Help please!! Thank you.

Question

A normal population has mean 

=μ60

 and standard deviation =σ17

.(a) What proportion of the population is greater than 

96

?(b) What is the probability that a randomly chosen value will be less than 

81

.Round answers to four decimal places.

Answer 1

Prob( x>96) = 1 - Prob(x<96) = 1 - Prob( z < (96-60)/17) = 1 - Prob( z < 36/17) = 1 - Prob( z < 2.117) =

1 - 0.98287 = 0.01713 = 1.713%

Prob (x < 80) = Prob ( z < (80-60)/17) = Prob ( z < 20/17) = Prob ( z < 1.1765) =0.8803 = 88.03%