Question

Flaws in a certain type of fabric are distributed as a Poisson distribution with the mean number of flaws equal to 1.000/square yard.

a. Find the probability that a random square yard of this fabric will contain more than 2 flaws.

b. Find the probability that a random square yard of this fabric will contain fewer than 2 flaws.

Answer 1

(a)

Probability Mass Function of Poisson Distribution with mean = 1 is given by:

,

     for x = 0, 1, 2..

P(X>2) = 1- [P(X=0) + P(X= 1)+P(X=2)}

So,

P(X>2) = 1 - (0.3679 + 0.3679+0.1839)

              = 1 - 0.9197

               = 0.0803

So,

Answer is:

0.0803

(b)

P(X<2) = P(X=0) + P(X= 1)

So,

P(X<2) = 0.3679 + 0.3679

               = 0.7358

So,

Answer is:

0.7358