findings mean, SD and approximate distribution

madgab · #1 $Old$ June 30th, 2009, 08:03 AM

Let X1,X2,....,X81 be a random sample from a distribution (not necessarily normal) with mean $\mu=59$ and SD $\sigma=8$ . Let $\bar{X}=\frac{1}{81}\Sigma^{81}_{i=1}X_{i}$

a) What is the mean of $\bar{X}$
b) What is the standard deviation of $\bar{X}$
c) What is the approximate distribution of $\bar{X}$

I'm not really sure what this question is asking. I know how to find means and standard deviations, but not from what is given in this question. Thank you.

Random Variable · #2 $Old$ June 30th, 2009, 09:15 AM

a) $\mu = 59$

b) $\frac {\sigma}{\sqrt{n}} = \frac{8}{9}$

c) $N\Big(\mu,\frac {\sigma^{2}}{n}\Big) = N\Big(59,\frac{64}{81}\Big)$

madgab · #3 $Old$ June 30th, 2009, 10:31 AM

The mean is the same but the SD is divided by $\sqrt{n}$ ? I never would have known that. What is that SD equation from? Also, how can you figure it is a normal distribution? It is probably obvious but I can't find info like that at all in my book and would like to learn.

Random Variable · #4 $Old$ June 30th, 2009, 11:28 AM

If $X_{1}, X_{2}, ... , X_{n}$ are indepedent random variables with respective means $\mu_{1}, \mu_{2}. ... , \mu_{n}$ and variances $\sigma^{2}_{1}, \sigma^{2}_{2}, ... , \sigma^{2}_{n}$

then the mean and variance of $Y = \sum^{n}_{i=1} a_{i} X_{i}$ are $\mu_{Y} = \sum^{n}_{i=1} a_{i} \mu_{i}$ and $\sigma^{2}_{Y} = \sum^{n}_{i=1} a_{i}^{2} \sigma^{2}_{i}$ respectively

You can prove the preceding directly from the definition.

Therefore, $\mu_{\bar{X}} = \sum^{n}_{i=1} \frac {1}{n} \mu = \frac {1}{n} (n \mu) = \mu$

and $\sigma^{2}_{\bar{x}} = \sum^{n}_{i=1} (\frac {1}{n})^{2} \sigma^{2} = \frac{1}{n^{2}} (n \sigma^{2}) = \frac {\sigma^{2}}{n}$

part(c) is the result of the central limit theorem