Question 270805
A population has a mean of 50 and a standard deviation of 7. You select a random sample of 36. Compute the following: 
The value for which there is a .025 probability that the sample mean will be greater than. 
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Find the t-score that has a 0.025 left-tail.
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invT(0.025,35) = -2.0301
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Find the corresponding "x-value":
x = ts + u
x = -2.0301*(7/sqrt(36)) + 50
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x = 47.63..
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Cheers,
Stan H.