Question

Defects in a product occur at random according to a Poisson distribution with parameter ???? = 0.04. What is the probability that a product has one or more defects? If the manufacturing process of the product is improved and the occurrence rate of defects is cut in half to ???? = 0.02. What effect does this have on the probability that the product has one or more defects?

Answer 1

Answer:

When the occurrence rate of defect is cut in half, the probability of a product having one or more defects drops drastically (0.0392 to 0.0198), to almost the half of its original value too.

Step-by-step explanation:

Poisson distribution formula

P(X=x) = f(x) = (λˣe^(-λ))/x!

λ = 0.04.

And the probability that a products one or more defects is the same thing as 1 minus the probability that a product has no defect.

P(X ≥ 1) = 1 - P(X = 0) = 1 - f(0)

P(X ≥ 1) = 1 - (0.04⁰e^(-0.04))/0! = 1 - 0.9608 = 0.0392

When the occurrence rate of defect is cut in half, that is, λ = 0.02,

P(X ≥ 1) = 1 - P(X = 0) = 1 - f(0)

P(X ≥ 1) = 1 - (0.02⁰e^(-0.02))/0! = 1 - 0.9802 = 0.0198

When the occurrence rate of defect is cut in half, the probability of a product having one or more defects drops drastically, to almost the half of its original value too.

Hope this helps!