Question

The mean life of a television set is 138 months with a variance of 324.

If a sample of 83 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by more than  5.4 months? Round your answer to four decimal places.

Answer 1

Solution :

Given that,

mean = = 138

standard deviation = = 18

n = 83

= = 138

= / n = 18 / 83 = 1.9758

P(132.6 < < 143.4) = P((132.6 - 138) / 1.9758<( - ) / < (143.4 - 138) / 1.9758))

= 1 - (P(-2.73 < Z < 2.73))

= 1 - (P(Z < 2.73) - P(Z < -2.73) ) Using z table,

= 1 - (0.9968 - 0.0032)

= 1 - 0.9936

= 0.0064

The probability that the sample mean would differ from the true mean by more than 5.4 months is 0.0064.