170\,\mathrm{cm} 的概率。

(3)

Molimba 有 60%60\% 的成年男性身高介于 170 cm170\,\mathrm{cm} 和 k cmk\,\mathrm{cm} 之间,其中 kk 大于 170170。

(b) 求 kk 的值,答案精确到 1 位小数。

(4)
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

170低于均值,标准化后求负标准分数右侧概率。

答题过程

展开

Let H∼N(176,4.82)H\sim\mathrm{N}(176,4.8^2). Then

P(H>170)= 1−Φ(170−1764.8)= 1−Φ(−1.25)=Φ(1.25)= 0.894350…≈ 0.894.\begin{align*} \mathrm{P}(H>170) =&\,1-\Phi\biggl(\frac{170-176}{4.8}\biggr)\\[4mm] =&\,1-\Phi(-1.25)=\Phi(1.25)\\[4mm] =&\,0.894350\ldots\\[4mm] \approx&\,\boxed{0.894}. \end{align*}

(b)

解法一

思路

展开

170以下的概率用(a)的补集,再加区间内0.6得k的左尾概率。

答题过程

展开

Using (a) without rounding,

P(H<170)=1−0.894350226…=0.105649773….\mathrm{P}(H<170)=1-0.894350226\ldots=0.105649773\ldots.

Since k>170k>170,

P(H<k)=0.105649773…+0.6=0.705649773….\mathrm{P}(H<k)=0.105649773\ldots+0.6=0.705649773\ldots.

Hence

k−1764.8=Φ−1(0.705649773…)=0.540720…,\frac{k-176}{4.8}=\Phi^{-1}(0.705649773\ldots)=0.540720\ldots,

and

k=176+4.8(0.540720…)=178.595456…≈178.6 cm.k=176+4.8(0.540720\ldots)=178.595456\ldots \approx\boxed{178.6\text{ cm}}.