Question: HW Problem here, not sure where I'm messing up.

Let $X$ be the amount won or lost in betting \$5 on red in roulette. Then $P(5) = \frac{18}{38}$ and $P(-5) = \frac{20}{38}$. If a gambler bets on red one hundred times, use the central limit theorem to estimate the probability that those wagers result in less than $50 in losses.

So I calculated:

$E(x) = 100\frac{18}{38} = \frac{900}{19}$

$var(x) = \frac{900}{19}\times\frac{20}{38} = \frac{9000}{361}$

$\sigma = \sqrt{\frac{9000}{361}} = 4.993...$

Then this is what I was thinking:

$P(-\infty \leq x \lt 50) \rightarrow P(\frac{x - 50}{4.993} \lt 0)$

Which would give

$\int\_{-\infty}^{0}{\left(\frac{1}{\sqrt{2\pi}}\right)\left(e^{\tfrac{-z^2}{2}}\right)}$

Which using the table should be $(0.0 - 3.09) = (.5 - .001) = .499$

However I know from the back of the book it should be $.6808$. So where am I going wrong?

EDIT: I see from the comments and from one answer that I have my $E(x)$ wrong, but I'm still stuck on how to finish it out. I would normally go into my teacher's office hours for this, but due to Thanksgiving, there won't be any.

Answer: 
By definition, the determinant is the sum of $60$ positive and $60$ negative terms, one of which is $\mbox{sgn}{\left(
\begin{array}{ccccc}
 1 & 2 & 3 & 4 & 5 \\
 3 & 1 & 5 & 2 & 4 \\
\end{array}
\right)}(1000)^5=\mbox{sgn}\left((13542)\right)(1000)^5=+(1000)^5$.

Each of the other terms has absolute value at most $9\cdot1000^4$, so the sum of the 60 negative terms is greater than or equal to $-60\cdot9\cdot1000^4$, and the determinant is greater than or equal to $(1000)^5-60\cdot9\cdot1000^4$, which is positive.
With an even number of adjacent row swaps, you can put the $1000$s on the diagonal. That determinant must be positive. Using the algorithm which looks at minors, the largest term is always positive.