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?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 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
feedback function of algebra.com is not working.
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