Question

A new restaurant with 134 seats is being planned. Studies show that 62​% of the customers demand a​ smoke-free area. How many seats should be in the​ non-smoking area in order to be very sure ​(mu plus 3 sigma​) of having enough seating​ there?

Answer 1

Answer: 100

Step-by-step explanation:

Given : The total number of seats planned in new restaurant =134

The percentage of customers demand a​ smoke-free area = 62%

It can also be written as $62\%=\dfrac{62}{100}=0.62$

The mean of this binomial distribution will be :-

$\mu=np\\\\\mu=(134)(0.62)=83.08$

Standard deviation:-

$\sigma=\sqrt{np(1-p)}\\\\\Rightarrow\sigma=\sqrt{134(0.62)(0.38)}\approx5.6$

Now, the number of seats should be in the​ non-smoking area in order to be very sure ​of having enough seating​ there :-

$\mu+3\sigma=83.08+3(5.6)=99.88\approx100$