# HELPPP: find λ

#### shaolu

##### New Member
find λ given that P(X>4) = 0.0527
thank you

#### obh

##### Active Member
What distribution? exponential?
What did you try to do? (hint, adding ln to both size of the equation, ln(x^y) = y ∙ ln(x))

#### shaolu

##### New Member
What distribution? exponential?
What did you try to do? (hint, adding ln to both size of the equation, ln(x^y) = y ∙ ln(x))
poissson distribution

#### obh

##### Active Member
Please show how did you try to solve.

#### katxt

##### Member
Calculate P(X<=4)
Then search through the Poisson cumulative tables at X = 4 to find the appropriate λ.

#### obh

##### Active Member
OMG tables, nobody today should use the prehistoric tables. you should use a computer...

#### katxt

##### Member
How? I'm thinking this is a statistics class assignment. You could use Excel GoalSeek in a more general case, but I think that probably the tables solution is the one wanted here because the answer comes out exactly to a nice round number.

#### obh

##### Active Member
Yes, it looks like assignment. But they should stop using old tables in assignments! it doesn't make sense.
like they don't use slide rules anymore ...

#### katxt

##### Member
It's not an easy problem without special software. Try to find λ given that P(X>4) = 0.1
So, Shaolu, are you working with tables?

#### katxt

##### Member
Fair enough, especially if you're lucky. Finding λ given that P(X>4) = 0.1 may take a little longer.

#### obh

##### Active Member
1 min for a reasonable answer with binary search #### katxt

##### Member
Excel GoalSeek is another easy and accurate method.

#### obh

##### Active Member
You may also do a loop in R, I just used brute force ... a binary search would be more elegant

q=0.0527
x=4
#-----
p=1
delta=0.01
p1=1-q

lambda=0.01
while (p>p1)
{
lambda=lambda+delta
p=ppois(x,lambda)
}
lambda-delta