Basic Statistics Question?

Question

I have an exam tomorrow, and for the life of me I can't figure out how to do problems like this:

An aircraft can seat 220 passengers, and each of the passengers booked on the flight has a
probability of .9 of actually arriving at the gate to board the plane, independently of the other
passengers.
1. Suppose the airline books 235 passengers on the flight. What is the probability that there
will be insuffcient seats to accommodate all of the passengers who wishes to board the
plane.
2. If the airplane wants to be 75% con¯dent that there will be no more than 220 passengers
who wish to board the plane, how many passengers can be booked on the flight?

For the first part, I don't know how to find the z-score - from looking at the solutions, I know the z-score is 1.957, but I don't know how to calculate that.

For the second part... I don't know where the confidence interval plays into it.

So it's more suitable for binomial, what does that mean in reference to, you know, solving it?

Answer 1

To use the binomial you call a sucess showing up, then you want to know the probability of more than 220 sucesses in 235 trials:

It's easier to get more than 15 faillures in 235 trials and subtract from 1

P( more than 220 sucesses in 235 trials) =

sum k=221 to 235 { 235Ck*p^k*q^(235-k)}

My TI-83 can do this:

1-0.9803 = 0.0197

So, there's about a 2% chance of insuffieient seats.



It is binomial but can be approximated using the normal.

To do so use:

mean = np = 235*0.9
sd = sqrt(np(1-p))

Answer 2

It's been a while since I've had statistics. However, I was reading on this earlier today. It seems to me that this situation will not have a normal distribution. It's either 'yes, they show up' or 'no they don't show up' and for simplicity it is assumed that these events are independent. So, I think this type of situation results in a binomial distribution. I know this will sound dumb, but are there z-scores for binomial distributions? If there is, you would need to know the mean and the standard deviation. The mean of a binomial distribution is given by the formula np, where p is the chance of a success: 90% and n is the number of trials, in this case 235. The standard deviation would be given by the sqrt [np(1-p)]. There is a formula for a binomial distribution (see your text or the link below) for calculating the chance that exactly x number of people show for their flight. However, you want to know the probability that 221 or more passengers will show up. I don't know of any shortcut. The direct way to calculate this is to use the formula for 221, 222, ...., 235. This really makes me wonder if there must be some shortcut.

I wish I had some advice for the second part of your question, but I can't even remember how to find a confidence interval.

Hope this helps. Good luck on your test tomorrow.

Answer 3

1. since n is sufficiently large: Z = (220.5 - E(X))/sigma(x)

E(X) = 0.9*235 = 211.5
sigma(x) = sqrt(0.9*0.1*235) = 4.59

The reason there is 220.5 there not 220 is because there is an extra 0.5 added for continuity and is commonly called the continuity correction

Z = (220.5-211.5)/4.59 = 1.59698

2. This is a little odd to me, because of the distribution the answer looks to be pretty small, but I'd do a normal confidence interval
220+z*sigma(x)/sqrt(n)

Answer 4

I had to take files in HS (it became an AP route which became brutal) and in college. you'll do nicely in case you listen, TAKE NOTES, study AND DO nicely on the try. files is like technology and math blended...you may want to gather files, diagnosis it, interpret it and understand what it skill. It became puzzling for me yet I surpassed with an A- because I listened and took time previous regulation with my professor after classification. you'll do advantageous!

Answer 5

A z score is a measure of the distance in standard deviations of a sample from the mean.
z = (x - mu) / s

Answer 6

that seems like a problem more suited for a binomial distribution