Forum Discussion
Standard Deviation without considering 0
Good afternoon capos,
I am trying to calculate the standard deviation based on the yield what happens is that it brings 0 (zeros) then the deviation comes out an absurd value I have tried to do it with if but I can not get it with the formula.
I am using a measure to calculate the yield (Volume/Hours), as I need to have the data for the overall total of the product at the end of the month.
| Product | Day | Volume (Tons) | Hours Worked (Hours) | Yield (Tons / Hours) |
| 1 | 1/9/2021 | 1784 | 55 | 32,44 |
| 1 | 02/09/2021 | 2015 | 33 | 61.06 |
| 1 | 9/03/2021 | 1021 | 34 | 30.03 |
| 1 | 4/9/2021 | 0 | 0 | - |
| 1 | 5/9/2021 | 2086 | 63 | 33.11 |
| 1 | 6/9/2021 | 2068 | 36 | 57,44 |
| 1 | 7/9/2021 | 2390 | 32 | 74,69 |
| 1 | 8/9/2021 | 1066 | 39 | 27,33 |
| 1 | 9/9/2021 | 2269 | sixty-five | 34,91 |
| 1 | 10/09/2021 | 2303 | 45 | 51,18 |
| 1 | 11/9/2021 | 2431 | 43 | 56,53 |
| 1 | 12/09/2021 | 0 | 0 | - |
| 1 | 13/9/2021 | 2091 | 44 | 47,52 |
| 1 | 14/9/2021 | 2967 | 49 | 60,55 |
| 1 | 15/9/2021 | 2125 | 55 | 38,64 |
| 1 | 16/9/2021 | 2039 | 88 | 23.17 |
| 1 | 17/9/2021 | 1525 | 80 | 19.06 |
Hi, Syndicate_Admin ;
Please try to create a measure to calculate the standard deviation.
standard deviation = STDEVX.P(ALL('Table'),DIVIDE([Volume (Tons)],[Hours Worked (Hours)]))The final output is shown below:
Best Regards,
Community Support Team_ Yalan Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
2 Replies
You should be able to solve this by applying a report level filter Volume>0
Hi, Syndicate_Admin ;
Please try to create a measure to calculate the standard deviation.
standard deviation = STDEVX.P(ALL('Table'),DIVIDE([Volume (Tons)],[Hours Worked (Hours)]))The final output is shown below:
Best Regards,
Community Support Team_ Yalan Wu
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
