Question

A textile manufacturing process finds that on average, five flaws occur per every 190 yards of material produced.

a. What is the probability of exactly four flaws in a 190-yard piece of material? (Do not round intermediate calculations. Round your final answers to 4 decimal places.) Probability . ......... b. What is the probability of no more than two flaws in a 190-yard piece of material? (Do not round intermediate calculations. Round your final answers to 4 decimal places.) Probability P c. What is the probability of no flaws in a 38-yard piece of material? (Do not round intermediate calculations. Round your final answers to 4 decimal places.) Probability

Answer 1

a)

Here, λ = 5 and x = 4
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 4)
P(X = 4) = 5^4 * e^-5/4!
P(X = 4) = 0.1755
Ans: 0.1755

b)

Here, λ = 5 and x = 2
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X < 2).
P(X < 2) = (5^0 * e^-5/0!) + (5^1 * e^-5/1!)
P(X < 2) = 0.0067 + 0.0337
P(X <2) = 0.0404

c)

Here, λ = 0.8 and x = 0
As per Poisson's distribution formula P(X = x) = λ^x * e^(-λ)/x!

We need to calculate P(X = 0)
P(X = 0) = 0.8^0 * e^-0.8/0!
P(X = 0) = 0.4493
Ans: 0.4493