You are Here: Home >< Maths

# Stats help needed watch

X-Po(2) Y-Po(3)
W=XY

Find P(W=4)
2. (Original post by Pinkdragon)

X-Po(2) Y-Po(3)
W=XY

Find P(W=4)
If we assume X and Y are two independent random variables we can see that the expected value of XY, i.e. E(XY) would be E(X)E(Y). In this case E(XY) or E(W) in other words would equal 6.

Therefore we can write W as being Poisson distributed with a distribution of W~Po(6).

From there we can use a table or calculator to evaluate: (e^-6)(6^4)/4! which gives:
0.1339 (4dp).
3. (Original post by Pinkdragon)

X-Po(2) Y-Po(3)
W=XY

Find P(W=4)
I don't get what is "Po" ? Is this teletubbies?
4. (Original post by hiq)
If we assume X and Y are two independent random variables we can see that the expected value of XY, i.e. E(XY) would be E(X)E(Y). In this case E(XY) or E(W) in other words would equal 6.

Therefore we can write W as being Poisson distributed with a distribution of W~Po(6).

From there we can use a table or calculator to evaluate: (e^-6)(6^4)/4! which gives:
0.1339 (4dp).
For part a) to this question, I worked out that E(XY)=6 and Var(XY)=36. Would this change the answer because for poison mean and variance are the same?
5. (Original post by Pinkdragon)
For part a) to this question, I worked out that E(XY)=6 and Var(XY)=36. Would this change the answer because for poison mean and variance are the same?
I think you've made a mistake calculating the combination variance:

Var(W) = Var(XY) = Var(X)Var(Y) = (2)(3) = 6

Therefore: Var(W) = E(W) and Poisson would be a justified model for this problem.
6. (Original post by hiq)
I think you've made a mistake calculating the combination variance:

Var(W) = Var(XY) = Var(X)Var(Y) = (2)(3) = 6

Therefore: Var(W) = E(W) and Poisson would be a justified model for this problem.
What exam board is this from?

I have edexcel s2 in 2 days and kinda worried never seen questions of this type.
7. (Original post by Hamzah249)
What exam board is this from?

I have edexcel s2 in 2 days and kinda worried never seen questions of this type.
I think the concept is first met in S1 (edexcel) with the idea of event independence with probability:

"If two events are independent P(AnB) = P(A) x P(B)"

You don't really get into combinations of distributions till S3 though, but even in S3 you're just combining normal distributions which is a little different.
8. (Original post by hiq)
I think the concept is first met in S1 (edexcel) with the idea of event independence with probability:

"If two events are independent P(AnB) = P(A) x P(B)"

You don't really get into combinations of distributions till S3 though, but even in S3 you're just combining normal distributions which is a little different.
Oh, that's a relief.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 22, 2017
Today on TSR

### Congratulations to Harry and Meghan!

But did you bother to watch?

### What do you actually do at University?

Poll
Useful resources

Can you help? Study help unanswered threadsStudy Help rules and posting guidelinesLaTex guide for writing equations on TSR

## Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE