SOLUTION: Find some variances and standard deviations. Suppose that X is a random variable with mean 20 and standarddeviation 3. Also suppose that Y is a random variable with mean 60 and sta

Algebra ->  Probability-and-statistics -> SOLUTION: Find some variances and standard deviations. Suppose that X is a random variable with mean 20 and standarddeviation 3. Also suppose that Y is a random variable with mean 60 and sta      Log On


   

Question 1191018: Find some variances and standard deviations. Suppose that X is a random variable with mean 20 and standarddeviation 3. Also suppose that Y is a random variable with mean 60 and standard deviation 2. 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. Be sure to show your work.
A)Z=33−8X.
B)Z=11X−6.
C)Z=X+Y.
D)Z=X−Y.
E)Z=−2X+2Y.
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Here's the breakdown of how to calculate the variance and standard deviation of Z for each case:
**Key Principles:**
* **Variance of a constant:** Var(c) = 0
* **Variance of a scalar multiple:** Var(aX) = a²Var(X)
* **Variance of a sum/difference (independent variables):** If X and Y are independent (correlation is 0), Var(X + Y) = Var(X) + Var(Y) and Var(X - Y) = Var(X) + Var(Y)
* **Standard Deviation:** SD(Z) = sqrt(Var(Z))
We're given Var(X) = 3² = 9 and Var(Y) = 2² = 4.
**A) Z = 33 - 8X**
* Var(Z) = Var(33 - 8X) = (-8)²Var(X) = 64 * 9 = 576
* SD(Z) = sqrt(576) = 24
**B) Z = 11X - 6**
* Var(Z) = Var(11X - 6) = 11²Var(X) = 121 * 9 = 1089
* SD(Z) = sqrt(1089) = 33
**C) Z = X + Y**
* Var(Z) = Var(X + Y) = Var(X) + Var(Y) = 9 + 4 = 13
* SD(Z) = sqrt(13) ≈ 3.61
**D) Z = X - Y**
* Var(Z) = Var(X - Y) = Var(X) + Var(Y) = 9 + 4 = 13
* SD(Z) = sqrt(13) ≈ 3.61
**E) Z = -2X + 2Y**
* Var(Z) = Var(-2X + 2Y) = (-2)²Var(X) + 2²Var(Y) = 4 * 9 + 4 * 4 = 36 + 16 = 52
* SD(Z) = sqrt(52) ≈ 7.21