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Resolver I

## Percentage of all Students spending more than 3 hours on their homework needed

This query gives me the nuber of students spending more tha 3 hours on heir home work.

How do I work out what the percentage is of those against the total student population?

homework_OT    - this is the main dataset

homework_OT[StudentID] -  all the students and their IDs are in [StudentID]

homework_OT[OT 3HRS PLUS] =1  - this is to filter out all those students that have a 1 in [OT 3HRS PLUS] to indicate that they spent more than 3 hours on their homework

What calculation will give me the % of [OT 3HRS PLUS] =1 from the total student population?

CNT Homework  > 3 HRS =
CALCULATE (
DISTINCTCOUNT( homework_OT[StudentID]),
FILTER ( homework_OT, homework_OT[OT 3HRS PLUS] =1 )
)
1 ACCEPTED SOLUTION
Super User

@acg , Try a measure like

CNT Homework > 3 HRS =
divide( CALCULATE (
DISTINCTCOUNT( homework_OT[StudentID]),
FILTER ( homework_OT, homework_OT[OT 3HRS PLUS] =1 )
), DISTINCTCOUNT( homework_OT[StudentID]))

or

CNT Homework > 3 HRS =
divide( Countx (FILTER (
values( homework_OT[StudentID]), calculate(sum(homework_OT[OT 3HRS PLUS])) >=1 ), [StudentID]), DISTINCTCOUNT( homework_OT[StudentID]))

Super User

@acg , Try a measure like

CNT Homework > 3 HRS =
divide( CALCULATE (
DISTINCTCOUNT( homework_OT[StudentID]),
FILTER ( homework_OT, homework_OT[OT 3HRS PLUS] =1 )
), DISTINCTCOUNT( homework_OT[StudentID]))

or

CNT Homework > 3 HRS =
divide( Countx (FILTER (
values( homework_OT[StudentID]), calculate(sum(homework_OT[OT 3HRS PLUS])) >=1 ), [StudentID]), DISTINCTCOUNT( homework_OT[StudentID]))

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