Question

Flaws in a certain type of fabric are distributed as a Poisson distribution with the mean number of flaws equal to 0.500/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

Solution :

Given that ,

mean = = 0.500

Using poisson probability formula,

P(X = x) = (e^- * ^x ) / x!

a)

P(X > 2) = 1 - P(X 2)

= 1 - P(X = 0) - P(X = 1) - P(X = 2)

= 1 - (e^-0.50 * 0.50⁰⁾ / 0! - (e^-0.50 * 0.50¹⁾ / 1! - (e^-0.50 * 0.50²⁾ / 2!

= 1 - 0.98561

= 0.01439

Probability = 0.01439

b)

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

= (e^-0.50 * 0.50⁰⁾ / 0! + (e^-0.50 * 0.50¹⁾ / 1!

= 0.90979

Probability = 0.90979