A small general practitioner clinic in rural Queensland has 7 appointment slots per day. On...

Question:

A small general practitioner clinic in rural Queensland has {eq}7 {/eq} appointment slots per day. On average, {eq}4 {/eq} patients require a {eq}GP {/eq} appointment per day. For the purposes of this question, assume that {eq}GP {/eq} appointments only occur in this clinic, and that each appointment is independent.

a) What is the probability that the clinic does not have enough appointment slots on a given day?

Poisson Distribution:

Poisson distribution is used to find out the probability of rare events. When n takes very large values and p takes small values then the binomial distribution is approximated by Poisson distribution with its parameter lambda given by:{eq}\lambda = np {/eq}

Given Information:

A small general practitioner clinic in rural Queensland has 7 appointment slots per day.

The average number of patients that requires a general practitioner appointment per day is 4.

Let X denotes the number of patients that require a general practitioner (GP) appointment per day.

{eq}\begin{align*} X &\sim Poisson\left( 4 \right)\\ P\left( {X = x} \right) &= \dfrac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}\\ n &= 7\\ \lambda &= 4 \end{align*} {/eq}

The probability that the clinic does not have enough appointment slots on a given day is possible only if all 7 slots are filled and no more slots is available and is given by:

{eq}\begin{align*} P\left( {X > 7} \right) &= 1 - P\left( {X \le 7} \right)\\ &= 1 - \sum\limits_{x = 0}^7 {\dfrac{{{e^{ - 4}}{4^x}}}{{x!}}} \\ &= 1 - 0.9489\\ &= 0.051133616 \end{align*} {/eq}

The probability that the clinic does not have enough appointment slots on a given day is 0.051133616.