Central limit problem. Calculate the probability that the flight will be overbooked. The problem said: 

An airplane has 120 seats. The probability that a ticketed passenger
  will show up for a flight is 0.95. Assume that all passengers act
  independently and that the airline has sold 130 tickets for a
  particular flight. Using the Normal approximation to the Binomial
  (with appropriate continuity correction), compute the approximate
  probability that the flight will be overbooked.

I know I need to use the central limit theorem, and I start by defining indicator variables:
Xi= 1 if passanger i show up for a fligth, 0 otherwise.

Then the approximate probability that the flight will overbooked, defined by even OB is:
P(OB)=P(sum from 1 to 130 of Xi > 120)  

I know also that P(xi)=0.95
But I'm missing something to finish this problem, and found the correct solution wich the book said is: 0.886 or 88.6%. Thanks for your help.
 A: Let $X_i \sim Ber(p)$, then $Y = \sum_{i=1}^n X_i \sim Bin(n,p)$.
From the CLT we have $$\frac{Y - E[Y]}{\sqrt{Var(Y)}} = \frac{Y - np} {\sqrt{np(1-p)}} \sim N(0,1),$$
Plugging the values you have:
$$\frac{120 - 130 \times 0,95} {\sqrt{130 \times 0,95 \times (1-0,95)}} =-1.40848,$$
So, $$\mathbb{P}(\text{overbook})=\mathbb{P}(Y>120) = \mathbb{P}(Z>1.40848),$$ where $Z \sim N(0,1)$. Hence, $\mathbb{P}(\text{overbook}) = 0.886$
A: First let's compute the variance of $X_i$:$$\sigma^2 = \mathrm{Var}[X_i] = 0.95-0.95^2 = \frac{14}{400}.$$
We now the define the random variable $S$ as the number of passengers that want to board the plane: $$S = \sum_{i=1}^n X_i.$$
The mean of $S$ is obvious: $$\mathrm{E}[S] = pn = 0.95 \times 130 = 123.5.$$ (This is bad already: we expect there to be more passengers than seats.)
The variance of $S$ can be estimated using the CLT:$$\mathrm{Var}[S]\rightarrow n\sigma^2 = 4.55.$$
Enter both values into a normal distribution calculator and you find the probability $\mathrm{Pr}[S>120]$.
