calculation of variance

QUESTION
The distribution of variable x is Normal with mean = 0 and sd =1; y is such that
for y=1 ; x<= Q1,
for y=2 ; Q1<= x<=Q2,
and for y=3 ; Q2<= x ,
then find the variance of y

I interpreted the problem as follows
If we consider that there exists N values of x arranged in ascending order then we can also consider that
for first N/4 values of x, y= 1,
for next N/4 values of x, y= 2,
and for last N/2 values of x, y= 3,
So we have N/4 no. of 1 , N/4 no. of 2 and N/2 no. of 3 as values of y. This helps the calculation as follows
E(y) = (1.N/4+2.N/4 +3.N/2)/N = 2.25
E(y^2) = (1.N/4+4.N/4 +9.N/2)/N = 5.75
Var(y) =E(y^2) - [(E(y)]^2 = 5.75 - (2.25)^2 =0.6875

Please comment whether the approach is acceptable or not