Chi-square distribution

Patriots326 · #1 $Old$ December 17th, 2008, 05:40 AM

If 15 observations are taken independently from a chi-square distribution with 15 degrees of freedom, find the probability that at least two of the sample items exceed 8.547.

Can anyone help with this?

meymathis · #2 $Old$ December 17th, 2008, 02:53 PM

Let $X$ be a random variable with distribution $\chi^2_{15}$ . A success is when $X>8.547$ . Now repeat 15 times (each trial is independent), denoting the $i^{th}$ trial as $X_i$ .

$\mathrm{P}(X_i>8.547)$ (i.e. the probabilities of success) are constant with respect to $i$ . In other words, you have 15 independent trials, success is constant ( $p=\mathrm{P}(X>8.547)$ ), and you are asked what is the probability of at least 2 successes. This smells like a binomial distribution if you let $N$ be the number of successes.

You use the $\chi^2_{15}$ distribution only to calculate $p$ .

$p=\mathrm{P}(X>8.547)=0.10$ (from $\chi^2$ table)

$\mathrm{P}(N>=2)=1-\mathrm{P}(N\leq 1)=1-\mathrm{P}(N= 0)-\mathrm{P}(N= 1)=\ldots$

Make sense?

Patriots326 · #3 $Old$ December 18th, 2008, 09:43 AM

Yea a little bit, I see if I can figure it out and if not, I can show you what I've done. Then we can figure it out.