1. ## 2 stats question

1.
suppose an airline sells 300 tickets when there is only 270 seats available. on avg, 15% of those with tickets do not show up. what is the probability that there will be enough seats for everybody who shows up?

2.
A special roulette wheel has seven equally likely outcomes: 0, 1, 2, 3, 4, 5, 6. If you bet that an odd number comes up, you win or lose \$10 according to whether or not that event occurs. If X denotes your ‘net’ gain, X=10 if we see 1, 3, or 5, and X = -10 otherwise. Suppose that you play this game 400 times. Let Y be your net gain after these 400 plays. The mean (expectation) and standard deviation of Y are...

i'd figured out the mean (its -571) but how do u figure out the stnd dev?

Thanks!!

2. 1. Pretty much the same as the other question. The probability that there will be enough seats, is the compliment of the probability that there WONT be enough seats - i.e. 271 or more people show up. This problem can be done one of two ways:

Using the Binomial distribution and calculating 29 binomial equations, where N is equal to 300 tickets and K is equal to going through those 300 tickets and choosing "29, 28, 27. . .0" that are going to no show;

Or you can remember that you can approximate the Binomial using the Normal as in the problem from your other thread (we can use the normal here since 5<np, and nq).

2. Remember:

$V(X)=\Sigma x^{2}P(X=x)-\mu^{2}_x$

3. Originally Posted by ANDS!
1. Pretty much the same as the other question. The probability that there will be enough seats, is the compliment of the probability that there WONT be enough seats - i.e. 271 or more people show up. This problem can be done one of two ways:

Using the Binomial distribution and calculating 29 binomial equations, where N is equal to 300 tickets and K is equal to going through those 300 tickets and choosing "29, 28, 27. . .0" that are going to no show;

Or you can remember that you can approximate the Binomial using the Normal as in the problem from your other thread (we can use the normal here since 5<np, and nq).

2. Remember:

$V(X)=\Sigma x^{2}P(X=x)-\mu^{2}_x$
so for 1. would this be correct?

$\mu=np=30(0.75)=22.5$
$\sigma= \sqrt{22.5(0.25)}=2.37$
$z=\frac{30-22.5}{2.37}=0.9992$

4. For approximations using Normal, we use the standard deviation of the Binomial as:

$\sigma_x=\sqrt{npq};$

$\sigma_x=\sqrt{(300)(.15)(.85)}$

Be careful what you set your mean to be. If you multiply it by the probability that someone shows up, then you would NOT be using 30 as the value you are comparing, you would be using 270 since that is the threshold for the number of people who are meant to show up.