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Frequent Visitor

## How to Calculate the Overall Passing Percentage for a centre when the Passing rate is 67%.

Hi peeps

Need your help in trying to  work out a calulation for the overall passing quality score where the number of assesments met score is greater than and equal to 67% ([% of assessements met]).

Trying to achieve this healine

1 ACCEPTED SOLUTION

Accepted Solutions
Community Support Team

## Re: How to Calculate the Overall Passing Percentage for a centre when the Passing rate is 67%.

It seems "staff passing quality" is a measure,

Then you could create a measure

```Measure =
COUNTX (
FILTER ( ALL ( Sheet1 ), [staff passing quality] <> 0 ),
[staff passing quality]
)
/ CALCULATE ( DISTINCTCOUNT ( Sheet1[css name] ), ALL ( Sheet1 ) )
```

Best Regards

Maggie

Community Support Team _ Maggie Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

3 REPLIES 3
Super User

## Re: How to Calculate the Overall Passing Percentage for a centre when the Passing rate is 67%.

Hi,

Why should it be 67%.  7/24 is 29.16%.

=[Total assessments met]/[Total assessments]

Hope this helps.

Frequent Visitor

## Re: How to Calculate the Overall Passing Percentage for a centre when the Passing rate is 67%.

Hi Ashish,

The 29.16% is correct for the 24 assesments that were carried out. But what i am after is how many CSS/staff passed quality ( in the table the is 5 people which should be 5/16 = 31.25% which is the total passing rate for the site)

Community Support Team

## Re: How to Calculate the Overall Passing Percentage for a centre when the Passing rate is 67%.

It seems "staff passing quality" is a measure,

Then you could create a measure

```Measure =
COUNTX (
FILTER ( ALL ( Sheet1 ), [staff passing quality] <> 0 ),
[staff passing quality]
)
/ CALCULATE ( DISTINCTCOUNT ( Sheet1[css name] ), ALL ( Sheet1 ) )
```

Best Regards

Maggie

Community Support Team _ Maggie Li
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.