Question

Suppose that X is a random variable with mean 30 and standard deviation 4. Also suppose that Y is a random variable with mean 50 and standard deviation of 8. Assume that the correlation between X and Y is zero. Find the variance and the standard deviation of the random variable Z for each of the following cases. Show work.
(a) Z = 35 -10X
(b)Z = 12X - 5
(c) Z = X +Y
(d) Z = X -Y
(e) Z = -2X + 2Y

Answer 1

Answer:

a) Var[z] = 1600

    D[z] = 40

b) Var[z] = 2304

    D[z] = 48

c) Var[z] = 80

    D[z] = 8.94

d) Var[z] = 80

    D[z] = 8.94

e) Var[z] = 320

    D[z] = 17.88

Step-by-step explanation:

In general

V([x+y] = V[x] + V[y] +2Cov[xy]

how in this problem Cov[XY] = 0, then

V[x+y] = V[x] + V[y]

Also we must use this properti of the variance  

V[ax+b] = $a^{2}$V[x]

and remember that

standard desviation = $\sqrt{Var[x]}$

a) z = 35-10x

   Var[z] = $10^{2}$ Var[x] = 100*16 = 1600

   D[z] = $\sqrt{1600}$ = 40

b) z = 12x -5

   Var[z] = $12^{2}$ Var[x] = 144*16 = 2304

   D[z] = $\sqrt{2304}$ = 48

c) z = x + y

   Var[z] =  Var[x+y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = $\sqrt{80}$ = 8.94  

d) z = x - y

   Var[z] =  Var[x-y] = Var[x] + Var[y] = 16 + 64 = 80

   D[z] = $\sqrt{80}$ = 8.94

e) z = -2x + 2y

   Var[z] = 4Var[x] + 4Var[y] = 4*16 + 4*64 = 320

  D[z] = $\sqrt{320}$ = 17.88