SOLUTION: According to the data published by the U.S. Dept. of Agriculture, the average annual amount of chicken eaten by American adults (excluding vegetarians) is normally distributed with
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Question 1031090: According to the data published by the U.S. Dept. of Agriculture, the average annual amount of chicken eaten by American adults (excluding vegetarians) is normally distributed with a mean of 55 pounds and a standard deviation of 9.2 pounds.
Suppose a trade organization randomly samples 50 American adults (excluding vegetarians) and records the annual chicken consumption of each. What is the probability that the sample mean is less than 52 pounds per year?
A.51
B.49
C.37
D.10
E.01 Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! need to calculate the z-value for the sample mean
:
z-value = (Sample mean - population mean) / standard deviation of the population
:
z-value = (52 - 55) / 9.2 = −0.326086957 approx -0.33
:
consult table of z-values for the probability
:
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Probability ( X < 52 ) = 0.3707
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