From past experience it is assumed that the number of flaws

From past experience, it is assumed that the number of flaws in rolls of grade-2 paper follows a Poisson distribution with an average of one flaw per five linear feet of paper.

a. What is the approximate probability that there will be at least five flaws in a 50 foot length of paper?

b. What is the approximate probability that there will be between five and fifteen flaws in a 50 foot length of paper?

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

Poisson distribution with an average of one flaw per five linear feet of paper
a)
in a 50 foot length of paper, count of flaws are = 50/5 = 10 flaw

P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-10 * 2 ^ 4 / 4! + e^-10 * ^ 3 / 3! + e^-10 * ^ 2 / 2! + e^-10 * ^ 1 / 1! + e^-10 * ^ 0 / 0!
= 0.0293

P( X > = 5 ) = 1 - P (X < 5) = 0.9707
b)
P( X < 5) = P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= e^-10 * 2 ^ 4 / 4! + e^-10 * ^ 3 / 3! + e^-10 * ^ 2 / 2! + e^-10 * ^ 1 / 1! + e^-10 * ^ 0 / 0!
= 0.0293
P( X < 15) = P(X=14) + P(X=13) + P(X=12)
= e^-10 * 2 ^ 14 / 14! + e^-10 * ^ 13 / 13! + e^-10 * ^ 12 / 12! + e^-10 * ^ 11 / 11! + ....
= 0.9165

P( X < 16) = P(X=15) + P(X=14) + P(X=13) + + ...P(X=0)
= e^-10 * 2 ^ 15 / 15! + e^-10 * ^ 14 / 14! + e^-10 * ^ 13 / 13! + ...
= 0.9513
P( 5 <= X <= 15) = P( X < 16) - P( X < 5) = 0.9165 - 0.0293 = 0.8872