1. ## anyone can help??

this is the question..:

yi ~ (µ,σ2) for I = 1,2,…,n. Then show that √n (ў - µ) ~ N (0,σ2)

2. Originally Posted by students
this is the question..:

yi ~ (µ,σ2) for I = 1,2,…,n. Then show that √n (ў - µ) ~ N (0,σ2)
What is ў meant to be ....? You need to be much clearer with your notation.

3. ## ..

i think it represent mean

4. Originally Posted by students
this is the question..:

yi ~ (µ,σ2) for I = 1,2,…,n. Then show that √n (ў - µ) ~ N (0,σ2)
First the sum of independent normals is normal and hence so is the mean, also the expected value of sample mean is the population mean. Then the variance of a sum of independent RVs is equal to the sum of the variances so the variance of the sum of $\displaystyle n$ iid normals of variance $\displaystyle \sigma^2$ is $\displaystyle n \sigma^2$.

RonL

5. Originally Posted by CaptainBlack
First the sum of independent normals is normal and hence so is the mean, also the expected value of sample mean is the population mean. Then the variance of a sum of independent RVs is equal to the sum of the variances so the variance of the sum of $\displaystyle n$ iid normals of variance $\displaystyle \sigma^2$ is $\displaystyle n \sigma^2$.

RonL
To the OP:
Originally Posted by CaptainBlack
.... the sum of independent normals is normal ....
is most easily proved using the moment generating function of the normal distribution.

In fact, the whole solution (normality, mean, variance) can be done in one hit using moment generating functions.