Question

(10 points) It is known that cars arrive at a drive-through at a rate of three cars per minute between 12 noon and 1:00 pm. Assuming the number of cars that arrive in any time interval follows a Poisson distribution, what is the probability that exactly 9 cars arrive between 12:10 and 12:20. (Note: the question gives the rate is

Answer 1

Answer:

0.000005 probability that exactly 9 cars arrive between 12:10 and 12:20.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

Mean of 3 cars per minute:

So, between 12:10 and 12:20, there is an interval of 10 minutes, which means that $\mu = 3*10 = 30$

What is the probability that exactly 9 cars arrive between 12:10 and 12:20?

This is $P(X = 9)$. So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 9) = \frac{e^{-30}*30^{9}}{(9)!} = 0.000005$

0.000005 probability that exactly 9 cars arrive between 12:10 and 12:20.