Question 752740: A variable X has a mean of 185.5 and a standard deviation of 21.4. A random sample of size 67 is selected. Assuming X is normally distributed, what is the probability, to two decimal places, that the sample mean exceeds 200?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A variable X has a mean of 185.5 and a standard deviation of 21.4. A random sample of size 67 is selected. Assuming X is normally distributed, what is the probability, to two decimal places, that the sample mean exceeds 200?
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z(200) = (200-185.5)/[21.4/sqrt(67)] = 5.5462
P(x-bar > 200) = P(z > 5.5462) = normalcdf(5.5462,100) = 0.0000000146
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Cheers,
Stan H.
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