Question 208486
In a normal distribution with a mean and a std deviation, how do you find the probability of selecting 3 random units, whose total average exceeds a certain value? 
For example, normal distribution with mean of 127, and SD of 3.1: what is the probability that if three random units are selected, their total average (add up the values and divide by three), is greater than 130?
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That average you are referring to is the sample mean.
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Since the population mean is 127, the mean of the sample means is 127.
Since the SD of the population is 3.1, the SD of the sample means is 3.1/sqrt(3)
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These facts are known because of the "Central Limit Theorem".
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So find the z-value of 130:
z(130) = (130-127)/[3.1/sqrt(3)] = 1.676
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This tells you that 130 is 1.676 standard deviations to the right of the
mean of the sample means, which is 127.
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Then P(a sample mean could be greater than 130) = P(z> 1.676) = 0.0469
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Cheers,
Stan H.
PS. If you have any response to this email stanbon@comcast.net as the 
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