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 [math]p[/math], and that the decisions are independent. What is [math]p[/math]?

• 0.04
• 0.055
• 0.063
• 0.07
• 0.075

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

Solution: C

Let [math]n[/math] be the number of members. We know that the number of members that show up, say [math]N[/math], is approximately normally distributed with mean [math]np[/math] and variance [math]np(1-p)[/math]. From the exercise, we know that

[[math]] P( N \geq 108 ) \approx 0.025, \, P( N \leq 72) \approx 0.025 [[/math]]

or

[[math]] P(Z \geq \frac{108-np}{\sqrt{np(1-p)}}) \approx 0.025, \, P(Z \leq \frac{72-np}{\sqrt{np(1-p)}}) \approx 0.025 [[/math]]

with [math]Z [/math] a standard normal variable. This gives:

[[math]] \frac{108-\mu}{\sigma} \approx 1.96, \frac{72-\mu}{\sigma} \approx -1.96 [[/math]]

with [math]\mu = np [/math] and [math]\sigma = \sqrt{np(1-p)}[/math].

We have

[[math]] \frac{108-\mu}{72-\mu} = -1 \implies \mu = np = 90. [[/math]]

And then we have

[[math]] \frac{18}{\sigma} = 1.96 \implies \sigma = \sqrt{np(1-p)} = \sqrt{90(1-p)} = 9.1837 [[/math]]

This immediately gives [math]p = 0.06289 [/math].