Heights of males are approximately normally distributed with

Heights of males are approximately normally distributed with mu = 175cm and sigma= 10cm. (a) What is the minimum height that a man can be if he is in the top 1.5% of heights in this population. (b) What proportion of males over 180cm tall are under 185cm tall? (c) A random sample of 300 males are taken as part of a medical study. What is the probability that at least 100 of them are over 180cm in height. (d) Now suppose that heights of females are also normally distributed with mean equal to 165cm and standard deviation equal to 10cm. Assuming that a random person is equally likely to be male or female, what proportion of person who are over 180cm are female?

Solution

(a)

P(X>c)=0.015

--> P((X-mean)/s <(c-175)/10)=1-0.015=0.985

--> P(Z<(c-175)/10)=0.985

--> (c-175)/10= 2.17 (from standard normal table)

So c= 175+2.17*10=196.7

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(b)P(180<X<185) = P((180-175)/10<Z<(185-175)/10)

=P(0.5<Z<1)

=0.1499 (from standard normal table)

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(c) P(X>180)= P(Z>(180-175)/10) = P(Z>0.5) =0.3085 (from standard normal table)

mean=n*p=300*0.3085 =92.55

standard deviation =sqrt(n*p*(1-p))=sqrt(300*0.3085*(1-0.3085))=7.999895

So the probability is

P(X>=100) = P(Z>(100-92.55)/7.999895)

=P(Z>0.93) =0.1762 (from standard normal table)

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(d)P(X>180) = P(Z>(180-165)/10)

=P(Z>1.5)

=0.0668 (from standard normal table)

i.e. 6.68%