# Binomial to Normal Approximation

• November 15th 2008, 10:14 PM
lllll
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:

$P\left(\frac{np-nr}{\sqrt{npq}}= \frac{8-7.75}{\sqrt{7.6}} = 0.90685\right)$

so I would have:

$P(Z\leq 0.90685) = 0.8186$

but the solution in the back of the book gives 0.8980 which would be $P(Z\leq 1.27) = 0.8980$
• November 16th 2008, 12:04 AM
CaptainBlack
Quote:

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:

$P\left(\frac{np-nr}{\sqrt{npq}}= \frac{8-7.75}{\sqrt{7.6}} = 0.90685\right)$

so I would have:

$P(Z\leq 0.90685) = 0.8186$

but the solution in the back of the book gives 0.8980 which would be $P(Z\leq 1.27) = 0.8980$

The mean number of passengers that arrive is $152$, the variance of this is $7.6$ and so the standard deviation is $\approx 2.757$.

Now to be overbooked we need to calculate the probability that more than $155$ passengers turn up. The z-score corresponding to this is:

$z=\frac{155.5-152}{2.757}\approx 1.269$

We use $155.5$ in this calculation as a continuity correction, $154.5$ to $155.5$ passengers round to 155 when we are forced to an integer.

CB