library(visualize)

Exercise 9.1.8

A club serves dinner to members only. They are seated at 12-seat tables. The manager observes over a long period of time that 95 percent of the time there are between six and nine full tables of members, and the remainder of the time the numbers are equally likely to fall above or below this range. Assume that each member decides to come with a given probability p, and that the decisions are independent. How many members are there? What is p?

Ans:

Let the total number of members to the club be n,

Knowing that 95 percent of the time there are between six and nine full tables of members and the remainder of the time the numbers are equally likely to fall above or below this range, we can calculate the mean, \(\mu\) , is actually the midpoint between 72 and 108, which is 90. As it’s a Bernoulli trial process, we know it’s \(\mu = np\), Thus we know \(p = \frac{90}{n}\)

p - 2\(\sigma\) = p - \(2\sqrt{\frac{pq}{n}} = \frac{72}{n}\)

\[ \begin{multline*} \begin{split} \frac{90}{n} - 2\sqrt{\frac{\frac{90}{n}\times\frac{n-90}{n}}{n}} &= \frac{72}{n} \\ \sqrt{\frac{\frac{90}{n}\times\frac{n-90}{n}}{n}} &= \frac{9}{n} \\ 90n^2 - 90^2n & = 81 \\ n & = 900 \\ \end{split} \end{multline*} \]

\(\therefore p = \frac{90}{900} = \frac{1}{10}\)

Appendix

As we know there are 2 standard deviations to the right of mean, \(\mu\), and 2 standard deviations to the left of \(\mu\), we can visualize the distribution of members being served dinner in the club as follows,

Distribution A - B is what we are targeting

# Distribution (A)
visualize.it(dist = 'binom', stat = 108,params = list(n=900,p=.1)) 

# Distribution (B)
visualize.it(dist = 'binom', stat = 72,params = list(n=900,p=.1))