Poisson Random Variable probability problem

[probhelp150]

Homework Statement

X is a Poisson Random Variable with rate of 1 per hour, following the Poisson arrival process
a. Find the probability of no arrivals during a 10 hour interval
b. Find the probability of X > 10 arrivals in 2 hours
c. Find the average interarrival time.
d. For an interval of 2 hours, let A = {10 > X > 6}, B = {5 < X < 9}, then find Pr(6 < X < 11|A, B)
e. Find E(X|A ∩ B) in the previous problem
f. If Y = exp(2X), then find E(Y |A ∪ B).
g. Find E(Y 2 |A ∪ B)
h. Find Var(Y |A ∪ B)

Homework Equations

The Attempt at a Solution

a.
$\lambda = \frac{1}{1 hour}$
$\frac{1}{1 hour} * \frac{10}{10}=\frac{10}{10 hours}$
$\lambda = 10$
$P = \frac{e^{-10}*10^x}{x!}$

b.
$\lambda = 2$
$P = \frac{e^{-2}*2^x}{x!}$
X>10 = 1 - Sum(P(0...10))
$\sum_{n=0}^{10} \frac{e^{-2}*2^x}{x!}= 0.99999$
1-0.99999=0.00001

c.
Average interarrival time = $\frac{1}{\lambda}=\frac{1}{2}=30 minutes$

Am I doing this right so far?

 

[Charles Link]

"a" and "b" look good. For part "a", you need to let x=0, and compute. $\\$ "c" I think is just 1/rate=1 hour. $\\$ In parts "a" and "b", $\lambda=rate \times time$. (The way you computed $\lambda$ was a little clumsy, but otherwise correct).

 

[probhelp150]

"a" and "b" look good. For part "a", you need to let x=0, and compute. $\\$ "c" I think is just 1/rate=1 hour. $\\$ In parts "a" and "b", $\lambda=rate \times time$. (The way you computed $\lambda$ was a little clumsy, but otherwise correct).

If part c were referring to part b, would the interarrival time be 30 minutes because $\lambda=2$?
I'm wondering about #2 because that is a really small number... However, if the average number of arrivals is 1 per hour, the chance 5 arrivals per hour (10 in 2 hours) would be fairly low. Is my method of thinking correct?

Do you have any tips for parts d to h?

 

[Charles Link]

If part c were referring to part b, would the interarrival time be 30 minutes because $\lambda=2$?
I'm wondering about #2 because that is a really small number... However, if the average number of arrivals is 1 per hour, the chance 5 arrivals per hour (10 in 2 hours) would be fairly low. Is my method of thinking correct?

Do you have any tips for parts d to h?

I believe part c is referring to any arbitrary measurement period. How far apart are the arrivals?

 

[probhelp150]

I believe part c is referring to any arbitrary measurement period. How far apart are the arrivals?

1 hour apart. I get it now, thanks.

 

[Charles Link]

1 hour apart. I get it now, thanks.

I don't understand the notation for part d. I'm not sure exactly what they are asking. Perhaps someone else might be able to help out on that part...

 

[probhelp150]

I don't understand the notation for part d. I'm not sure exactly what they are asking. Perhaps someone else might be able to help out on that part...

Ok, thank you anyways!