# Find variance of a sample X and Y

I've taking a test and got fail in this exercise. It says:

"Let $$X_1,X_2,X_3$$ and $$X_4$$ be independent, identically distributed random variables such taht each of them has a normal distribution with mean 0 and variance 1. We have $$Y_1,Y_2$$ and $$Y_3$$ which are independent identically distributed random variables with mean 2 and variance 3. Firther, all $$X_i$$ and $$Y_i$$ are independent. Let $$\overline{X}$$ and $$\overline{Y}$$ denote the sample means of $$X_i$$ and $$Y_j$$ respectively. Calculate $$V(\overline{X}-2\overline{Y})$$"

(a) 1.25

(b) 4.25

(c) -5

(d) 7

I know that $$V(\overline{X}-2\overline{Y})=V(\overline{X})-2V(\overline{Y})$$ and then I know $$V(\overline{X})=1$$ because of the information in the exercise, similar for $$V(\overline{Y})=3$$ hence $$V(\overline{X}-2\overline{Y})=1-2\cdot 3=-5$$" but it's not true.

Any help would be nice

Suppose $$n$$ iid random variables share a variance $$v$$, then the variance of their mean is not $$v$$ but $$v/n$$. It follows that $$V(\overline X)=\frac14$$ and $$V(\overline Y)=1$$. Finally, multiplying a random variable by $$a$$ multiplies its variance by $$a^2$$, and variances always add. Thus the answer is $$\frac14+2^2×1=4.25$$.
You should have rejected $$-5$$ immediately, since variances are always nonnegative.