Question

The Poisson distribution gives the probability for the number of occurrences for a "rare" event. Now, let x be a random variable that represents the waiting time between rare events. Using some mathematics, it can be shown that x has an exponential distribution. Let be a random variable and let o be a constant. Thenis a curve representing the exponential distribution. Areas under this curve give us exponential probabilities. If a and b are any numbers such that .n using some extra mathematics, it can be shown that the area under the curve above the interval t, e) is the following. Notice that by definition, x cannot be negative, so, P- The random variable x is called an exponential random variable. Using some more mathematics, it can be shown that the mean and standard deviation of x are the following Note: The number -2.7182. is used throughout probability, statistics, and mathematics. The key is conveniently located on most calculators. Comment: The Poisson and exponential distributions have a special relationship. Specifically, it can be shown that the waiting time between successive Poisson arrivals (i.e., successes or rare events) has an exponential distribution with s-u, where λ is the average number of Poisson successes (rare events) per unit of time. Fatal accidents on scheduled domestic passenger flights are rare events. In fact, airlines do all they possibly can to prevent such accidents. However, around the world such fatal accidents do occur. Let x be a random variable representing the waiting time between fatal airline accidents. Research has shown that x has an exponential distribution with a mean of approximately 44 days.t We take the point of view that x (measured in days as units) is a continuous random variable. Suppose a fatal airline accident has just been reported on the news. What is the probability that the waiting time to the next reported fatal airline accident is the following? (a) less than 40 daysosx4 (Round your answer to four decimal places.) (b) ore than 70 days e, 0 four decimal places.) Hint: (Round your answer to

Hello, need help solving the rest. I might be doing it wrong and cannot figure it out. Thank you.

Answer 1

$X\sim exponential~distribution~with~mean=44~days\\ The~pdf~of~X~is\\ f_X(x)=\frac{1}{44}\exp(-x/44);~x>0\\ =0~otherwise\\ (a)~P(X<40)=\int_0^{40}\frac{1}{44}\exp(-x/44)dx=1-\exp(-40/44)=0.5971\\ (b)~P(X>70)=\int_{70}^{\infty}\frac{1}{44}\exp(-x/44)dx=\exp(-70/44)=0.2037\\ (c)~P(30<X<60)=\int_{30}^{60}\frac{1}{44}\exp(-x/44)dx\\ =\exp(-30/44)-\exp(-60/44)=0.2500\\ (d)~E(X)=\int_{0}^\infty x\times \frac{1}{44}\exp(-x/44)dx=44~days\\ ~E(X^2)=\int_{0}^\infty x^2\times \frac{1}{44}\exp(-x/44)dx=44^2\Gamma(3)=2*44^2\\ Var(X)=E(X^2)-E^2(X)=2*44^2-44^2=44^2\\ Standard~deviation=\sqrt{Var(X)}=44~days.$