1. The problem statement, all variables and given/known data X refers to score distribution in Math and Y refers to score distribution in Stat in a certain degree course exam. It is known that X~N(mean = 62, sigma=7) while Y~N(mean = 68, sigma=10). If X and Y are independent, find (i) P[X+Y>120]; (ii) P[X<Y]; (iii) P[X+Y>140]; (iv)P[100 < X+Y < 150]. 2. Relevant equations 3. The attempt at a solution P(X+Y>120)... So I have to combine X & Y into one Norm Distribution by find there combined E(x) & V(x) E(X+Y)=E(X)+E(Y)=62+68=130 V(X+Y)=V(X)+V(Y)=7^2 + 10^2 = 149, Standard Dev = sqrt(149) So where Z = X + Y > 120, Z~N(130,sqrt(149)) P(X+Y>120) = P(Z>120)=1-PHI((130-120)/sqrt(149)) Is this correct so far.... P(X<Y)=P(X-Y<0) E(X-Y)=E(X)-E(Y) = 62-68=-6 V(X-Y) = V(X)+V(Y)=sqrt(149) W~N(-6,sqrt(149)) P(X-Y<0)=P(W<0)=PHI((-6-0)\sqrt(149)) How does this look. The rest are basically the same I think if I got these two conceptualy correct. any issues?