[radiation]

Hi.. This is a question I found in a text that I am unable to solve.. Help would be appreciated..
"Because not all airline passengers show up for their reservation, an airline sells 125 tickets for a flight that holds 120 passengers.The probability that a passenger does not show up is 0.1, and the passengers behave independently
a) What is the probability that every passenger who shows up can take the flight?
b) What is the probability that the flight departs with empty seats?"

The problem I think is based on binomial distribution of random variable.

Thanks.. :confused:

[Unknown008]

That problem sound familiar to me somehow :confused:

Anyway, this is a binomial distribution all right.

Let X be the even that a passenger shows up.
X~B(0.9, 120)

Find P(X = 120) which is also

You can visualise it on a probability tree? The probability that the first passenger shows up is 0.9, then for the second to show up is 0.9 too, together being 0.9^2. The third is 0.9 again, making a total of 0.9^3, etc.

When it departs with empty seats, that means that it is not full. That is X is everything from 0 to 119 but not 120. So, take the probability you got in (a) from 1 to get the probability of having empty seats.

[radiation]

I did do the same thing, but the answer is not consistent with the solution given.. For a) the solution given is 0.9961...

[Unknown008]

Hmm.. there might be something I overlooked... now that '125 tickets' bothers me. I'll have to think about it more. I just hope I'll not forget.

If by Sunday I didn't posted, you can post so that I get a notification.

[galactus]

You can use as binonial with the

probability a passenger shows is

.90

X is the # of passenegers who show.



There are 125 passengers and we look at each one as a Bernoulli trial.

Because the numbers are large, we can use the binomial with continuity correction. If you have a calculator, it wouldn't matter though.







Look up in the z table to find the probability.

[Unknown008]

So it has also a little of the normal distribution curve?

I'm having some trouble understanding how you got 120.5 for x for P(x<120), or the reason you used less than or equal to. :confused:

[morgaine300]

I can explain the less than or equal to 120 part. It only holds 120 passengers so in order to make sure everyone can have a seat, no more than 120 people can show up. But 120 can show up, hence the equal to. In other words, 1 can show, 2 can show, all the way to 120. It doesn't say the probability that exactly all seats are taken.

I don't know what the 120.5 is. Maybe he's just using that as the point at which it goes over and everyone doesn't get a seat? I don't know what a continuity correction is. I know where the 112.5 came from and also the standard deviation. I don't know why it can be used in a z equation.

What I do know is that z of 2.385 doesn't equal the .9961 that's supposed to be the answer. A P of .9961 is z = 2.66. So, I work backwards. 2.66 * 3.354 = 8.92161 + 112.5 = 121.4.

Hmm. Don't know.

Unky, yours doesn't work for 2 reasons. One, you have to account for the probabilities of the other 5 customers not showing. You also have to account for the less than or equal to - you're doing exactly 120, instead of no more than 120. OK, 3 reasons. Also, you're not accounting for the number of ways those people can show/not show. i.e. how many ways are there for 120 to show and 5 to not show?

[morgaine300]

Well, I got the answer of .9961. But I had to do it the long way, it took a while, and I was getting nasty numbers that I had to stick in memory to work with them. I just basically did it the old fashioned way that I originally learned.

I have a hard time believing that's what expected of them for that problem, but it got me exactly .9961.

radiation, do you know what Galactus is talking about? Were you just taught:
?
Or do you just not understand what the problem is asking for to begin with?

[galactus]

What I used is the binomial for the normal distribution. It comes in handy when the numbers are large. The 120.5 comes from the continuity correction.

[Unknown008]

Yes, I asked my teacher today and she explained why you used the 120.5 part, how the binomial distribution becomes like the normal distribution curve for very large numbers of x.

But now, why did you choose exactly 120.5? Is it determined from any calculation?

[radiation]

Well, I got the answer of .9961. But I had to do it the long way, it took a while, and I was getting nasty numbers that I had to stick in memory to work with them. I just basically did it the old fashioned way that I originally learned.

I have a hard time believing that's what expected of them for that problem, but it got me exactly .9961.

radiation, do you know what Galactus is talking about? Were you just taught:
?
Or do you just not understand what the problem is asking for to begin with?

Yeah, I was just given the formula you specified...

[Unknown008]

No worries for my query, I checked in my stats book and I understood the continuity correction. Now, I'm getting the probability as 0.9914 from my z table...

Or is it because using the normal curve while p is quite large (0.9) makes the approximation quite far from the real probability? :confused:

[femiogunjumo]

Using excel:
=BINOMIAL(120,125,0.9,1)
Answer of first question is 0.996141
For 2nd question,
=BINOMIAL(119,125,0.9,1)
Answer is 0.988568


Using Excel:
First question:
=BINOMDIST(120,125,0.9,1)
Answer is 0.996141
2nd question:
=BINOMDIST(119,125,0.9,1)
Answer is 0.988568

Sorry, I meant =BINOMDIST when you use excel, not =BINOMIAL.


Using excel:
=BINOMIAL(120,125,0.9,1)
Answer of first question is 0.996141
For 2nd question,
=BINOMIAL(119,125,0.9,1)
Answer is 0.988568