Question 420498
The diameters of oranges in a certain orchard are normally distributed with a mean of 5.30 inches and a standard deviation of 0.40 inches. Show all work. 
(A) What percentage of the oranges in this orchard have diameters less than 4.7 inches?
Find the z of 4.7:
z(4.7) = (4.7-5.3)/0.4 = -1.5
P(x < 4.7) = P(z < -1.5) = 6.68%

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(B) What percentage of the oranges in this orchard are larger than 5.10 inches?
z(5.10) = (5.10-5.3)/0.4 = -0.2/0.4 = -1/2

P(x > 5.10) = P(z > -1/2) = 69.15%

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Cheers,
Stan H.
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