Question 696933
The diameters of grapefruits in a certain orchard are normally distributed with a mean of 6.10 inches and a standard deviation of 0.40 inches. Show all work. 
(A) What percentage of the grapefruits in this orchard have diameters less than 6.8 inches? 
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z(6.8) = (6.8-6.1)/0.4 = 0.7/0.4 = 1.75
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P(x < 6.8) = P(z < 1.75) = 0.9599
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(B) What percentage of the grapefruits in this orchard are larger than 6.7 inches?
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z(6.7) = (6.7-6.1)/0.4 = 0.6/0.4 = 3/2
P(x > 6.7) = P(z > 1.5) = normalcdf(1.5,100) = 0.0401
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Cheers,
Stan H.
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