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# S2 help ~ Poisson. ♥ watch

1. Question uno.
Given that lamda = 2.2 and P(X=x) = 0.1967 fiind the value of x.

So I subed it in the equation to work out Px and I got:
(2.2^x)/x! = 0.1967/e^-2.2

But I can't seem to solve this would I just have to do it trial and error?

Question dos.
Suppose that for a positive integer x,
P(X=x-1)<=P(X=x) and P(X=x+1)<P(X=x)
Determine in terms of x, the possible range of values for lambda.
=/ Like I'm using the formula thing: Px=lamda/P(x-1)

But it's not really working. It seems like the question is wrong....
2. Bump
3. (Original post by SmileyGurl13)
Question uno.
Given that lamda = 2.2 and P(X=x) = 0.1967 fiind the value of x.

So I subed it in the equation to work out Px and I got:
(2.2^x)/x! = 0.1967/e^-2.2

But I can't seem to solve this would I just have to do it trial and error?
You are not going to get an analytic solution. You just need to try values for x and see which one fits the bill.

Question dos.
Suppose that for a positive integer x,
P(X=x-1)<=P(X=x) and P(X=x+1)<P(X=x)
Determine in terms of x, the possible range of values for lambda.
=/ Like I'm using the formula thing: Px=lamda/P(x-1)

But it's not really working. It seems like the question is wrong....
So what have you done?

PS: Don't recognize that formula; you might want to check it.
4. (Original post by ghostwalker)
You are not going to get an analytic solution. You just need to try values for x and see which one fits the bill.

So what have you done?

PS: Don't recognize that formula; you might want to check it.
Okie for 1

And the formula is:

P(X=x)=lambda/x * P(X=x-1)
5. The formula for a Poisson distribution is:

$P(X=r)&space;=&space;e^{-\lambda}&space;\frac{\lambda^r}{r!}$

It has an r! in it - r is what you want to find, and because of the r! you know r is a positive integer. So just start at one and go up. It didn't take me too long is your only hint.
6. (Original post by SmileyGurl13)
And the formula is:

P(X=x)=lambda/x * P(X=x-1)
That looks more like it. So, what have you done with it?
7. (Original post by ghostwalker)
That looks more like it. So, what have you done with it?

Well I said P(X=x-1) = x/lamda P(X=x)

so x/lambda P(X=x) <= P(x=x)
therefore x/lamda<=1
so lamda>=x

And P(x=x+1) = lamda/xP(x=x)

So lamda/xP(x=x)<P(X=x)
so lamda/x<1

so lamda<x

8. (Original post by tcb1992)
The formula for a Poisson distribution is:

$P(X=r)&space;=&space;e^{-\lambda}&space;\frac{\lambda^r}{r!}$

It has an r! in it - r is what you want to find, and because of the r! you know r is a positive integer. So just start at one and go up. It didn't take me too long is your only hint.
Yeah I've got it now thanks
9. (Original post by SmileyGurl13)
And P(x=x+1) = lamda/xP(x=x)

So lamda/xP(x=x)<P(X=x)
so lamda/x<1

so lamda<x

That's not correct, but from your latter post, I presume you've sorted it now.

But just for the record.

P(x=x+1) = lamda/(x+1) * P(x=x)

And you have
10. (Original post by ghostwalker)
That's not correct, but from your latter post, I presume you've sorted it now.

But just for the record.

P(x=x+1) = lamda/(x+1) * P(x=x)

And you have
No I haven't got it :P I meant I'd got the first question, Can you explain how you got P(x=x+1) = lamda/(x+1) * P(x=x) thanks
11. Poisson is French for fish. If that's the help you're after.
12. (Original post by SmileyGurl13)
No I haven't got it :P I meant I'd got the first question, Can you explain how you got P(x=x+1) = lamda/(x+1) * P(x=x) thanks
That's just the original equation you had evaluated at "x+1" rather than at "x".

Alternatively, you would write down P(x+1) from the definition of the poisson distribution, and similarly P(x), then work out what P(x+1)/P(x) equals, and you will get that equation.
13. (Original post by ghostwalker)
That's just the original equation you had evaluated at "x+1" rather than at "x".

Alternatively, you would write down P(x+1) from the definition of the poisson distribution, and similarly P(x), then work out what P(x+1)/P(x) equals, and you will get that equation.
thank you very much you've helped a lot. +ve rep coming your way

I may be asking for help in the future, cos I can't ask teachers as I'm self-studying :| is tht okay?
14. (Original post by SmileyGurl13)
thank you very much you've helped a lot. +ve rep coming your way
Cheers.

I may be asking for help in the future, cos I can't ask teachers as I'm self-studying :| is tht okay?
That's what the forum is here for.

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