Math Help Forum - View Single Post - Chi-square distribution

meymathis · #2 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?

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