The number of fire alarms went out in ZSH has a Poisson dist

The number of fire alarms went out in ZSH has a Poisson distribution with a mean of 2 outages per year.
(a) What is the probability that there will be no more than one fire alarm went out in a year
(b) What is the probability that there will be two fire alarms went out in a year
(c) What is the probability that more than one and less than four fire alarms went out in a year
(d) What is the expected number E(X) and standard deviation(X) of fire alarms went out in a year

Solution

Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials
mean of 2 outages per year.
a)
P( X < = 1) = P(X=1) + P(X=0) +   
= e^-2 * 2 ^ 1 / 1! + e^-2 * 25 ^ 0 / 0! +
= 0.406
b)
P( X = 2 ) = e ^-2 * 2^2 / 2! = 0.2707
c)
P( X = 3 ) = e ^-2 * 2^3 / 3! = 0.1804

P( 1 < X < 4 ) = P( X = 2 ) + P( X = 3 ) = 0.2707 + 0.1804 = 0.4511