# Airline Report Problem

• April 18th 2009, 10:54 AM
Allenge
Airline Report Problem
Hello everybody,
I have solved the way my professor tought us in class but I'm getting a different answer. If somebody can help me I will really Appreciate.....

An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows. Use the normal distribution to approximate the binomial distribution.

Thanks....
• April 18th 2009, 12:00 PM
Twig
hi
hi

The normal approximation

Binomial distribution - Wikipedia, the free encyclopedia

$N(np, np(1-p))$ (1)

$0.15 \cdot 150 = 22.5$

Now use the formula given, (1). I gotta go know, in a hurry, I´ll check in later. GL
• April 19th 2009, 12:51 AM
Twig
$np = 150 \cdot 0.15 = 22.5$
$np(1-p) = 19.125$

$\frac{Z-\mu}{\sigma} = \frac{30 - 22.5}{4.37} \approx 1.72$

$P(Z\leq 1.72) \approx 0.9573$

To be honest, I have never done this before, but I hope it´s right, otherwise let me know!
• April 19th 2009, 06:44 AM
Allenge
Quote:

Originally Posted by Twig
$np = 150 \cdot 0.15 = 22.5$
$np(1-p) = 19.125$

$\frac{Z-\mu}{\sigma} = \frac{30 - 22.5}{4.37} \approx 1.72$

$P(Z\leq 1.72) \approx 0.9573$

To be honest, I have never done this before, but I hope it´s right, otherwise let me know!

The answer does not match with the ones I have; I see you have used 30; I dont know how you got that? I tried to use 20 as given in the question, still it doesn't give the right answer.
• April 19th 2009, 06:53 AM
mr fantastic
Quote:

Originally Posted by Allenge
Hello everybody,
I have solved the way my professor tought us in class but I'm getting a different answer. If somebody can help me I will really Appreciate.....

An airline reports that it has been experiencing a 15% rate of no-shows on advanced reservations. Among 150 advanced reservations, find the probability that there will be fewer than 20 no-shows. Use the normal distribution to approximate the binomial distribution.

Thanks....

Let X be the random variable number of no shows.

X ~ Binomial(n = 150, p = 0.15).

Calculate Pr(X < 20) using the normal approximation.

Please say where you're stuck if you are still stuck.
• April 19th 2009, 07:06 AM
Allenge
Quote:

Originally Posted by mr fantastic
Let X be the random variable number of no shows.

X ~ Binomial(n = 150, p = 0.15).

Calculate Pr(X < 20) using the normal approximation.

Please say where you're stuck if you are still stuck.

Still stuck; tried to use the normal approximation but I'm not getting the right answer.

p(x < 20) => p(z< {20-22.5}/4.37) = -0.5721

p(z< -0.5721) = 0.2843

This is what I get but it is not one of the answers....
• April 19th 2009, 12:33 PM
Twig
Mr fantastic, where is the fault in my reasoning?

thanks
• April 19th 2009, 03:15 PM
Allenge
• April 19th 2009, 09:50 PM
mr fantastic
Quote:

Originally Posted by Allenge
Still stuck; tried to use the normal approximation but I'm not getting the right answer.

p(x < 20) => p(z< {20-22.5}/4.37) = -0.5721

p(z< -0.5721) = 0.2843

This is what I get but it is not one of the answers....

Did you use the continuity correction?

Quote:

Originally Posted by Twig
Mr fantastic, where is the fault in my reasoning?

thanks

Quote:

Originally Posted by Twig
$np = 150 \cdot 0.15 = 22.5$

$np(1-p) = 19.125$

$\frac{Z-\mu}{\sigma} = \frac{{\color{red}30} - 22.5}{4.37} \approx 1.72$

$P(Z\leq 1.72) \approx 0.9573$

To be honest, I have never done this before, but I hope it´s right, otherwise let me know!

Where did the 30 come from?