If X is a normal random variable XN22 and Y43X Find P Soluti

If X is a normal random variable X~N(2,2) and Y=4-3X Find P

Solution

X is a normal variable with X~N(2,2)

so mean of X is E[X]=2 and standard deviation of X is SD(X)=2

now Y=4-3X

so E[Y]=4-3E[X]=4-3*2=-2

and SD(Y)=SD(4-3X)=|-3|*SD(X)=3*2=6

as Y is a linear combination of X and X is a normal random variable so Y is also a normal random variable

with mean E[Y]=-2 and SD(Y)=6

so Y~N(-2,6)

a) P[X>1]=1-P[X<1]=1-P[(X-2)/2<(1-2)/2]=1-P[Z<-0.5] where Z~N(0,1)

             =1-0.308538=0.691462   [answer]   [using minitab]

b) P[-2<Y<1]=P[(-2+2)/6<(Y+2)/6<(1+2)/6]=P[0<Z<0.5]=P[Z<0.5]-P[Z<0] where Z~N(0,1)

                   =0.691462-0.5=0.191462 [answer] [using minitab]