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.

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, $\mathbb{P}(\text{overbook}) = 0.886$