A manufacturer knows that their items have a lengths that are skewed right, with a mean of 17.3 inches, and standard deviation of 4.3 inches.

Question

If 50 items are chosen at random, what is the probability that their mean length is greater than 18.3 inches?

(Round answer to four decimal places)

Answer 1

xbar = mean length

sample size = 50

population mean = 17.3

standard deviation = 4.3

want to find P(xbar > 18.3)

z-statistic: (xbar - pop mean)/(SD/square root(sample size))

compute z-statistic for xbar = 18.3

z-statistic = (18.3 - 17.3)/(4.3/sqrt(50)) = 1.64

P(xbar > 18.3) = P(z > 1.64) = 1 - P(z < 1.64) = 1 - 0.9495 = 0.0505