Question

A textile manufacturing process finds that on average, 3 flaws occur per every 260 yards of material produced. a. What is the probability of exactly 1 flaws in a 260-yard piece of material? (Round your intermediate calculations and final answer to 4 decimal places.) Probability b. What is the probability of no more than 2 flaws in a 260-yard piece of material? (Round your intermediate calculations and final answer to 4 decimal places.) Probability c. What is the probability of no flaws in a 130-yard piece of material? (Round your intermediate calculations and final answer to 4 decimal places.) Probability When answering could you please explain how you derive at the answer? When it is shown for example, as P(X=5)=(6^5)*exp(-6)/5! = 0.1606231 I am not sure how to compute that to get that answer of 0.1606231. Thank you.

Answer 1

Let X :- Number of flaws occur in textile manufacturing process

X poisson ( ) average 3 flaws occur per every 260 yards of material produced

Probability Mass Function of Poisson distribution is as follows

P(X = x) = ( e^- *  ^X ) / X!

Part a) To find the probability  of exactly 1 flaws in a 260-yard piece of material

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

P(X = 1) = ( e^-3 * 3¹ ) / 1!

P(X = 1) = (0.04979 * 3) / 1 since, 1! = 1

P(X = 1) = 0.1494

Part 2) What is the probability of no more than 2 flaws in a 260-yard piece of material?

Solution :- No more than 2 flaws it means there could be 0,1 or 2 flaws in a material

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

P(X = x) = ( e^- *  ^X ) / X!

P(X = 0) = ( e^-3 * 3⁰ ) / 0!

P(X = 0) = ( 0.04979 * 1) / 1

P(X = 0) = 0.0498

P(X = 1) = ( e^-3 * 3¹ ) / 1!

P(X =1) = ( 0.04979 * 3 ) / 1

P(X = 1) = 0.1494

P(X = 2) = ( e^-3 * 3² ) / 2!

P(X = 2) = ( 0.04979 * 9) / 2

P(X = 2) = 0.2240

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

P ( X <= 2) = 0.0498 + 0.1494 + 0.2240

P ( X <= 2) = 0.4232

Part c) What is the probability of no flaws in a 130-yard piece of material

Solution :- There are on an average 3 flaws in 260 yard of material, if price of material goes to half than number of flaws should be half of 260yards of material i.e

On an average 1.5 flaws in 130 yard of material

P(X = x) = ( e^- *  ^X ) / X!

P ( X = 0 ) = ( e^-1.5 * 1.5⁰ ) / 0! since,   

P ( X = 0 ) = ( 0.2231 * 1 ) / 1

P ( X = 0 ) = 0.2231