Given information:
- Var(X) = 0.2
- Var(Y) = 0.5
- Z = 5X - 2Y

To find: Var(Z)

Solution:
We know that the variance of a linear combination of random variables can be calculated using the following formula:
Var(aX + bY) = a^2Var(X) + b^2Var(Y) + 2abCov(X,Y)

where Cov(X,Y) is the covariance between X and Y.

Using this formula, we can find the variance of Z as follows:

Var(Z) = Var(5X - 2Y)
= 5^2Var(X) + (-2)^2Var(Y) + 2(5)(-2)Cov(X,Y) (substituting values)
= 25(0.2) + 4(0.5) - 20Cov(X,Y)
= 5 + 2 - 20Cov(X,Y)
= 7 - 20Cov(X,Y)

So, to find Var(Z), we need to find the covariance between X and Y.

We know that the covariance between two random variables X and Y is defined as:

Cov(X,Y) = E[XY] - E[X]E[Y]

where E[XY] is the expected value of their product, and E[X] and E[Y] are their respective expected values.

We don't have enough information to find E[XY], so we can use the following inequality:

|Cov(X,Y)| <=>

Using this inequality, we can obtain an upper bound on |Cov(X,Y)| as follows:

|Cov(X,Y)| <= sqrt(0.2*0.5)="">

So,

- If Cov(X,Y) = 0, then Var(Z) = 7
- If Cov(X,Y) = 0.316, then Var(Z) = 0 (minimum value)
- If Cov(X,Y) = -0.316, then Var(Z) = 14

Since the covariance can take any value between -0.316 and 0.316, the variance of Z can take any value between 0 and 7. However, the maximum value of Var(Z) occurs when the covariance is 0, so the answer is (A) 7.

Therefore, the correct answer is option (A) 7.