SOLUTION: The diameter of grapefruit in a certain orchard are normally distributed with a mean of 5.73 and a standard deviation of 0.48 inches.show all work.
What % of the grapefruit in t
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-> SOLUTION: The diameter of grapefruit in a certain orchard are normally distributed with a mean of 5.73 and a standard deviation of 0.48 inches.show all work.
What % of the grapefruit in t
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Question 424703: The diameter of grapefruit in a certain orchard are normally distributed with a mean of 5.73 and a standard deviation of 0.48 inches.show all work.
What % of the grapefruit in this orchard is larger than 5.93?
A random sample of 100 grapefruit is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.93?
Thank You
Becky
You can put this solution on YOUR website! The diameter of grapefruit in a certain orchard are normally distributed with a mean of 5.73 and a standard deviation of 0.48 inches.show all work.
What % of the grapefruit in this orchard is larger than 5.93?
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z(5.93) = (5.93-5.73)/0.48 = 0.4167
P(x > 5.93) = P(z > 0.4167) = 0.3385
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A random sample of 100 grapefruit is gathered and the mean diameter is calculated. What is the probability that the sample mean is greater than 5.93?
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z(5.93) = (5.93-5.73)/[0.48/sqrt(100)] = 4.167
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P(x-bar > 5.93) = P(z > 4.1667) = 0.00001546..
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Cheers,
Stan H.
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