Q1:

An airline finds that 5% of the persons who make reservations on a certain flight do not show up for the flight. If the airline sells 160 tickets for a flight with only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

A1: Let p=0.05, n=160, and $\displaystyle W~Normal(np,\sqrt{np(1-p)}$.

Then, $\displaystyle P(W\leq\\155.5)=P(\frac{W-np}{\sqrt{np(1-p)}}$$\displaystyle \leq\frac{155.5-8}{\sqrt{7.6}}$)

$\displaystyle =P(Z\leq\\53.50$, which cannot be correct. The correct answer is 0.898, which I am nowhere close to. Where am I going wrong?

Q2: (b) A rancom sample of $\displaystyle n$ items is to be selected from a large lot, and the number of defectives $\displaystyle Y$ is to be observed. What value of $\displaystyle n$ guarantees that $\displaystyle \frac{Y}{n}$ will be within .1 of the true fraction of defectives, with probability 0.95?

Note: Part (a) of this question showed that the variance of $\displaystyle \frac{Y}{n}$, where $\displaystyle Y$ has a binomial distribution with n trials and success probability of p, has a maximum at p=0.5, which I was able to show.

A2: Symbolically, I think I need to solve:

$\displaystyle P(|\frac{Y}{n}-p|\leq\\0.1)=0.95$, so I want to use the normal approximation to some how come up with a z-value, which I can use to solve for n. Although, when I go down this path I end up with a false equation for n.

Thanks for you help