Let $X_1\dots X_{500}$ be i.i.d. random variables with expected value $2$ and variance $4.$ The random variables $Y_1\dots Y_{500}$ are independent of the $X_i$ variables and also i.i.d., but they have expected value $2$ and variance $2.$ Estimate $P\left(\sum_{i=1}^{500}X_i>\sum_{i=1}^{500}Y_i+50\right)$ using the CLT.
Outline:
$$P\left(\sum_{i=1}^{500}X_i>\sum_{i=1}^{500}Y_i+50\right)=P\left(\sum_{i=1}^{500}X_i-\sum_{i=1}^{500}Y_i>50\right)=P\left(\sum_{i=1}^{500}[X_i-Y_i]>50\right)$$
Now, let $S_n=\sum_{i=1}^{500}[X_i-Y_i].$ Then we have $\color{red}{(1)}$ $E[S_n]=500\cdot 2-500\cdot 2=0$ and $Var(S_n)=500\cdot 4+500\cdot 2=3000$ (by independence). So, by CLT we have, $$ \begin{align} P\left(\sum_{i=1}^{500}[X_i-Y_i]>50\right)=P\left(S_n>50\right)&=P\left(\frac{S_n-E[S_n]}{\sqrt{Var(S_n)}}>\frac{50-0}{\sqrt{3000}}\right)\\ &=P\left(\frac{S_n-E[S_n]}{\sqrt{Var(S_n)}}>0.91287\right)\\ &=1-\Phi(0.91287)\\ &=0.1814 \end{align} $$ According to the answer key, the answer is $0.0202.$ I'm not sure where I made a mistake but my best guess would be setting up the expected value and variance in $\color{red}{(1)}.$ Any help would be appreciated.