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##### What is the expected value of the sample mean weight? Standard deviation?

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A particular variety of watermelon weighs on average 23.1 pounds with a standard deviation of 1.47 pounds. Consider the sample mean weight of 95 watermelons of this variety. Assume the individual watermelon weights are independent.
Mar 8th, 2015

The expected value of the sample mean is just the population mean: 23.1 pounds.

The standard error around this sample mean will be the standard deviation / sqrt( sample size ):

S.E.M = 1.47/sqrt(95)

= 0.151 pounds

So the expected sample mean is 23.1 with a standard error of 0.151.

Mar 8th, 2015

Thank you for helping, could you please explain how to find the approximate probability the sample mean weight will be less than 23.21 using the standard deviation you got above?

Mar 8th, 2015

Sure, you can assume the sample mean will be normally distributed (as you have quite a high n, and your population standard deviation is known) with a mean of 23.1 and a sd = your standard error = 0.151. So the probability that you'll get <23.21 from this distribution is given by the Normal cdf. For example, in Excel:

NORM.DIST(x, mean, sd, 0=pdf / 1 = cdf)

= NORM.DIST(23.21, 23.1, 0.151, 1)

0.767

Mar 8th, 2015

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Mar 8th, 2015
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Mar 8th, 2015
Sep 21st, 2017
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