Question

Arrivals of cars at a gas station follow a Poisson distribution. During a given 5-minute period, one car arrived at the station. Find the probability that it arrived during the last 30 seconds of the 5-minute period g.

Answer 1

Answer:

0.9 = 90% probability that it arrived during the last 30 seconds of the 5-minute period.

Step-by-step explanation:

The car is equally as likely to arrive during each second of the interval, which means that the uniform distribution is used to solve this question.

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distribution has two bounds, a and b, and the probability of finding a value higher than x is given by:

$P(X \geq x) = \frac{b - x}{b - a}$

5-minute period

This means that $a = 0, b = 5*60 = 300$

Find the probability that it arrived during the last 30 seconds of the 5-minute period.

300 - 30 = 270. So

$P(X \geq 270) = \frac{300 - 270}{300 - 0} = 0.9$

0.9 = 90% probability that it arrived during the last 30 seconds of the 5-minute period.