1. ## AP stats

About 60% of customers at a restaurant want to sit in a non-smoking section. A new restaurant with 120 seats is being planned. How many seats should be in the non-smoking area in order to be very sure of having enough seating?

How would I find this?

2. Originally Posted by Morgan82
About 60% of customers at a restaurant want to sit in a non-smoking section. A new restaurant with 120 seats is being planned. How many seats should be in the non-smoking area in order to be very sure of having enough seating?

How would I find this?
The simple-minded approach would be to calculate the expected number of non-smokers: (n)(p) = (120)(0.6) = 72.

But how sure is 'very sure'? 99% sure? Having 72 seats wouldn't make you 99% sure ...... (in fact, it would only make you about 53% sure!).

Let X be the random variable number of non-smoking seats required.

Then X ~ Binomial (n = 120, p = 0.6)

To be say, 99% sure, you need to find the smallest value of x such that $\Pr(X \leq x) > 0.99$. This value of x is the number of seats required.

You can use technology to very easily solve $\Pr(X \leq x) > 0.99$ for x ..... I get x = 84.

3. Originally Posted by mr fantastic

You can use technology to very easily solve $\Pr(X \leq x) > 0.99$ for x ..... I get x = 84.
Technology Assistance for Morgan82:

Binomial Distribution

For a 95% confidence ratio, use x = 81