Question 980415
P(5) with Poisson parameter 5.  It is e^(-lambda) *Lambda^5 (or whatever number x is)/x!
exp(-5)*5^5/5!=0.1755  This is the expected value. Also, the variance is 5 as well.  
exp(-5)5^4/4!=0.1755
exp(-5)5^3/3!=0.1404
exp(-5)5^6/6!=0.1462
That is 1 and 0
exp(-5)*5=0.0337 + exp(-5)=0.0067 and that sum is 0.0404.
!=1  0!=1