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    Need some statistics/math help

    If a group is 88% successful in completing all assigned tasks with 24 employees, is there a way to guesstimate the success rate percentage if you add 6 additional employees (assuming all other factors such as production/skill/etc are equal)? Thanks.

    #2
    The percentage will be the same assuming all other factors equal.

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      #3
      I'm getting at workload. If they can achieve 88% success with 24 employees (assuming they can only work a fixed amount of hours per day), how much with the decreased workload and additional help affect the success rate.

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        #4
        .88 x (24+6) = 26.4 successful employees out of 30

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          #5
          Is it as simple as saying each employee contributes x percentage of success so 6 additional employees would result in a combined XX percentage of success?

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            #6
            Let me try again.

            There were 133 projects last year with 24 employees working on them. The group was able to achieve an 88% completion rate.

            How can I compute the completion rate if there had been 30 employees (all other factors being equal) with the same 133 projects.

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              #7
              110%

              ...you would need a 27-28 employees to equal 100%
              Last edited by RedYote; 10-27-2021, 09:06 AM.

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                #8
                Originally posted by RedYote View Post
                110%
                Ha. That's what I just came up with. I guess that ain't going to work for my application. Thanks.

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                  #9
                  Originally posted by Chew View Post
                  Let me try again.

                  There were 133 projects last year with 24 employees working on them. The group was able to achieve an 88% completion rate.

                  How can I compute the completion rate if there had been 30 employees (all other factors being equal) with the same 133 projects.
                  24 employees completed 117 projects equaling 4.875 completed projects per employee.

                  If you had 30 employees with the same rate you would have completed 146 projects.

                  RedYote was right with 110%

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                    #10
                    so for this you have 133/24 that gives you 5.54 per person.

                    the 88% is not needed for what you are asking

                    so 5.54 per person and you add 6 more you get 33.25 so you would be able to solve 33 more cases in a year

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                      #11
                      I'll have to rethink it. Not gonna work for my current application. Thank yall for the help.

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                        #12
                        This is what I get... 24 employees working on 133 projects and completed 88% of them. So 88% of 133 is 117. Each employee did 4.876 %. There were 16 projects that didnt get done, so if you add 3.2 employees with each doing 4.86% (like the others), you should be able to get 100% completion...........maybe

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                          #13
                          Y'all are correct in your math...but just not going to work for my application right now. Too many other factors.

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                            #14
                            Originally posted by Chew View Post
                            Y'all are correct in your math...but just not going to work for my application right now. Too many other factors.
                            Did you factor in F+J+B?

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                              #15
                              Unless your achievements are directly proportional to effort with no outside influences (like assembling Ikea furniture or consuming jelly donuts), you cannot use a simple calculation. Sounds like you need some sort of statistical analysis, with variable inputs and simulated outputs. I occasionally use software (based on Monte Carlo simulations) to evaluate risks and calculate contingencies, but the extent of my understanding is the inputs and outputs. The calculations taking place in the background are beyond my comprehension, but probably exciting stuff for a nerdy statistician.

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