SOLUTION: 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 t

Algebra ->  Probability-and-statistics -> SOLUTION: 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 t      Log On


   



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?
(B) What percentage of the oranges in this orchard are larger than 5.10 inches?

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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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