# Probability Distributions

The attached deals with gamma and chi-square.

If 15 observations are taken independently from a chi-square distribution with four degrees of freedom, find the probability that at most 3 of the 15 observations exceed 7.779.

If 10 observations are taken independently from a chi-square distribution with 19 degrees of freedom, find the probability that exactly 2 of the 10 sample items exceed 30.14.

Cars arrive at a tool booth at a mean rate of five cars every 10 minutes according to a Poisson process. Find the probability that the toll collector will have to wait longer than 26.30 minutes before collecting the eighth toll.

In a medical experiment, a rat has been exposed to some radiation. The experimenters believe that the rat's survival time X ha s the p.d.f.

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Solution 5-14

Let X denote the number of cars that arrive in 26.3 minutes. Then clearly X follows a Poisson distribution with parameter , where is the expected number of occurrences over the particular interval of interest. Since the expected rate of occurrence is 5/10 = 0.5 cars per minute, the expected number of cars in 26.3 minutes is =26.3*0.5 = 13.15.

That is, X follows a Poisson distribution with parameter =13.15 and the p.m.f. of X is given by,

, x = 0, 1, 2, ...

Now, the probability that the toll collector will have ...

#### Solution Summary

The solution contains the determination of various probabilities using Gamma and Chi-Square distribution.