Question 614104
The diameters in oranges in a certain orchard are normally distributed with a mean of 5.30 inches and a standard deviation of 0.40.
What percentage of the oranges in this orchard have diameters of less than 4.7 inches?
z(4.7) = (4.7-5.3)/0.4 = -0.6/0.4 = -3/2
P(x < 4.7) = P(z < -3/2) = normalcdf(-100,-3/2) = 0.0668 
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What percentage of the oranges in this orchard are larger than 5.10? 
z(5.1) = (5.1-5.3)/0.4 = -0.2/0.4 = -1/2
P(x> 5.1) = P(Z > -1/2) = normalcdf(-1/2,100) = 0.6917
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Cheers,
Stan H.
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