Question 626694
The diameters of apples in a certain orchard are normally distributed with a mean of 4.45 inches and a standard deviation of 0.42 inches. Show all work. 
(A) What percentage of the apples in this orchard is larger than 4.38 inches?
z(4.38) = (4.38-4.45)/0.42 = -0.17
P(x > 4.38) = P(z > -0.17) = normalcdf(-0.17,100) = 0.5662 = 56.62%
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(B) A random sample of 100 apples is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 4.38 inches?
z(4.38) = (4.38-4.45)/[0.42/sqrt(100)) = -1.67
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P(x-bar > 4.38) = P(z > -1.67) = normalcdf(-1.67,100) = 0.9525 = 95.25%
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Cheers,
Stan H.
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