Question

Given a normal population whose mean is 540 and whose standard deviation is 66, find each of the following:

A. The probability that a random sample of 6 has a mean between 554 and 568.

Probability =

B. The probability that a random sample of 16 has a mean between 554 and 568.

Probability =

C. The probability that a random sample of 21 has a mean between 554 and 568.

Probability =

Answer 1

Solution :

Given that ,

mean =   = 540

standard deviation = = 66

n = 6

= 540

=  / n= 66 / 6=26.94

P(554<     <568 ) = P[(554-540) / 26.94< ( - ) /   < (568-540) / 26.94)]

= P( 0.52< Z <1.04 )

= P(Z <1.04 ) - P(Z <0.52 )

Using z table

=0.8508 - 0.6985

=0.1523

probability= 0.1523

b.

mean =   = 540

standard deviation = = 66

n = 16

= 540

=  / n= 66 / 16=16.5

P(554<     <568 ) = P[(554-540) / 16.5< ( - ) /   < (568-540) / 16.5)]

= P( 0.85< Z <1.70 )

= P(Z <1.70 ) - P(Z <0.85 )

Using z table

=0.9554 - 0.8023

=0.1531

probability= 0.1531

c.

n = 21

= 540

=  / n= 66 / 21=14.40

P(554<     <568 ) = P[(554-540) / 14.40< ( - ) /   < (568-540) / 14.40)]

= P( 0.97< Z <1.94 )

= P(Z <1.94 ) - P(Z <0.97)

Using z table

=0.9738 - 0.834

=0.1398

probability= 0.1398