SOLUTION: Find some means. Suppose that X is a random variable with mean 20 and standard deviation 2. Also suppose that Y is a random variable with mean 40 and standard deviation 7. Assume t
Algebra ->
Statistics
-> Describing-distributions-with-numbers
-> SOLUTION: Find some means. Suppose that X is a random variable with mean 20 and standard deviation 2. Also suppose that Y is a random variable with mean 40 and standard deviation 7. Assume t
Log On
Question 1191145: Find some means. Suppose that X is a random variable with mean 20 and standard deviation 2. Also suppose that Y is a random variable with mean 40 and standard deviation 7. Assume that the correlation between X and Y is zero. Find the mean of the random variable Z for each of the following cases. Be sure to show your work.
A)Z=25−12X.
B)Z=13X−8.
C)Z=X+Y.
D)Z=X−Y.
E)Z=−3X+3Y
You can put this solution on YOUR website! Here's how to find the mean of the random variable Z for each case:
**Key Principle:**
The mean of a linear combination of random variables is equal to the linear combination of their means. That is:
E(aX + bY) = aE(X) + bE(Y)
Where 'a' and 'b' are constants.
We are given E(X) = 20 and E(Y) = 40.
**A) Z = 25 - (1/2)X**
E(Z) = E(25 - (1/2)X)
E(Z) = 25 - (1/2)E(X)
E(Z) = 25 - (1/2)(20)
E(Z) = 25 - 10
E(Z) = 15
**B) Z = (1/3)X - 8**
E(Z) = E((1/3)X - 8)
E(Z) = (1/3)E(X) - 8
E(Z) = (1/3)(20) - 8
E(Z) = (20/3) - (24/3)
E(Z) = -4/3 or -1.33 (approximately)
**C) Z = X + Y**
E(Z) = E(X + Y)
E(Z) = E(X) + E(Y)
E(Z) = 20 + 40
E(Z) = 60
**D) Z = X - Y**
E(Z) = E(X - Y)
E(Z) = E(X) - E(Y)
E(Z) = 20 - 40
E(Z) = -20
**E) Z = -3X + 3Y**
E(Z) = E(-3X + 3Y)
E(Z) = -3E(X) + 3E(Y)
E(Z) = -3(20) + 3(40)
E(Z) = -60 + 120
E(Z) = 60
