Question

A normal population has a mean of 19 and a standard deviation of 5.

a. Compute the z value associated with 22.

b. What proportion of the population is between 19 and 22?

c. What proportion of the population is less than 14?

Answer 1

Solution :

Given that ,

mean = = 19

standard deviation = = 5

a.using z score formula

z= (x - ) / =(22-19) /5=0.6

b.

P(19< x <22 ) = P[(19-19) /5 < (x - ) / < (22-19) /5 )]

= P( 0< Z <0.6 )

= P(Z <0.6 ) - P(Z <0 )

Using z table   

= 0.7257-0.5

probability= 0.2257

c.

P(X< 14) = P[(X- ) / < (14-19) /5 ]

= P(z <-1 )

Using z table

= 0.1587