Thread: Binomial to Normal Approximation

Assume an airline manages to sell 160 seats for a plane that is only capable of seating 155. Assume that 0.05 of ticket holders cancel their reservation. What is the probability that the place will be overbooked?

I found these http://www.mathhelpforum.com/math-he...obability.html and http://www.mathhelpforum.com/math-he...rs-flight.html that deal with the exact same question, although they focus on solving the problem through the binomial method, which I have no trouble doing, nevertheless I would like to know how to solve it using the Normal Approximation. I am not sure what variables I have to use to get the result...

I would think that my variance would be npq = 160*0.05*0.95=7.6

then I would have:



so I would have:



but the solution in the back of the book gives 0.8980 which would be

Originally Posted by lllll

Assume an airline manages to sell 160 seats for a plane that is only capable of seating 155. Assume that 0.05 of ticket holders cancel their reservation. What is the probability that the place will be overbooked?

I found these http://www.mathhelpforum.com/math-he...obability.html and http://www.mathhelpforum.com/math-he...rs-flight.html that deal with the exact same question, although they focus on solving the problem through the binomial method, which I have no trouble doing, nevertheless I would like to know how to solve it using the Normal Approximation. I am not sure what variables I have to use to get the result...

I would think that my variance would be npq = 160*0.05*0.95=7.6

then I would have:



so I would have:



but the solution in the back of the book gives 0.8980 which would be

The mean number of passengers that arrive is , the variance of this is and so the standard deviation is .

Now to be overbooked we need to calculate the probability that more than passengers turn up. The z-score corresponding to this is:



We use in this calculation as a continuity correction, to passengers round to 155 when we are forced to an integer.

CB