Question

A population has a mean of 74 with a standard deviation of 9.8.

a) What is the probability that one element of the population selected at random is between 70 and 91??

b) What is the probability that a random sample of 36 from this population has a sample mean between 73 and 79?

Answer 1

Solution :

Given that ,

mean = = 74

standard deviation = = 9.8

P(70< x < 91) = P[(70-74) / 9.8< (x - ) / < (91-74) / 9.8)]

= P(-0.41 < Z <1.73 )

= P(Z < 1.73) - P(Z < -0.41)

Using z table   

= 0.9582-0.3409

probability= 0.6173

b.

Given that ,

mean = = 74

standard deviation = = 9.8

n = 36

= 74

=  / n= 9.8/ 36=1.63

P(73<     < 79) = P[(73-74) /1.63 < ( - ) /   < (79-74) /1.63 )]

= P( -0.61< Z <3.07 )

= P(Z < 3.07) - P(Z < -0.61)

Using z table

=0.9989 -0.2709

=0.7280