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.
This is a slightly tricky question in that $X_i$, our random variable is the dollar win or loss. Therefore, we can't just apply the Binomial formula for $EX$ and $\sigma$. Note that $X = X_1 + X_2 + \cdots X_{100}$ are each independent but have the same distribution, and so the Central Limit Theorem applies.
First, we want our losses to be less than \$50, so we have $P\left(Z > \frac{-50 - 100\mu}{10\sigma}\right) = 1 - P\left(Z < \frac{-50 - 100\mu}{10\sigma}\right)$, but we need to identify $\mu$ and $\sigma$.
As Suzu answered, for one trial, we have $\mu = EX = (5)\frac{18}{38} - (5)\frac{20}{38} \approx -0.2632$.
Now, recall that $Var(X) = E(X^2) - (EX)^2$, so for one trial, $E(X^2) = 25$ (do you see why?). Therefore, $Var(X) = 25 - (-0.2632)^2 \approx 24.931$ and so $\sigma = 4.99$.
Now we may apply the CLT. Therefore, $P\left(Z < \frac{-50 - 100\mu}{10\sigma}\right) = P\left(Z < \frac{-50 + 26.32}{10(4.99)}\right) = P(Z < -0.473)$. Hence, $P(X > -50) = 1- P(Z< -0.47) = 1- 0.3192 = 0.6808$ as desired.
Above, $P(Z< -0.47) = 0.3192$ can be looked up in the back of any statistics book (or, calculated numerically or for the very bored, calculated by hand using techniques from complex analysis).