Rank: Super forum user
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achrn wrote:Stern wrote:
This is still driving me insane as both look like legitimate methods but both give different answers.
Option 1 is correct.
Option 2 is wrong. It isn't the average number employed each day during the year.
Use this example:
Job 1 - 250 days, 2500 sign-ins (ie, 10 people working every working day)
Job 2 - 2 days, 3 sign-ins (ie, almost negligible)
Job 3 - 2 days, 2 sign-ins (ie, even smaller)
Obviously, the average employed in the year is very slightly over 10, because every single day you had at least 10 people working.
Method 1: (2500+3+2)/250 = 10.02
Method 2: (10+1.5+1)/3 = 4.17
Method 2 is obviously wrong. The average of a set of numbers cannot be lower than the lowest number.
Look at it on a day-by day basis. To save my keyboard do just four-weeks worth of data:
Job 1 - 20 days, 200 sign-ins
Job 2 - 2 days, 3 sign-ins
Job 3 - 2 days, 2 sign-ins
The number employed each day in the period
week 1: 10, 10, 10, 10, 10
week 2: 10, 10, 12, 11, 10
week 3: 10, 10, 10, 10, 10
week 4: 10, 10, 11, 11, 10
What is the average? Method 2 says the average of all those 10s, 11s and 12 is 4.17. It's obviously not right that the average of a load of 10s, 11s and 12s is 4, isn't it?
SOLD! Thanks achrn, you're a life saver, that makes complete sense now. I knew it would be something simple but I was just hitting a brick wall with it!
Thanks for your help (and everybody else too!)
Right, back to sanity...
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