Thread: Help with finding probability of normal function...

1. Help with finding probability of normal function...

Let $\displaystyle Y_{1},Y_{2},...,Y_{n}$ be independent, normal random variables, each with mean $\displaystyle \mu$ and variance $\displaystyle \sigma^{2}$.

a) Find the density function of $\displaystyle U=Y^{bar}=\frac{1}{n}\sum_{i=1}^{n}Y_{i}$.

Solution to (a): So, U is normally distributed with mean $\displaystyle \mu$ and variance $\displaystyle \frac{\sigma^{2}}{n}$.

b) If $\displaystyle \sigma^{2}=16$ and $\displaystyle n=25$, what is the probability that the sample mean, $\displaystyle Y^{bar}$, takes on a value that is within one unit of the population mean, $\displaystyle \mu$? That is, find $\displaystyle P(|Y^{bar}-\mu|\leq\\1)$.

So, the way I have been trying to solve it is by finding the probability $\displaystyle P(\mu-1\leq\\Y^{bar}\leq\mu+1)$, but I have had no luck. I get the feeling it is right in front of me. Should I use Tchebysheff's theorem?

Thanks for the help

oh, and $\displaystyle Y^{bar}$ is supposed to represent y with a bar (_) over it...not sure how to do that.

2. No, the point is that chebyshev's is a very weak inequality in this setting.

$\displaystyle P(|\bar Y-\mu|< 1) = P(-1<\bar Y-\mu< 1)$

$\displaystyle = P(-1/(4/5)<{\bar Y-\mu\over \sigma/\sqrt{n}}< 1/(4/5))$

$\displaystyle = P(-1.25<Z< 1.25)$