Question #d03b4

1 Answer
Dec 12, 2017

#X~P_o(5)#

#P(X=k)=e^(-k)(5^k)/(k!)#

Explanation:

For a Poisson distribution with mean #lambda#

#X~P_o(lambda)#

#P(X=k)=e^(-lambda)(lambda^k)/(k!)#

in this case for #lambda=5#

#X~P_o(5)#

#P(X=k)=e^(-k)(5^k)/(k!)#