# Stuck on this probability problem

stuck on this probability problem....

this using normal distribution
so an airline on average has 90% of ticket holders will not showup. So the airline sells more then seats it has to make it up. suppose the airline sold 260 tickets, and only has 240 seats.

use the normal curve to approximate the probability that not all tickets holder who show up can be accommodated.

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ok, so writing the formula, it is pretty much the summation from 240 to 260 of (260 C K)*.9 * .1^k. for k = 241 to 260.

that part is easy. now when I calculate it, here is what I did.

basically, need to know P(240 < z <= 260).

n = 260
p = .9
by continuity correction, we have P(239.5 < z <= 260.5)

np = 234
np(1-p) = 23.4

I have P(Z - np)/(sqrt(np(1-p)). So using 239.5 for Z, I get my answer is 1.14.

now, I do the same thing, with Z = 260.5, and I get 5.5.

now, I dont feel good about the 5.5, because I"m suppose to use my table in my textbook to solve this, but the textbook table only goes to 3.0.

the first answer, I look at the table and get .9099.

how do I get the answer for the 5.5??

the back of the book says the answer is .0901 as the whole answer....so I dont know what to do w/ the 5.5

anybody help?

EDIT: ok, so I got the answer, but dont understand it. basically, once they got the answer for the 240.5, they took that, and did the integeral from 240.5 to infinity of the function, and 1 - whatever that is, becomes .90901.

but what I dont get is why did we not have to do anything with the 260? I mean, we solved from 240 to infinity....but what about the 260?
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because right now, i'm doing another problem and the question specificaly asks what is the probability that it is above A, but also below B. so how do I handle that?

for above A, is it a< Z < inf, and B is -inf<Z<B, and take the difference?

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semidevil said:
this using normal distribution
so an airline on average has 90% of ticket holders will not showup. So the airline sells more then seats it has to make it up. suppose the airline sold 260 tickets, and only has 240 seats.
Just want to make sure I understand the problem. Are you sure it is 90% of ticket holders will NOT show up? Only 10% show up?

use the normal curve to approximate the probability that not all tickets holder who show up can be accommodated.

--------------------------------------------------
semidevil said:
ok, so writing the formula, it is pretty much the summation from 240 to 260 of (260 C K)*.9 * .1^k. for k = 241 to 260.

that part is easy. now when I calculate it, here is what I did.

basically, need to know P(240 < z <= 260).

n = 260
p = .9
by continuity correction, we have P(239.5 < z <= 260.5)

np = 234
np(1-p) = 23.4

I have P(Z - np)/(sqrt(np(1-p)). So using 239.5 for Z, I get my answer is 1.14.

now, I do the same thing, with Z = 260.5, and I get 5.5.

now, I dont feel good about the 5.5, because I"m suppose to use my table in my textbook to solve this, but the textbook table only goes to 3.0.

the first answer, I look at the table and get .9099.

how do I get the answer for the 5.5??

the back of the book says the answer is .0901 as the whole answer....so I dont know what to do w/ the 5.5

anybody help?
Your normal approximation already takes into account that n=260, and the z score cannot go beyond 260. ie. P(z<=260) = 1. If you use the table (if it went long enough), you won't get this exact value, since it is only an approximation but it will be close. But you know that P(z<=260)=1 as an exact value... you don't need the table for this part.

P(240.5<z<=260.5)=P(z<=260.5)-P(z<=240.5)=1-P(z<=240.5)

or another way to think about it:
P(240.5<z<=260.5)=P(z>240.5) (since z can never be greater than 260.5)
= 1-P(z<=240.5)

Remember that your normal distribution takes into account sigma=np, and the npq... and n=260.