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# Simple Stats watch

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1. Here's the scenario.

Production line: 1000 widgets/day (24 hrs)

I have a p.d.f. of a faulty widget being produced, at t hrs after startup. i.e. p = p(t).

Question: what is the rate (widgets/hr, say) at which faulty widgets are being produced.

Can anyone tell me how I use a pdf to work out that rate ??

I have no idea
2. Hi

My stats is a little rusty but I did a first year course on probability at uni and what you've got is a random variable, the number of faulty widgets produced in a day with p.d.f p(t) therefore the probability of a faulty widget being produced between the times t1 and t2 after startup is the integral between t1 and t2 of p(t) with respect to t. The expected number of faulty widgets produced between t1 and t2 is the integral between t1 and t2 of t*p(t) with respect to t, wouldn't the rate of faulty widgets produced per hour then be d(Expectation)/dt = t*p(t) where t is time in hours? Hope this helps
3. Thanks a lot. That's it I think.

I missed a bit as well which you filled in. I had to show that there were approx so many failures per hour after 12 hrs, say.
So I should do int t*p(t) from t = 12 to t = 13, yes??

I knew about probabliity of faulty widget between t1 and t2 etc, but it was the other bit, t*p(t), I needed. Thanks.
4. Doing int t*p(t) from t = 12 to t = 13 will give you the expected number of failures in an hour which should be almost the same as doing int t*p(t) from t = 14 to t = 15 say if the function t*p(t) is approx constant after 12 hours which i think is what you are saying
5. What I wrote was a paraphrase for my actual problem.
The pdf is a decaying exponential.
And I have to find the number of "faults" per week after 6 years into the "program".
So I think the above approach is what I'm supposed to do.

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