Airline Report Problem

**Allenge** · April 18th 2009, 09:54 AM

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....

**Twig** · April 18th 2009, 11:00 AM

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

**Twig** · April 18th 2009, 11:51 PM

$np = 150 \cdot 0.15 = 22.5$
$np(1-p) = 19.125$

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

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

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

**Allenge** · April 19th 2009, 05:44 AM

Originally Posted by Twig

$np = 150 \cdot 0.15 = 22.5$
$np(1-p) = 19.125$

$\frac{Z-\mu}{\sigma} = \frac{<b>30</b> - 22.5}{4.37} \approx 1.72$

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

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.

**mr fantastic** · April 19th 2009, 05:53 AM

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.

**Allenge** · April 19th 2009, 06:06 AM

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....

**Twig** · April 19th 2009, 11:33 AM

Mr fantastic, where is the fault in my reasoning?

thanks

**Allenge** · April 19th 2009, 02:15 PM

Any more Suggestions please...?

**mr fantastic** · April 19th 2009, 08:50 PM

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?

Originally Posted by Twig

Mr fantastic, where is the fault in my reasoning?

thanks

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?

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