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 $mu$?
normal-distribution sampling
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up vote
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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 $mu$?
normal-distribution sampling
You are right. The definition of the sample mean here is wrong.
â 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!
â statsguy21
Sep 7 at 8:48
add a comment |Â
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 $mu$?
normal-distribution sampling
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
normal-distribution sampling
asked Sep 7 at 8:23
statsguy21
353
353
You are right. The definition of the sample mean here is wrong.
â 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!
â statsguy21
Sep 7 at 8:48
add a comment |Â
You are right. The definition of the sample mean here is wrong.
â 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!
â statsguy21
Sep 7 at 8:48
You are right. The definition of the sample mean here is wrong.
â Kavi Rama Murthy
Sep 7 at 8:30
You are right. The definition of the sample mean here is wrong.
â 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!
â 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!
â statsguy21
Sep 7 at 8:48
add a comment |Â
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You are right. The definition of the sample mean here is wrong.
â 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!
â statsguy21
Sep 7 at 8:48