Math Is Fun Forum

Yes that is right but job C is a quick job that takes say 1 hour where job B takes on average 3 hours. So you are right we need to find out how long it takes to efficiently complete all the jobs over one year.  The average is useful to forecast what would be likely to occur next year.

So we can forecast that the jobs will decrease by say 5% next year we can then with the average forecast how many staff will be required to complete the tasks.  So really it is not only historical but future orientated based on outcomes (if that makes sense)

ok staying on line.  fyi the incidents are actually events that have taken place and i am trying to work out how many people it would take to efficiently complete those tasks. On average a worker would spend 8 hrs a day, 38 hours a week for 45 weeks on average per year. this figure is where I get the 1710 from

Hi Bob,

Thanks for that, I understand your formula and the theory thereof. If I divide 22.10 by 347 I get an extraordinarily small answer 0.0637, however, if I divide 347 by 22.10 I get 15.7 which seems more believable I think.

I have one more part to this. When I get my weighted value by the supplied formula I need to correlate that to the amount of hours worked in one year. In this case that amount is 1710. Is it simply a case of multiplying the weighted average answer by the number of occurrences?

Thank you very much for your help thus far.

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