Finding Mean and Distribution of Normal Random Variables

Assume that $X_1$, and $X_2$ are i.i.d. normal random variables with mean $0$ and variance $1$. Let $Y_1$ and $Y_2$ be defined as $Y_1 =8X_1+6X_2$ and $Y_2 = X_1$.

1. $E[Y_1]= 0$ correct? because it's just $8 \times 0 + 6 \times 0$?
2. $Var(Y_1) = 100$? I'm getting this because $Var(x) = a^2Var(X)$ So each equal $64($variance of $X_1 = 1) + 36($variance of $X_2 = 1) = 100$.
3. $P(Y_1 \ge 12)= 1 - P(Y_1 < 12) =$Do I use the standard normal table for this and shift the standard normal over or something like that?
4. $Cov(Y_1,Y_2) =?$. I know $$Cov(X,Y) = E[(X-E[X])·(Y- E[Y])] = E[XY ]- E[X]E[Y ]$$ but I'm not sure how I apply it in this situation.

Any help solving these problems and offering suggestions would be greatly appreciated. Thank you all very much! I'm happy to also offer more clarification.

1. Correct.

2. Correct.

3. If you use tables for your normal CDF calculations then yes, recast to a standard normal with $Z=\frac{Y_1-\mu}{\sigma}$ and read off the number you need.

4. You're a fair bit of the way there. The next step is to write $E[Y_1 Y_2]$ in terms of expectations involving $X_1,X_2$, specifically: $E[Y_1 Y_2]=E[(8X_1+6X_2)X_1]=E[8X_1^2+6X_2X_1]$. Any ideas for the next step?

• I distribute X1 and plug in al the parts where I know mean for the variables? Do I need the formula Var(X) = E[X2]−(E[X])2 ? Also for part 3 do I say z = 12-0/1? That doesn't seem right Mar 29, 2016 at 11:49
• @user2598152 In part 3 you use the mean and variance of $Y_1$, the normal variable of interest. As for part 4, technically you will need that formula, though you also know $E[X_1]=0$ so...
– Ian
Mar 29, 2016 at 12:41
• Is the answer to four 0? lol Mar 29, 2016 at 12:45
• @user2598152 Nope.
– Ian
Mar 29, 2016 at 12:56
• I figured it out haha thank you so much! Mar 29, 2016 at 12:59