Here is the question: Assume that 20 parts are checked each hour and that X deno
ID: 2918590 • Letter: H
Question
Here is the question: Assume that 20 parts are checked each hour and that X denotesthe number of parts in the sample of 20 that require rework. Partsare assumed to be independent with respect to rework. a) If the percentage that requre rework remains at 1%, what isthe probability that hour 10 is the first sample at which X exceeds1? I think I need to use negative binomial distribution but Idon't know how to solve it. Please help me !! Here is the question: Assume that 20 parts are checked each hour and that X denotesthe number of parts in the sample of 20 that require rework. Partsare assumed to be independent with respect to rework. a) If the percentage that requre rework remains at 1%, what isthe probability that hour 10 is the first sample at which X exceeds1? I think I need to use negative binomial distribution but Idon't know how to solve it. Please help me !!Explanation / Answer
We can solve this by using binomial distributiononly.
We need P( hour 10 is the first hour that X exceeds 1) =P(first 9 hours x doesn't exceed 1 and on 10th hour x exceeds1)
Let p = probability that in any hour X exceed 1
P( hour 10 is the first hour that X exceeds 1) = P(first9 hours x doesn't exceed 1 and on 10th hour x exceeds1) = (1-p)9 p
Now we have to calculate p = probability that in any hourX exceed 1 = 1- probability that x doesn't exceed 1
= 1- P(x=0)-P(x=1)
we know that number of reworking in an hour among 20 piecesfollows binomial distribution with n =20 and p = 1%
therefore P(x=0) = 20c0 1%0 * 99%20 = 81.79%
P(x=1) = 20c1 1%1 * 99%19 = 16.52%
therefore p = 1- 81.79% - 16.52% =1.69%
therfore P( hour 10 is the first hourthat X exceeds 1) = (1-p)9 p =1.44%
Hope it helps.