Question 488457
Need Help!!!!The diameters of oranges in a certain orchard are normally distributed with a mean of 4.85 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 6.3 inches?
z(6.3) = (6.3-4.85)/0.4 = 3.625
P(x < 6.3) = P(z < 3.625) = 0.9999
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(B) What percentage of the oranges in this orchard are larger than 4.95 inches? 
z(4.95) = (4.95-4.85)/0.4 = 0.25
P(x > 4.95) = P(z > 0.25) = 0.4013
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Cheers,
Stan H.