Stuck on this probability problem

[semidevil]

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.

--------------------------------------------------

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 don't 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 don't know what to do w/ the 5.5

anybody help?

EDIT: ok, so I got the answer, but don't 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 don't 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?
---------------------------------------------------------------------
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?

 

[learningphysics]

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.

--------------------------------------------------

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 don't 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 don't 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.

 

[wais]

It seems like you are on the right track with using the normal distribution to solve this probability problem. The formula you mentioned, P(Z - np)/(sqrt(np(1-p)), is the correct formula to use for this type of problem. However, I can see where the confusion lies with the values of 239.5 and 260.5 and why the textbook table only goes up to 3.0.

First, let's break down the formula a bit more. The Z in the formula represents the standard normal distribution, which has a mean of 0 and a standard deviation of 1. In this case, we are trying to find the probability that Z lies between 240 and 260, which can be written as P(240 < Z < 260).

To solve this, we need to use the continuity correction, which is why you are using 239.5 and 260.5 instead of just 240 and 260. This correction takes into account that we are dealing with a continuous distribution and not a discrete one.

Now, to find the probability for 239.5, we need to use the z-score formula: z = (x - μ)/σ, where x is the value we are looking for, μ is the mean, and σ is the standard deviation. In this case, μ = 234 and σ = sqrt(np(1-p)) = sqrt(234*0.1*0.9) = 4.297.

Plugging in these values, we get z = (239.5 - 234)/4.297 = 1.29. Using the z-score table, we can find that the probability of z being less than 1.29 is 0.9015. This means that the probability of z being greater than 1.29 is 1 - 0.9015 = 0.0985.

Now, to find the probability for 260.5, we use the same formula but with x = 260.5. This gives us z = (260.5 - 234)/4.297 = 6.12. However, as you mentioned, the z-score table only goes up to 3.0. In this case, we can use the fact that the normal distribution is symmetric around the mean to our advantage. This means that the probability of z being above 3.0 is the same as the probability