Question 693969
The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.83 inches and a standard deviation of 0.46 inches. Show all work. 
(A) What percentage of the grapefruits in this orchard is larger than 5.92 inches?
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z(5.92) = (5.92-5.83)/0.46 = 0.1957
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P(x > 5.92) = P(z > 0.1957) = 0.4224
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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 5.92 inches?
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the std will be 0.46/sqrt(100) = 0.046
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z(x-bar = 5.92) = (5.92-5.83)/0.046 = 1.9565
P(x-bar > 5.92) = normalcdf(1.9565,100) = 0.0252
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Cheers,
Stan H.
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