Question 483103
The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.93 inches and a standard deviation of 0.59 inches. Show all work. 
(A) What percentage of the grapefruits in this orchard is larger than 5.88 inches? 
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z(5.88) = (5.88-5.93)/0.59 = -0.0847
P(x > 5.88) = P(z > -0.0847) = 0.5338
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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 5.88 inches?
z(5.88) = (5.88-5.93)/[0.59/sqrt(100)] = -0.8475
P(x-bar > 5.88) = P(z > -0.8475) = 0.8016
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Cheers,
Stan H.