Question

An assembly line worker averages µ = 1 error per week of work (population characteristic). The errors occur on a random basis. What is the probability of two errors in a specific 50 hours of work?

Answer 1

Let X be the the number of error committed by worker in a time interval t.

And X follows Poisson distribution process

Now it is given that on average a ramdom worker commits one mistake or error per week of work

Number of hours in a week of work = 7 × 24 = 168 hours of work

So, lambda = parameter = 1/168 error per hour of work

The probability distribution of X for k number of events or success in a time interval t is given by

Putting k =2 , t =50 and lambda = 1/168 , we get

probability or two errors occuring in a 50 hour work

= (( 1/168 × 50)2 × e-(1/168 ×50) ) / 2!

= 0.032888

= 0.0329

So, there is 3.29% chance that there will be two errors in a 50 hours work by a random worker.

(don't get confuse that the worker is working 24×7 as he or she is not working all the time..

1 week of work may be spanned over any amount of time. It means that the total productive hours of work put by worker is in total equal to 1 week of time.. It is just a conversion)