a factory produces plastic wrap On average the factory produ

a factory produces plastic wrap. On average, the factory produces two defects every eight hours. Assume the number of defects follows a Poisson distribution. In 24 hours of operation, what is the probability of exactly three defects? Ten or more?

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
the factory produces two defects every eight hours
defect rate for 24 hours is = 24 * 2/8 = 6
a)
P( X = 3 ) = e ^-6 * 6^3 / 3! = 0.0892
b)

P( X < 10) = P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-6 * 3 ^ 9 / 9! + e^-6 * ^ 8 / 8! + e^-6 * ^ 7 / 7! + e^-6 * ^ 6 / 6! + e^-6 * ^ 5 / 5! + e^-6 * ^ 4 / 4! + e^-6 * ^ 3 / 3! + e^-6 * ^ 2 / 2! + e^-6 * ^ 1 / 1! + e^-6 * ^ 0 / 0!
= 0.9161
P( X > = 10 ) = 1 - P (X < 10) = 0.0839