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?
Let be a random variable with distribution . A success is when . Now repeat 15 times (each trial is independent), denoting the trial as .
(i.e. the probabilities of success) are constant with respect to . In other words, you have 15 independent trials, success is constant ( ), and you are asked what is the probability of at least 2 successes. This smells like a binomial distribution if you let be the number of successes.
You use the distribution only to calculate .
(from table)
Make sense?