Question 424703
The diameter of grapefruit in a certain orchard are normally distributed with a mean of 5.73 and a standard deviation of 0.48 inches.show all work. 
What % of the grapefruit in this orchard is larger than 5.93?
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z(5.93) = (5.93-5.73)/0.48 = 0.4167
P(x > 5.93) = P(z > 0.4167) = 0.3385
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A random sample of 100 grapefruit is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.93?
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z(5.93) = (5.93-5.73)/[0.48/sqrt(100)] = 4.167
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P(x-bar > 5.93) = P(z > 4.1667) = 0.00001546..
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Cheers,
Stan H.
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