Question

A normal population has a mean of 68 and a standard deviation of 6. You select a sample of 52. Compute the probability that the sample mean is (round z score 2 decimals and final answer 4): 1) Less than 67 2) Between 67 and 69.

Answer 1

Solution:

Given that,

mean = = 68

standard deviation = = 6

Sample size = n = 52

So,

   = 68

=  ( /n) = (6 / 52 ) = 0.8321

1 ) ( <  67 )

= p ( - /) <  (67 - 68 /0.8321 )

= p( z < - 1 / 0.8321 )

= p ( z < - 1.20 )   

Using z table

= 0.1151

Probability = 0.1151

2 ) (67 <   <  69 )

= p( 67 - 68 /0.8321 ) < ( - /) <  (69 - 68 /0.8321 )

= p(-1 /0.8321 < z < 1 / 0.8321 )

= p ( - 1.20 < z < 1.20

= p (z < 1.20) - p (z < - 1.20)   

Using z table

= 0.8849 - 0.1151

= 0.7698

Probability = 0.7698