Question

The heights of female students at a particular University are distributed with mean of 169 cm

and a standard deviation of 9 cm.

(a) Given that 80% of these female students have a height less than b cm, find the of b.

(b) Given than 60% of these female students have height greater than 5 cm, find the value of s.

Answer 1

$X$~$N(169, 9^2)$

a) $P(X<b)=0.8\implies P(169+9N(0,1)<b)=0.8\implies P(N(0,1)<{\frac {b-169} 9})=0.8\implies {\frac {b-169} 9}=0.84\implies b=176.56$

b) As i understand, there must be s cm not 5 cm, otherwise it doesn't make much sense

$P(X>s)=0.6\implies P(169+9N(0,1)>s)=0.6\implies P(N(0,1)>{\frac {s-169} 9})=0.6\implies P(N(0,1)<{\frac {s-169} 9})=0.4\implies {\frac {s-169} 9}=-0.253\implies s=166.72$

All of the probabilities you can find in the table of standard normal distribution, it is pretty easy to find in the internet