SOLUTION: The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.93 inches and a standard deviation of 0.59 inches. Show all work.
(A) What percentag
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-> SOLUTION: The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.93 inches and a standard deviation of 0.59 inches. Show all work.
(A) What percentag
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Question 483103: The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.93 inches and a standard deviation of 0.59 inches. Show all work.
(A) What percentage of the grapefruits in this orchard is larger than 5.88 inches?
(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.88 inches?
Thanks much Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The diameters of grapefruits in a certain orchard are normally distributed with a mean of 5.93 inches and a standard deviation of 0.59 inches. Show all work.
(A) What percentage of the grapefruits in this orchard is larger than 5.88 inches?
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z(5.88) = (5.88-5.93)/0.59 = -0.0847
P(x > 5.88) = P(z > -0.0847) = 0.5338
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(B) A random sample of 100 grapefruits is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.88 inches?
z(5.88) = (5.88-5.93)/[0.59/sqrt(100)] = -0.8475
P(x-bar > 5.88) = P(z > -0.8475) = 0.8016
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Cheers,
Stan H.
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