Tom K. answered 06/04/20
Knowledgeable and Friendly Math and Statistics Tutor
z = (17.7-15.1)/(2.8/sqrt(7)) = 2.45676907455998
P(Z <= 2.45676907455998) = , from Excel, normsdist(2.45676907455998) = 0.99299036144302, which can round to .9930.
Alexis C.
asked 06/02/20A manufacturer knows that their items have a normally distributed length, with a mean of 15.1 inches and standard deviation of 2.8 inches
if seven items are chosen at random, what is the probability that they are mean length is less than 17.7 inches
Tom K. answered 06/04/20
Knowledgeable and Friendly Math and Statistics Tutor
z = (17.7-15.1)/(2.8/sqrt(7)) = 2.45676907455998
P(Z <= 2.45676907455998) = , from Excel, normsdist(2.45676907455998) = 0.99299036144302, which can round to .9930.
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