题目
Problem
In Molimba, the heights, in cm, of adult males are normally distributed with mean 176cm and standard deviation 4.8cm.
(a) Find the probability that a randomly chosen adult male in Molimba has a height greater than 170cm.
(3)
60% of adult males in Molimba have a height between 170cm and kcm, where k is greater than 170.
(b) Find the value of k, 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 为单位)服从均值为 176cm、标准差为 4.8cm 的正态分布。
(a) 求从 Molimba 随机选出的一名成年男性身高大于 170cm 的概率。
(3)
Molimba 有 60% 的成年男性身高介于 170cm 和 kcm 之间,其中 k 大于 170。
(b) 求 k 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
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170低于均值,标准化后求负标准分数右侧概率。
答题过程
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Let H∼N(176,4.82). Then
P(H>170)===≈1−Φ(4.8170−176)1−Φ(−1.25)=Φ(1.25)0.894350…0.894.
(b)
解法一
思路
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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
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Using (a) without rounding,
P(H<170)=1−0.894350226…=0.105649773….
Since k>170,
P(H<k)=0.105649773…+0.6=0.705649773….
Hence
4.8k−176=Φ−1(0.705649773…)=0.540720…,
and
k=176+4.8(0.540720…)=178.595456…≈178.6 cm.