sampling distributions

Question

A particular fruit's weights are normally distributed, with a mean of 613 grams and a standard deviation of 33 grams.

If you pick 18 fruit at random, what is the probability that their mean weight will be between 591 grams and 594 grams?

Answer 1

xbar = mean weight, N(613,33). Pop mean = 613, pop std dev = 33, Sample size = n = 18

want to find P(591 < xbar < 594)

z-statistic = (xbar - pop mean)/(pop std dev/sqrt(n)) transforms xbar to N(0,1).

z=statistic for our problem is (xbar - 613)/(33/sqrt(18)) = (xbar - 613)/7.78

Compute z-statistic for xbar = 591 and xbar = 594:

z = (591 - 613)/7.78 = -2.83

z - (594 - 613)/7.78 = -2.44

so P(591 < xbar < 594) = P(-2.83 < z < -2.44) = P(z < - 2.44) - P(z < -2.83) = 0.0073 - 0.0023 = 0.005