Question 263379
Mean: 1.31
Standard deviation: 0.04
random Sample: 16 
What is the probablity that the sample is between 1.28 and 1.32
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t(1.28) = (1.28-1.31)/[0.04/sqrt(16)] = -3
t(1.32) = (1.32-1.31)/[0.04/sqrt(16)] = 1
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P(1.28< x < 1.32) = P(-3< t < 1 when df = 15) = 0.8289..
Cheers,
Stan H.