Chapter 5.3, Problem 5.40E

(a)

To determine

The value of the probability $P\left(x\ge 5\right)$ .

Answer to Problem 5.40E

  $P\left(x\ge 5\right)=\left(10.853\right)e^{-2.5}$

Explanation of Solution

Given, $\mu =2.5$ and ‘X’ is a Poisson random variable

Now, distribution of ‘X’ will be as follows

  $P\left(x=k\right)=\frac{{\mu }^k{e}^{-\mu }}{k!}$  $\begin{array}{l}k=0,1,2,\mathrm{..........}\\ \mu >0\end{array}$

We are asked to find the value of $P\left(x\ge 5\right)$ .

  $\begin{array}{l}P\left(x\ge 5\right)=1-P\left(x<5\right)\\ P\left(x\ge 5\right)=1-\left\{P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)+P\left(x=4\right)\right\}\\ P\left(x\ge 5\right)=1-\left\{\frac{e^{-2.5}.{\left(2.5\right)}^0}{0!}+\frac{e^{-2.5}.{\left(2.5\right)}^1}{1!}+\frac{e^{-2.5}.{\left(2.5\right)}^2}{2!}+\frac{e^{-2.5}.{\left(2.5\right)}^3}{3!}+\frac{e^{-2.5}.{\left(2.5\right)}^4}{4!}\right\}\\ P\left(x\ge 5\right)=e^{-2.5}\left\{1+2.5+3.125+2.601+1.627\right\}\\ P\left(x\ge 5\right)=\left(10.853\right)e^{-2.5}\end{array}$

(b)

To determine

The value of the probability $P\left(x<6\right)$ under given conditions.

Answer to Problem 5.40E

  $P\left(x\ge 5\right)=\left(11.666\right)e^{-2.5}$

Explanation of Solution

Given, $\mu =2.5$ and ‘X’ is a Poisson random variable

Now, distribution of ‘X’ will be as follows

  $P\left(x=k\right)=\frac{{\mu }^k{e}^{-\mu }}{k!}$  $\begin{array}{l}k=0,1,2,\mathrm{..........}\\ \mu >0\end{array}$

We are asked to find the value of $P\left(x<6\right)$ .

  $\begin{array}{l}P(x<6)=1-P\left(x<6\right)\\ P\left(x\ge 5\right)=1-P\left(x<5\right)\\ P\left(x\ge 5\right)=1-\left\{P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)+P\left(x=4\right)+P\left(x=5\right)\right\}\\ P\left(x\ge 5\right)=1-\left\{\frac{e^{-2.5}.{\left(2.5\right)}^0}{0!}+\frac{e^{-2.5}.{\left(2.5\right)}^1}{1!}+\frac{e^{-2.5}.{\left(2.5\right)}^2}{2!}+\frac{e^{-2.5}.{\left(2.5\right)}^3}{3!}+\frac{e^{-2.5}.{\left(2.5\right)}^4}{4!}+\frac{e^{-2.5}.{\left(2.5\right)}^5}{5!}\right\}\\ P\left(x\ge 5\right)=e^{-2.5}\left\{1+2.5+3.125+2.601+1.627+0.813\right\}\\ P\left(x\ge 5\right)=\left(11.666\right)e^{-2.5}\end{array}$

(c)

To determine

The value of the probability $P\left(x=2\right)$ for the given conditions.

Answer to Problem 5.40E

  $P\left(x=2\right)=3.125e^{-2.5}$

Explanation of Solution

Given, $\mu =2.5$ and ‘X’ is a Poisson random variable

Now, distribution of ‘X’ will be as follows

  $P\left(x=k\right)=\frac{{\mu }^k{e}^{-\mu }}{k!}$  $\begin{array}{l}k=0,1,2,\mathrm{..........}\\ \mu >0\end{array}$

We are asked to find the value of $P\left(x=2\right)$ .Putting $x=2$ and $\mu =2.5$ in above equation .

  $\begin{array}{l}P\left(x=2\right)=\frac{e^{-2.5}.{\left(2.5\right)}^2}{2!}\\ P\left(x=2\right)=3.125e^{-2.5}\end{array}$

(d)

To determine

The value of the probability $P\left(1\le x\le 4\right)$ for the given conditions.

Answer to Problem 5.40E

  $P\left(1\le x\le 4\right)=\left(9.853\right)e^{-2.5}$

Explanation of Solution

Given, $\mu =2.5$ and ‘X’ is a Poisson random variable

Now, distribution of ‘X’ will be as follows

  $P\left(x=k\right)=\frac{{\mu }^k{e}^{-\mu }}{k!}$  $\begin{array}{l}k=0,1,2,\mathrm{..........}\\ \mu >0\end{array}$

We are asked to find the value of $P\left(1\le x\le 4\right)$ .

  $\begin{array}{l}P\left(1\le x\le 4\right)=P\left(x=1\right)+P\left(x=2\right)+P\left(x=3\right)+P\left(x=4\right)\\ P\left(1\le x\le 4\right)=\frac{e^{-2.5}.{\left(2.5\right)}^1}{1!}+\frac{e^{-2.5}.{\left(2.5\right)}^2}{2!}+\frac{e^{-2.5}.{\left(2.5\right)}^3}{3!}+\frac{e^{-2.5}.{\left(2.5\right)}^4}{4!}\\ P\left(1\le x\le 4\right)=e^{-2.5}\left\{2.5+3.125+2.601+1.627\right\}\\ P\left(1\le x\le 4\right)=\left(9.853\right)e^{-2.5}\end{array}$