Multivariate probability distributions?

[kingwinner]

Homework Statement

Let Y be the number of customers entering a ABC bank in a day. It is known that Y has a Poisson distribution with some unknown mean lambda. Suppose that 1% of the customers entering the branch in a day open a new ABC bank account. Find the mean and variance of the number of customers who open a new ABC bank account in a day.

Homework Equations

The Attempt at a Solution

Just based on past experience, I THINK this is related to multivariate distributions and PROBABLY would use the fact E[E(Y|X)]=E(Y). But I am not sure how to define the random variables properly...

Can someone explain?

Thanks for any help!

 

[HallsofIvy]

No, it is NOT a multivariate distribution- there is only the single variable, the number of customers. You are asked for the mean and variance of 0.01Y where Y is Poisson distributed.

If $\sum yP(y)= \lambda$, what is $\sum 0.01 yP(y)$?

 

[kingwinner]

OK, so this is actually a univariate problem...

Let Z=number of customers who open a new ABC bank account in a day
Z~Poisson(0.01*lambda)
Then our job is to find E(Z) and Var(Z)?

Or is our job to find E(0.01Y) and Var(0.01Y) where Y~Poisson(lambda)?

Thank you!

 

[kingwinner]

So Z = 0.01 Y, we need to find E(Z) and Var(Z)
E(Z)=E(0.01Y)=0.01E(Y)=0.01*lambda
Var(Z)=Var(0.01Y)=(0.01^2) Var(Y)=0.0001*lambda

Am I right? Is to any way to find lambda?