Question

Suppose cattle in a large herd have a mean weight of 1158lbs and a standard deviation of 92lbs. What is the probability that the mean weight of the sample of cows would differ from the population mean by more than 12lbs if 55 cows are sampled at random from the herd? Round your answer to four decimal places.

Answer 1

Solution:

Given that ,

= 1158

= 92

A sample of size n = 55 is taken from this population.

Let be the mean of sample.

The sampling distribution of the is approximately normal with

Mean = = 1158

SD =      = 92/​55 =   12.4052774693

P(Sample mean would differ from population mean by more than 12lbs)

= P( will differ from μ by more than 12)

= 1 - P( will differ from μ by less than 12)

= 1 - P( - 12< < + 12)

= 1 - P( 1158- 12 < < 1158 + 12)

= 1 - P( 1146 < < 1170)

= 1 - { P( < 1170) - P( < 1146) }

= 1 - { P[( - )/ < ( 1170 - 1158)/ 12.4052774693] - P[( - )/ < (1146 - 1158)/ 12.4052774693] }

= 1 - { P[Z <0.97] - P[Z < -0.97] }

= 1 - { 0.8340 - 0.1660} .. (use z table)

= 1 - 0.6680

= 0.3320

P(Sample mean would differ from population mean by more than 12lbs) = 0.3320