Answer :
Answer:
a) [tex] P(X=1) = \frac{e^{-10} 10^1}{1!}= 0.000454[/tex]
b) For this case the new parameter [tex]\lambda[/tex] would be:
[tex] \lambda = 10 \frac{cars}{hour} * 3 hours =30[/tex]
And we want to calculate the following probability:
[tex] P(X >29)[/tex]
And we can use the complement rule for this case:
[tex] P(X >29) =1-P(X\leq 29)[/tex]
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: [tex] P(X >29) =1-P(X\leq 29)=0.5243[/tex]
Step-by-step explanation:
Part a
Let X the random variable that represent the number of cars arriving for gasoline at Shell station. We know that [tex]X \sim Poisson(\lambda= 10)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And for this case we want this probability:
[tex] P(X=1)[/tex]
Using the probability mass function we got:
[tex] P(X=1) = \frac{e^{-10} 10^1}{1!}= 0.000454[/tex]
Part b
For this case the new parameter [tex]\lambda[/tex] would be:
[tex] \lambda = 10 \frac{cars}{hour} * 3 hours =30[/tex]
And we want to calculate the following probability:
[tex] P(X >29)[/tex]
And we can use the complement rule for this case:
[tex] P(X >29) =1-P(X\leq 29)[/tex]
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: [tex] P(X >29) =1-P(X\leq 29)=0.5243[/tex]