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?

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- Dec 17th 2008, 03:40 AMPatriots326Chi-square distribution
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? - Dec 17th 2008, 12:53 PMmeymathis
Let $\displaystyle X$ be a random variable with distribution $\displaystyle \chi^2_{15}$. A success is when $\displaystyle X>8.547$. Now repeat 15 times (each trial is independent), denoting the $\displaystyle i^{th}$ trial as $\displaystyle X_i$.

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

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

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

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

Make sense? - Dec 18th 2008, 07:43 AMPatriots326
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.