Non classic average calculation

[al1981]

Dear colleagues

I have one basic question with possible answer already

As due to complexiti of the old model, i cant swith model to calculate averega activiy of the employee, so i found one workarrond

Example for  June 2024 (Acitivity is pretty long calculation of different contacts with customer, depends also on the channel and type of event)

Employee name	Activity	DaysOff	UserActiveDays
Employee 1	7.06	1.000	18.000
Employee 2	7.00	3.000	16.000
Employee 3	6.24	0.250	18.750

Question i have: if i sum Activity and devide by number of Employees i get average 6.77, which i think its not ok, but if i remove Employees from the rows, i get Average activitity 6.75. My basic knowledge explains me, 2nd one is more correct as rows are not equal/does not have same weight: Employee 3 did in more days less activity and therefore id decrease average for all?

Thank you for your feedback!

[al1981]

Dear @dharmendars007 , dear @Anonymous , thank you to both! If i understand: both of yours are "same" or at least are taking same solution. And i do agree: weighted average is something i need in my model! Many tnx to both of you!

[Anonymous]

Hi @al1981 ,

@dharmendars007 @lbendlin , thanks for your concern about this case. I tried to create sample data to perform weighted calculations.Please check if there is anything that can be improved. Here is my solution:

1\My data source (Table)

2\Create a measure

Result = SUMX('Table','Table'[Activity]*'Table'[UserActiveDays])/SUM('Table'[UserActiveDays])

Best Regards,

Bof

[dharmendars007 · Accepted solution]

Hello @al1981 , 

To calculate a more accurate "average activity per active day" across employees, you should use a weighted average formula:

Weighted Average = Activity*Active Days /Active Days

This will account for differences in the number of days employees were active and therefore provide a more balanced view of the average activity.

In your example:

• Employee 1: 7.06×18.00=127.087.06 \times 18.00 = 127.087.06×18.00=127.08
• Employee 2: 7.00×16.00=112.007.00 \times 16.00 = 112.007.00×16.00=112.00
• Employee 3: 6.24×18.75=117.006.24 \times 18.75 = 117.006.24×18.75=117.00

Sum of weighted activities: 127.08+112.00+117.00=356.08

Sum of active days: 18.00+16.00+18.75=52.75

So, the weighted average activity would be= 356.08/52.75 = 6.75

This is closer to the second value you mentioned, and it's more accurate because it takes into account the different active days for each employee.

If you find this helpful , please mark it as solution which will be helpful for others and Your Kudos/Likes 👍 are much appreciated!

Thank You

Dharmendar S

LinkedIN 

[lbendlin]

You need to decide what is more meaningful in your scenario - the average of averages, the average over all values, or a weighted average.