SOLUTION: A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting 30 values and getting a mean greater than 3.6.
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Question 941685: A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting 30 values and getting a mean greater than 3.6. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting 30 values and getting a mean greater than 3.6.
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Note: Standard deviation for the mean of the sample means = s/sqrt(n)
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Your Problem::
z(3.6) = (3.6-3.2)/[0.7/sqrt(30)] = 0.4*sqrt(30)/0.7 = 3.1300
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P(x-bar > 3.6) = normalcdf(3.13,100) = 0.0008750
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Cheers,
Stan H.
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