Question

Suppose a batch of steel rods produced at a steel plant have a mean length of 151 millimeters, and a variance of 64.

If 116 rods are sampled at random from the batch, what is the probability that the length of the sample rods would differ from the population mean by more than 0.81 millimeters? Round your answer to four decimal places.

Answer 1

Solution:

Given that ,

= 151

² = 64

So , = 8

n = 116

Let be the mean of sample.

= 151

=  /n = 8/116 = 0.7427813527

What is the approximate probability that will differ from μ by more than 0.7?

P( will differ from μ by more than 0.81 )

= 1 - P( will differ from μ by less than 0.81 )

= 1 - P( - 0.81 < < + 0.81 )

= 1 - P( 151 - 0.7 < < 151 + 0.7)

= 1 - P( 150.3 < < 151.7)

= 1 - { P( < 151.7) - P( < 150.3 ) }

= 1 - { P[( - )/ < ( 151.7 - 151)/ 0.7427813527] - P[( - )/ < (150.3 - 151)/ 0.7427813527] }

= 1 - { P[Z < 0.94] - P[Z < -0.94] }

= 1 - { 0.8264 - 0.1736} .. (use z table)

= 0.3472

Required probability = 0.3472