Question

A normal population has mean =μ9 and standard deviation =σ5.

(a) What proportion of the population is less than 20?

(b) What is the probability that a randomly chosen value will be greater than 5?

Round the answers to four decimal places.

Answer 1

A) Given data

Mean value(X')=9

Standard deviation (S)=5

a)We need to determine the proportion of population that is less than 20 is

i.e P(X<20)

First determine z score value

Z score=(X-X')/S

=(20-9)/5=2.2

P(X<20)=P(Z<2.2)

=0.9861( From normal area tables)

The proportion of the population is less than 20 =0.9861

b) We need to determine the probability that a randomly choosen Value will be greater than 5 is

i.e P(X>5)

Z score Value=(X-X')/S

=(5-9)/5

=-0.8

P(X>5)=P(Z>-0.8)

=0.7881(From normal area tables)

So,the probability that a randomly selected Value greater than5 is=0.7881