-
O mais eficiente (usando over()).
select Grade, count(*) * 100.0 / sum(count(*)) over() from MyTable group by Grade
-
Universal (qualquer versão SQL).
select Grade, count(*) * 100.0 / (select count(*) from MyTable) from MyTable group by Grade;
-
Com CTE, o menos eficiente.
with t(Grade, GradeCount) as ( select Grade, count(*) from MyTable group by Grade ) select Grade, GradeCount * 100.0/(select sum(GradeCount) from t) from t;