Question

Suppose that X is a random variable with mean 21 and standard deviation 4 . Also suppose that Y is a random variable with mean 42 and standard deviation 8 . Find the mean of the random variable Z for each of the following cases

(Give your answer to three decimal places.)

a) Z = 3 + 10X

b) Z = 3X − 10

c) Z = X + Y

d) Z = X − Y

e) Z = −4X − 3Y

Answer 1

Solution:

We have the a random variable X with mean 21 and standard deviation 4.

Mean = E(X) = 4

We have another random variable Y with mean 42 and standard deviation 8.

Mean = E(Y) = 42

a) Z = 3 + 10X

We have to obtain mean of Z (i.e. E(Z))

We know that if X is a random variable and a and b are two constants then,

E(a + bX) = a + b E(X)

Hence, E(Z) = E(3 + 10X)

E(Z) = 3 + 10 E(X)

E(Z) = 3 + (10 × 21)

E(Z) = 213

Hence, mean of Z is 213.

b) Z = 3X - 10

We have to obtain mean of Z (i.e. E(Z))

We know that if X is a random variable and a and b are two constants then,

E(aX - b) = a E(X) - b

Hence, E(Z) = 3 E(X) - 10

E(Z) = 3 E(X) - 10

E(Z) = (3 × 21) - 10

E(Z) = 53

Hence, mean of Z is 53.

c) Z = X + Y

We have to obtain mean of Z ( i.e. E(Z))

We know that if X and Y are two random variables then

E(X + Y) = E(X) + E(Y)

Hence, E(Z) = E(X + Y)

E(Z) = E(X) + E(Y)

E(Z) = 21 + 42

E(Z) = 63

Hence, mean of Z is 63.

d) Z = X - Y

We have to obtain mean of Z ( i.e. E(Z))

We know that if X and Y are two random variables then

E(X - Y) = E(X) - E(Y)

Hence, E(Z) = E(X - Y)

E(Z) = E(X) - E(Y)

E(Z) = 21 - 42

E(Z) = -21

Hence, mean of Z is -21.

e) Z = -4X - 3Y

We have to obtain mean of Z ( i.e. E(Z))

We know that if X and Y are two random variables and a and b are two constants then,

E(aX + bY) = a E(X) + b E(Y)

Hence, E(Z) = E(-4X - 3Y)

E(Z) = (-4) E(X) + (-3) E(Y)

E(Z) = (-4 × 21) + (-3 × 42)

E(Z) = -84 - 126 = -210

Hence, mean of Z is -210.

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