Question 270623
Can you help me with this question: 
Determine the approximate probability that a sample average (x-bar) taken from some population with the mean = 50 and standard deviation = 4 exceeds 50.6, where the sample size is 100?

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The Central Limit Theory tells you something about the 
distribution of the means of samples of a given size, n.
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It says the mean of the sample means is the same as the
mean of the population.
It says the standard deviation of the sample means is 
(standard deviation of the population)/(sqrt(sample size).
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Your Problem:
The z-score of 50.6 = (50.6-50)/[4/sqrt(100)] = 0.6*10/4 = 1.5
P(x-bar > 50.6) = P(z > 1.5) = normalcdf(1.5,100) = 0.0668
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Cheers,
Stan H.