Question

The number of orders that come into a mail-order sales office each month is normally distributed with a population mean of 298 and a population standard deviation of 15.4. For a particular sample size, the probability is 0.2 that the sample mean exceeds 300. How big must the sample be?

Answer 1

P(z<Z) table :

SD sample = SD/(n^0.5)

= 15.4/(n^0.5)

P(x>300) = 0.2

P(x<300) = 0.8

P(z<Z) = 0.8

Z from table : Z = 0.84

0.84 = (x-mean)/SD

= (300-298)/(SD sample)

SD sample = (300-298)/0.84 = 2.38

15.4/(n^0.5) = 2.38

n = (15.4/2.38)^2 = 41.87

n = 42 {rounded to integer}

the sampl size must be 42

(please UPVOTE)