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So I have this table
| Variety | Size | Min | Max | Max MM | Min MM | ProbA |
| Grapefruit | 18 | 4.83 | 6.06 | 153.924 | 122.682 | 0.993578 |
| Grapefruit | 23 | 4.38 | 4.82 | 122.428 | 111.252 | 0.844918 |
| Grapefruit | 27 | 4.23 | 4.37 | 110.998 | 107.442 | 0.699794 |
| Grapefruit | 32 | 4.01 | 4.22 | 107.188 | 101.854 | 0.422138 |
| Grapefruit | 36 | 3.92 | 4.00 | 101.6 | 99.568 | 0.311689 |
| Grapefruit | 40 | 3.72 | 3.91 | 99.314 | 94.488 | 0.125933 |
| Grapefruit | 48 | 3.59 | 3.71 | 94.234 | 91.186 | 0.058042 |
| Grapefruit | 56 | 3.30 | 3.58 | 90.932 | 83.82 | 0.005854 |
| Grapefruit | 64 | 3.02 | 3.29 | 83.566 | 76.708 | 0.000294 |
| Grapefruit | 64 | 0.00 | 3.00 | 76.2 | 0 | 8.36E-41 |
ProbA formula is
Hi @Anonymous ,
Or this.
Measure =
NORM.DIST(
//MAX represents the value of the current row here
MAX([Min MM]),
[Middle of this Range],
[Average Standard Dev],
TRUE()
)
Best regards,
Lionel Chen
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
@Anonymous
I can't test this, but according to the documentation, the first expression of the NORM.DIST function must be a number value (you are feeding it a column).
so try an aggregation function like AVERAGE(Variety sizes[Min MM]) as the first expression:
NORM.DIST(AVERAGE(Variety sizes[Min MM]), ...etc
Proud to be a Super User!
Paul on Linkedin.
@Anonymous Have you tried using
Respectfully,
Zoe Douglas (DataZoe)
Follow me on LinkedIn at https://www.linkedin.com/in/zoedouglas-data
See my reports and blog at https://www.datazoepowerbi.com/
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