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: 
Expand the determinant as a sum of $120$ terms, each being a product of $5$ factors. (That is, *imagine* you expand it; don't write it out. . . )

All these terms will be positive; then $60$ of them will be added and $60$ will be subtracted. Just one of the terms will be $1000^5$, and by checking the permutation you can see that it will have a positive coefficient.

Every other term will be at most $1000^3\times8\times9$; since exactly $60$ of these terms will be subtracted, we have
$$\eqalign{D
 &\ge1000^5-60\times1000^3\times8\times9\cr
 &>1000^5-100\times1000^3\times10\times10\cr
 &=990\times1000^4\ .\cr}$$
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.