A manufacturer knows that their items have a lengths that are approximately normally distributed, with a mean of 9.8 inches, and standard deviation of 1.6 inches. If 48 items are chosen at random, what is the probability that their mean length is greater than 9.2 inches

0.6462

Step-by-step explanation:

Given that

Mean (m) = 9.8

Standard deviation (s) = 1.6

Probability that mean length is greater than 9.2

Z = (x - m) / s

Z = (9.2 - 9.8) / 1.6

Z = - 0.6 / 1.6

Zscore = - 0.375 = - 0.375

P(Z >- 0.375) = 1 - P(Z < - 0.375) ;

P(Z < - 0.375) = 0.3538

1 - P(Z < - 0.38) = 1 - 0.3538 = 0.6462

ok so what's the question?

step-by-step explanation:

is the anwser b. (-13)?