Question 494801
A normal population has a mean of 75 and a standard deviation of 5. A sample of size 40 has been randomly selected. The probability that the sample mean is less than 74 is equal to:
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t(74) = (74-75)/[5/sqrt(40)] = -1.2649
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P(x-bar < 74) = P(t < -1.2649 when df = 39) = 0.1067
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If you use the z-distribution you would get:
P(x-bar < 74) = P(z <-1.2649) = 0.1030
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Cheers,
Stan H.
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1. 0.4207
2. 0.0020
3. 0.1038
4. 0.0548
5. 0.5793