Question

Historical evidence indicates that times between fatal accidents on scheduled American domestic passenger flights have an approximately exponential distribution. Assume that the mean time between accidents is 52 days. (a) If one of the accidents occurred at 12:00 am on October 1 of a randomly selected year in the study period, what is the probability that another accident occurred that same month

Answer 1

Answer:

The probability that another accident will occur that same month = 0.44907

Step-by-step explanation:

Let consider X to be the random variable that follows an exponential distribution with a mean of 52.

The mean of the exponential distribution $\dfrac{1}{\lambda } = 52 \implies \lambda = \dfrac{1}{52}$

Thus, the probability distribution of X is:

$f(x) = \lambda e ^{-\lambda x}$

$f(x) =\dfrac{1}{52} e ^{-x/52} \ \ \ \ x \geq 0$

Suppose X represents the time interval between the accident that happened in October 1  and the following accident.

Then, the probability that another accident will occur in the same month is the same as the event that the time between the two accidents are less than 31 (given that the month of October have 31 days)

Thus, we can calculate the required probability as follows:

$P(X \leq 31) = \int ^{31}_{0} \ f(x) \ dx$

$P(X \leq 31) = \int ^{31}_{0} \ \dfrac{1}{52} \ e^{-x/52} \ dx$

$P(X \leq 31) = 1 - e^{-31/52}$

$\mathbf{P(X \leq 31) = 0.44907}$

Thus, the probability that another accident will occur that same month = 0.44907