# Thread: Normal approx to the binomial dist. questions

1. ## Normal approx to the binomial dist. questions

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 $W~Normal(np,\sqrt{np(1-p)}$.

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

$=P(Z\leq\\53.50$, 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 $n$ items is to be selected from a large lot, and the number of defectives $Y$ is to be observed. What value of $n$ guarantees that $\frac{Y}{n}$ 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 $\frac{Y}{n}$, where $Y$ 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:

$P(|\frac{Y}{n}-p|\leq\\0.1)=0.95$, 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

2. Q1: I would venture to guess that the most common error with a binomial distribution is forgetting what we mean by "success". You really just have to write down what you mean. Try p = 0.95. In this way, you will get an expected number of people appearing, not an expected number failing to show up. 53.5 is fine if your are trying to measure to probability of only 8 people appearing at the gate.