Sample mean of normal approximation to sampling distribution of $barX$

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As I was reading through my notes, I came across this formula. It states that for independent and iid random variables $X_1,X_2...X_n...$ with the same mean $mu$ and variance $sigma$,



$$barX_n = sum_i = 1^nX_i$$



However, this is not in line with what I read for other formulae for the mean, which normally has a $frac1n$ before the entire summation.



It also doesn't seem to make any logical sense to me. If I add up all the random variables which have mean $mu$, wouldn't that just make it $nmu$? So shouldn't I have a $frac1n$ at the front to make it $mu$?










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  • You are right. The definition of the sample mean here is wrong.
    – Kavi Rama Murthy
    Sep 7 at 8:30










  • Ok thanks a lot. I went to check against the library textbooks as well, and the formula in the lecture slides was indeed wrong. Thanks!
    – statsguy21
    Sep 7 at 8:48














up vote
0
down vote

favorite












As I was reading through my notes, I came across this formula. It states that for independent and iid random variables $X_1,X_2...X_n...$ with the same mean $mu$ and variance $sigma$,



$$barX_n = sum_i = 1^nX_i$$



However, this is not in line with what I read for other formulae for the mean, which normally has a $frac1n$ before the entire summation.



It also doesn't seem to make any logical sense to me. If I add up all the random variables which have mean $mu$, wouldn't that just make it $nmu$? So shouldn't I have a $frac1n$ at the front to make it $mu$?










share|cite|improve this question





















  • You are right. The definition of the sample mean here is wrong.
    – Kavi Rama Murthy
    Sep 7 at 8:30










  • Ok thanks a lot. I went to check against the library textbooks as well, and the formula in the lecture slides was indeed wrong. Thanks!
    – statsguy21
    Sep 7 at 8:48












up vote
0
down vote

favorite









up vote
0
down vote

favorite











As I was reading through my notes, I came across this formula. It states that for independent and iid random variables $X_1,X_2...X_n...$ with the same mean $mu$ and variance $sigma$,



$$barX_n = sum_i = 1^nX_i$$



However, this is not in line with what I read for other formulae for the mean, which normally has a $frac1n$ before the entire summation.



It also doesn't seem to make any logical sense to me. If I add up all the random variables which have mean $mu$, wouldn't that just make it $nmu$? So shouldn't I have a $frac1n$ at the front to make it $mu$?










share|cite|improve this question













As I was reading through my notes, I came across this formula. It states that for independent and iid random variables $X_1,X_2...X_n...$ with the same mean $mu$ and variance $sigma$,



$$barX_n = sum_i = 1^nX_i$$



However, this is not in line with what I read for other formulae for the mean, which normally has a $frac1n$ before the entire summation.



It also doesn't seem to make any logical sense to me. If I add up all the random variables which have mean $mu$, wouldn't that just make it $nmu$? So shouldn't I have a $frac1n$ at the front to make it $mu$?







normal-distribution sampling






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asked Sep 7 at 8:23









statsguy21

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  • You are right. The definition of the sample mean here is wrong.
    – Kavi Rama Murthy
    Sep 7 at 8:30










  • Ok thanks a lot. I went to check against the library textbooks as well, and the formula in the lecture slides was indeed wrong. Thanks!
    – statsguy21
    Sep 7 at 8:48
















  • You are right. The definition of the sample mean here is wrong.
    – Kavi Rama Murthy
    Sep 7 at 8:30










  • Ok thanks a lot. I went to check against the library textbooks as well, and the formula in the lecture slides was indeed wrong. Thanks!
    – statsguy21
    Sep 7 at 8:48















You are right. The definition of the sample mean here is wrong.
– Kavi Rama Murthy
Sep 7 at 8:30




You are right. The definition of the sample mean here is wrong.
– Kavi Rama Murthy
Sep 7 at 8:30












Ok thanks a lot. I went to check against the library textbooks as well, and the formula in the lecture slides was indeed wrong. Thanks!
– statsguy21
Sep 7 at 8:48




Ok thanks a lot. I went to check against the library textbooks as well, and the formula in the lecture slides was indeed wrong. Thanks!
– statsguy21
Sep 7 at 8:48















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