Poisson Distribution Mean & SD: Solving for Y

[Mo]

I am attempting a past paper question from school, i don't have the answer (and it doesn't look like i will anytime soon!)

The question:
"The Random variable X has a poisson distribution with mean 4. The random variable Y is defined by

Y = 4X + 1

Find the mean and standard deviation of Y"

So .. where do i begin?

I understand that the mean and variance of a poisson distribution is lambda.I know that the square root of the variance is the SD.They are telling us that this R.V X has a mean and variance of 4 right?

Am i right in thinking that the mean of Y is the same as the mean of (4X + 1)?

So is this right .. (at least to start with)?

E(4X + 1) = 4

or is it ...

4E(X) + 1 where E(X) is 4?


Help is very much appreciated!

Regards,
Mo

PS: Stats is not something that i understand all that easy

 

[kleinwolf]

The 2nd : E(4X+1)=4*E(X)+1...and then i think it's like :

var(Y)=E(Y^2)-E(Y)^2=E(16X^2+8X+1)-(4*E(X)+1)^2=16(E(X^2)-E(X)^2)=16*var(X)...but I'm not sure

 

[mathman]

Note to Kleinwolf: The variance calculation is correct. In general, adding a constant leaves the variance unchanged, while multiplying by a constant changes the variance by multiplying by the constant squared.