1. Help with Poisson Distribution!

The mean of a Poisson random variable X is μ = 9. Compute
P(μ-2σ < X < μ + 2σ).

My reasoning is as follows:
P(9-2(9) < X < 9 + 2(9)) = P(-9 < X <27) . This is all I can think about doing at the moment. Feedback would be appreciated.

2. Originally Posted by MathRules!
The mean of a Poisson random variable X is μ = 9. Compute
P(μ-2σ < X < μ + 2σ).

My reasoning is as follows:
P(9-2(9) < X < 9 + 2(9)) = P(-9 < X <27) . This is all I can think about doing at the moment. Feedback would be appreciated.
$\displaystyle \sigma^2 = 9 \Rightarrow \sigma = \sqrt{9} = 3$.

Calculate $\displaystyle \Pr(3 < X < 15) = \Pr(4 \leq X \leq 14)$. There's no easy way of doing this by hand - you just have to calculate each probability and then add them all up.

You might be able to use the rule of thumb .... It will depend on what accuracy is required.

3. Most books have tables of the Possion.
This would be F(14)-F(3),
where you need to find the $\displaystyle \lambda=9$ table.

4. Originally Posted by mr fantastic
$\displaystyle \sigma^2 = 9 \Rightarrow \sigma = \sqrt{9} = 3$.

Calculate $\displaystyle \Pr(3 < X < 15) = \Pr(4 \leq X \leq 14)$. There's no easy way of doing this by hand - you just have to calculate each probability and then add them all up.

You might be able to use the rule of thumb .... It will depend on what accuracy is required.
Ohh, I see.

so I added the summation from 4 to 14 and have obtained that it is about 0.9373

5. That's correct to 4 places. F(14)=.9585 and F(3)=.0212.
The difference is .9373. Note that
$\displaystyle F(14)=P(X=0)+P(X=1)+\cdots +P(X=14)$
and
$\displaystyle F(3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$.
If you want to sum these 11 probabilities, one thing you should do is factor out the common term $\displaystyle e^{-\lambda}$

6. Originally Posted by matheagle
That's correct to 4 places. F(14)=.9585 and F(3)=.0212.
The difference is .9373. Note that
$\displaystyle F(14)=P(X=0)+P(X=1)+\cdots +P(X=14)$
and
$\displaystyle F(3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)$.
If you want to sum these 11 probabilities, one thing you should do is factor out the common term $\displaystyle e^{-\lambda}$
Why didn't I think of that? ..... That is very smart! I will do that from now on.

7. It's more experience than smart.
That's why I make my students do the homework.
I assign a lot, but I put those same problems on the exams and quizzes.
They do their homework, they get good grades and I know they are learning. It works.