Question

if X~N(3,3) and Y~N(-0.5,0.25) are independent, find the probability that W=X+2Y is negative.

Answer 1

In the given problem X follows normal distribution with mean 3 and standard deviation 3

and Y be anothe independent normal variate with mean -0.5 and standard deviation 0.25

now W be a new variable it is a linear combination of X and Y

W= X+2Y

W is also a normal variate with mean= 3 - 0.5 =2.5 and standard deviation= (V(X)+V(2Y) =9+4X0.0625 = 9.25 =3.04138   since linear combination of two independent normal variates is also a normal variate

Hence W is also follows normal distribution with mean 2.5 and standard deviation 3.04138

now probability that W Is negative means P( W<0) = P ( Z < -0.822) where Z is a standard normal variate defined on W and is given as Z = (0 - 2.5) / 3.04138 = - 0.822

P(W<0) = P( Z< --0.822) = 0.5 - P(-0.822<Z<0)

= 0.5 - P 0<Z<0.822) since p(-0.822<Z<0) =P(0<Z<0.822)

= 0.5 - 0.2939 =0.2061 it is approximately equal to 0.21

Hence we conclude that the probabilitythat W= X+2Y is negative is 0.21

i.e P( W=X+2Y <0) = 0.21