count distinct user with above 100% efficiency

[Heny]

I have this expression  Sum(norm*total_pages_count)/sum(total_time) that give me efficiency percentage

now I would like to count users on a monthly scale that are above 100%

I've tried 

count({<user={"=Sum(norm*total_pages_count)/sum(total_time)>0.1"}>} distinct user)

but I guess I'm totally off with my expression

any help?

[Kushal_Chawda]

@Heny  please mark appropriate reply as solution

[Heny]

I have a problem with my browser, it marks my solution as accepted

trying to change it