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put this solution on YOUR website!The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.44 inches and a standard deviation of 0.41 inches. Show all work.
(A) What percentage of the grapefruits in this orchard is larger than 5.36 inches?
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z(5.36) = (5.36-5.44)/0.41 = -0.1951
P(x > 5.36) = P(z > -0.1951) = 0.5774
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(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.36 inches?
t(5.36) = (5.36-5.44)/[0.41/sqrt(100)] = -1.9512
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P(x-bar > 5.36) = P(t > -1.9512 when df = 99) = 0.9731
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Cheers,
Stan H.
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