P(xbar < 20.6) = P(z < (xbar - mu)/(STD/sqrt(sample size))
P(z < (20.6 - 20)/(1.8/sqrt(25)) = P(z < 1.66) = 0.9515
Nate G.
asked • 04/21/22Statistics CLT Homework
A manufacturer knows that their items have a normally distributed length, with a mean of 20 inches, and standard deviation of 1.8 inches.
If 25 items are chosen at random, what is the probability that their mean length is less than 20.6 inches?
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