Question

Cars arriving for gasoline at a Shell station follow a Poisson distribution with a mean of 1010 per hour. (Enter all answers rounded to at least 3 decimal places.) (a) Determine the probability that over the next hour, only one car will arrive. (b) Compute the probability that in the next 3 hours, more than 29 cars will arrive.

Answer 1

a)
`P(X=1) = (e^(-10) 10^1)/(1!)= 0.000454`

b) For this case the new parameter
`\lambda` would be:

`\lambda = 10 (cars)/(hour) * 3 hours =30`

And we want to calculate the following probability:

`P(X >29)`

And we can use the complement rule for this case:

`P(X >29) =1-P(X\leq 29)`

And for this case we can use the following excel code in order to find the required probability:

"=1-POISSON.DIST(29,30,TRUE)"

And we got:
`P(X >29) =1-P(X\leq 29)=0.5243`

Explanation:

Part a

Let X the random variable that represent the number of cars arriving for gasoline at Shell station. We know that
`X \sim Poisson(\lambda= 10)`

The probability mass function for the random variable is given by:

`f(x)=(e^(-\lambda) \lambda^x)/(x!) , x=0,1,2,3,4,...`

And for this case we want this probability:

`P(X=1)`

Using the probability mass function we got:

`P(X=1) = (e^(-10) 10^1)/(1!)= 0.000454`

Part b

For this case the new parameter
`\lambda` would be:

`\lambda = 10 (cars)/(hour) * 3 hours =30`

And we want to calculate the following probability:

`P(X >29)`

And we can use the complement rule for this case:

`P(X >29) =1-P(X\leq 29)`

And for this case we can use the following excel code in order to find the required probability:

"=1-POISSON.DIST(29,30,TRUE)"

And we got:
`P(X >29) =1-P(X\leq 29)=0.5243`