math statics and probality

Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 120 months, with a variance of 100.

If he is correct, what is the probability that the mean of a sample of 57 computers would differ from the population mean by less than 0.9 months? Round your answer to four decimal places.

Answer 1

z= (x-mean)/s = .9/10= .09,

s=sqrvar= sqr100=10

use a z calculator to find

P(-.09<z<09)= 0.0717

zs/sqrn= .09(10)/sqr57=.1192

CI =confidence interval= 120 +/- zs/sqrn = (119.8808, 120.1192)