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

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] = V[x]

and remember that

standard desviation =

a) z = 35-10x

   Var[z] = Var[x] = 100*16 = 1600

   D[z] = = 40

b) z = 12x -5

   Var[z] = Var[x] = 144*16 = 2304

   D[z] = = 48

c) z = x + y

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

   D[z] = = 8.94  

d) z = x - y

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

   D[z] = = 8.94

e) z = -2x + 2y

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

  D[z] = = 17.88

one set of data has more variation than the other. hope this !

answer:  

step-by-step explanation:

odd numbers are only divisible by 1 and itself.

even numbers are divisible by more than 1 and itself.

for example 6 is even because it is divisible by more than 1 and its self. 1 x 6= 6 ; 2 x 3= 6

and 7 is odd because only 7 x 1 equals seveb