So we have the following
A B
mean 0.1 0.12
std dev 0.2 0.15
Correlation AB = -0.33
What we want is P(0.5A + 0.5B > 0.6A +0.4B) which reduces to
P(0.5A +0.5B – 0.6A – 0.4B)>0 which futher reduces to
P(0.1B – 0.1A)>0
Because the difference 0.1B – 0.1A is a linear combination of a random variable,it is itself normally distributed.
Its mean is therefore E(0.1B -0.1A) = 0.1*0.12 – 0.1*0.1 = -0.002
Its standard dev is SQRT((0.1^2*0.15^2 + 0.1^2*0.2^2 -2(0.1)(0.1)(0.2)(0.15)(-0.33))=0.028687976
The probability we are calculating is therefore P(Z> 0-(-0.002)/0.028687976) which reduces to
P(Z>0.0697156)= 47.21%
I AM NOT SURE WHAT HAPPENS WITH THE Z VALUE IN THE EXAM. MAYBE SOMEONE WHO HAS SAT FOR THE EXAM WILL KNOW IF THEY DO GIVE A RANGE OF Z VALUES TO CHOOSE FROM.