A metalstrip is produced by continuous casting process. The inspection of the st
ID: 3041930 • Letter: A
Question
A metalstrip is produced by continuous casting process. The inspection of the strip revealed that the number of inclusions per 1 meter of metal is represented by discrete probability function: f(x) = What is the probability that there will be 2 inclusion in 1 meter What is the probability that there will be at least 3 inclusions in one meter Ifon average there is 1 inclusion per meter, what is the probability that there will be at most 2 inclusion in 3 meters. If10 strips of metals each with 3 meter length are selected cut from the continuous strip. Knowing that a 3 meter strip is considered a defect if more than 2 inclusions are present within, what is the probability that at most 80% of the 10 strips are not defected? (lps) a) b) c) d)Explanation / Answer
THe probability is poission distribution with parameter = 1 inclusion per 1 meter of metal.
(a) Here if X is the number of inclusion
f(2) = e-1 /2! = 0.1839
(b) Pr(X > = 3) = 1 - [ Pr(X = 0) + Pr(X =1) + Pr(X = 2)] = 1 - [POISSON (0 ; 1) + POISSON (1, 1) + POISSON (2,1)]
= 1 - [e-1 /0! + e-1 /1! + e-1 /2!]
= 1 - (0.3679 + 0.3679 + 0.1839) = 0.0803
(c) Expected number of inclusion in 3 meters = 3 * 1= 3
so, if X is the number of inclusions in 3 meter
Pr(X < = 2; 3) = POISSON (X = 0, 3) + POISSON (X = 1, 3) + POISSON ( X= 2, 3)
= 0.0498 + 0.1494 + 0.2240 = 0.4232
(d) Pr(more than 2 inclusion are present in a 3 meter strip) = 1 - 0.4232 = 0.5768
Pr(defect) = 0.5768
Pr(not defected) = 0.4232
Here n = 10 and p = 0.5768
Here at most 80% of the 10 strips are not defected that means at most 8 strips out of 10 are not defected.
Pr(X <= 8 ; n = 10 ; p = 0.4232) = 0.9973