Q. 8.12 - A First Course in Probability

Question

A clinic is equally likely to have 2, 3, or 4 doctors volunteer for service on a given day. No matter how many volunteer doctors there are on a given day, the numbers of patients seen by these doctors are independent Poisson random variables with a mean of $30$. Let X denote the number of patients seen in the clinic on a given day.

(a) Find $E\left[X\right]$

(b) Find Var$\left(X\right)$

(c) Use a table of the standard normal probability distribution to approximate P$P\{X>65\}$.

Step-by-Step Solution

Answer

a) The value of $E\left[X\right]=90$

b) The value of $Var\left(X\right)=690$

c) The approximate value of P using the standard normal probability distribution is $.7447.$

Part (a) Step 1: Given information

The number of patients seen by these doctors is independent of Poisson random variables with a mean of $30$

Let $X$ be the number of patients

Find the value of $E\left\{X\right\},Var\left[X\right]$ and the approximate value of P.

Part (a) Step 2: Explanation

Let's see the next three events

$D_2=\text{'}\text{'}$there are 2 volunteer doctors on a given day$\text{'}\text{'}$,

$D_3=\text{'}\text{'}$there are 3 volunteer doctors on a given day$\text{'}\text{'}$,

$D_4=\text{'}\text{'}$there are 4volunteer doctors on a given day$\text{'}\text{'}$,

Since there are 2,3 or 4 doctors, we have that

$p=P\left\{D_2\right\}=P\left\{D_3\right\}=P\left\{D_4\right\}=\frac{1}{3}$

The number of doctors in the clinic on any particular day is denoted by N. This is clearly a discrete random variable that accepts values 2,3 or 4.

$P\{N=2\}=P\{N=3\}=P\{N=4\}=p$

Then the mean is

$$\begin{gathered} E\left[N\right]=2\left(\frac{1}{3}\right)+3\left(\frac{1}{3}\right)+4\left(\frac{1}{3}\right)=3 \\ E\left[N^2\right]=2^2\left(\frac{1}{3}\right)+3^2\left(\frac{1}{3}\right)+4^2\left(\frac{1}{3}\right)=\frac{29}{3} \end{gathered}$$

The variance is

$Var\left(N\right)=\frac{29}{3}-3^2=\frac{2}{3}$

Part (a) Step 3: Calculation

Let X stand for the number of patients seen in the clinic on any given day. The numbers of patients seen by these doctors are independent Poisson random variables with a mean of $30$ no matter how many volunteer doctors there are on any particular day. Hence,

$X\mid N=n,n=2,3,4$

$\lambda =30n$

$E[X\mid N=n]=Var(X\mid N=n)=30n$

$E[X\mid N]=Var(X\mid N)=30N$

By seeing proposition $5.1$

$E\left[X\right]=E\left[E\right[X\mid N\left]\right]=E\left[30N\right]=30E\left[N\right]=30\left(3\right)=\mathbf{90}.$

Part (b) Step 1: Given information

The number of patients seen by these doctors is independent of Poisson random variables with a mean of $30$

Let X be the number of patients

Find the value of $E\left\{X\right\},Var\left[X\right]$ and the approximate value of P.

Part (b) Step 2: Explanation

By using the proposition $5.2$

$$\begin{gathered} Var\left(X\right)=E[Var(X\mid N\left)\right]+Var\left(E\right[X\mid N\left]\right)=E\left[30N\right]+Var\left(30N\right) \\ =30E\left[N\right]+{30}^{2}Var\left(N\right)=\mathbf{690}\mathbf{.} \end{gathered}$$

Part (c) Step 1: Given information

The number of patients seen by these doctors is independent of Poisson random variables with a mean of $30$

Let X be the number of patients

Find the value of $E\left\{X\right\},Var\left[X\right]$ and the approximate value of P.

Part (c) Step 2: Explanation

Already we know, $X\mid N=n,n=2,3,4$ is a Poisson random variable having $\lambda =30n$

$$\begin{gathered} *P\{X>65\mid N=2{\}}^{\text{the continuity correction }}\stackrel{=}{=}\{X>65.5\mid N=2\}= \\ P\left\{\frac{(X\mid N=2)-60}{\sqrt{60}}>\frac{65.5-60}{\sqrt{60}}\right\}=1-\Phi (.71)=1-.7611=.2389 \\ *P\{X>65\mid N=3{\}}^{\text{the continuity correction }}=P\{X>65.5\mid N=3\}= \\ P\left\{\frac{(X\mid N-3)-90}{\sqrt{90}}>\frac{65.5-90}{\sqrt{90}}\right\}=1-\Phi (-2.58)=\Phi (2.58)=.9951 \\ \left.*P\{X>65\mid N=3{\}}^{\text{the continuity correction }}\stackrel{=}{=}PX>65.5\mid N=3\right\}= \\ P\left\{\frac{(X\mid N=4)-120}{\sqrt{120}}>\frac{65.5-120}{\sqrt{120}}\right\}=1-\Phi (-4.98)=\Phi (4.98)\approx 1. \end{gathered}$$

Part (c) Step 3: Calculation

The wanted probability is 

$$\begin{gathered} P\{X>65\}=\sum _{n=2}^4 P\{X>65\mid N=n\}P\{N=n\} \\ =\frac{1}{3}(.2389+.9951+1)=.7447 \end{gathered}$$