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.
