Question 662745: The diameters of apples in a certain orchard are normally distributed with a mean of 4.77 inches and a standard deviation of 0.43 inches. Show all work.
(A) What percentage of apples in this orchard is larger than 4.71 inches?
(B) A random sample of 100 apples is gathers and the mean diameter is calculated. What is the probability that the samply that the sample mean is greater than 4.71 inches?
Please, I need help!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The diameters of apples in a certain orchard are normally distributed with a mean of 4.77 inches and a standard deviation of 0.43 inches. Show all work.
(A) What percentage of apples in this orchard is larger than 4.71 inches?
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z(4.71) = (4.71-4.77)/0.43 = -0.06/0.43 = -0.1395
P(x > 4.71) = P(z > -0.1395) = normalcdf(-0.1395,100) = 0.5555
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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 4.71 inches?
x-bar = 4.71
std for the distribution of sample means when n = 100 = 0.43/sqrt(100)
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z(4.71) = (4.71-4.77)/(0.43/sqrt(100)) = -1.3953
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P(x-bar > 4.71) = P(z > -1.3953) = normalcdf(-1.3953,100) = 0.9185
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Cheers,
Stan H.
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