Question

The variance of the scores on a skill evaluation test is 123,201 with a mean of 1528 points. If 343 tests are sampled, what is the probability that the mean of the sample would differ from the population mean by greater than 29 points? Round your answer to four decimal places.

Answer 1

Solution:

Given that ,

= 1528

² = 123,201

=  123,201 = 351

A sample of size n = 343 is taken from this population.

Let be the mean of sample.

The sampling distribution of the is approximately normal with

Mean() = =  1528

SD() =    = 351/​343 = 18.9522185752

P( will differ from μ by more than 29 )

= 1 - P( will differ from μ by less than 29)

= 1 - P( - 29 < < + 29 )

= 1 - P( 1528 - 29 < < 1528 + 29)

= 1 - P( 1499 < < 1557)

= 1 - { P( < 1557) - P( < 1499 ) }

= 1 - { P[( - )/ < ( 1557 - 1528)/ 18.9522185752] - P[( - )/ < (1499 - 1528)/ 18.9522185752] }

= 1 - { P[Z <1.53] - P[Z < -1.53] }

= 1 - { 0.9370 - 0.0630} .. (use z table)

= 1 - 0.8740

= 0.1260

Answer : Required probability is 0.1260