Chapter 4.8, Problem 4.83E

a. $p\left(3\right)$

To determine

To find:theprobability as in the given table.

Answer to Problem 4.83E

  $p\left(3\right)=0.2$.

Explanation of Solution

Given:

A random variable can assume five values. i.e. $0,1,2,3,4,5$ .

A portion of the probability distribution is shown below:

  

Calculation:

Since, ${\displaystyle \sum _{x=0}^{4}p\left(x\right)}=1$ , then

  $\begin{array}{l}0.1+0.3+0.3+p\left(3\right)+0.1=1\\ p\left(3\right)+0.8=1\\ p\left(3\right)=0.2\end{array}$

Hence, the value of $p\left(3\right)=0.2$ .

b.

To determine

To calculate: the mean, variance and standard deviation.

Answer to Problem 4.83E

  $0.18$

Explanation of Solution

Given:

A random variable can assume five values. i.e. $0,1,2,3,4,5$ .

A portion of the probability distribution is shown below:

  

Calculation:

Following is the probability histogram to describe $p\left(x\right)$ .

c.

To determine

To calculate: the mean, variance and standard deviation.

Answer to Problem 4.83E

  $\mu =1.9;{\sigma }^2=1.29;\sigma =1.136$

Explanation of Solution

Given:

A random variable can assume five values. i.e. $0,1,2,3,4,5$ .

A portion of the probability distribution is shown below:

  

Calculation:

According to question,

  $\begin{array}{l}\text{Mean, }\mu ={\displaystyle \sum _{x=0}^{4}p\left(x\right)}x\\ =0.1\left(0\right)+0.3\left(1\right)+0.3\left(2\right)+0.2\left(3\right)+0.1\left(4\right)\\ =0+0.3+0.6+0.6+0.4\\ =1.9\end{array}$

  $\begin{array}{l}\text{Variance, }{\sigma }^2={\displaystyle \sum _{x=0}^{5}p\left(x\right)}{x}^2-{\mu }^2\\ =0.1{\left(0\right)}^2+0.3{\left(1\right)}^2+0.3{\left(2\right)}^2+0.2{\left(3\right)}^2+0.1{\left(4\right)}^2\\ =1.29\end{array}$

  $\begin{array}{l}\text{Standard deviation, }\sigma =\sqrt{{\sigma }^2}\\ \sigma =\sqrt{1.29}\\ \sigma =1.136\end{array}$

Hence, the mean, variance and standard deviationis $\mu =1.9;{\sigma }^2=1.29;\sigma =1.136$ .

d.

To determine

To find: the probability that $x$ is greater than 2.

Answer to Problem 4.83E

  $P\left(x>2\right)=0.3$

Explanation of Solution

Given:

A random variable can assume five values. i.e. $0,1,2,3,4,5$ .

A portion of the probability distribution is shown below:

  

Calculation:

According to question,

  $\begin{array}{l}P\left(x>2\right)=p\left(3\right)+p\left(4\right)\\ P\left(x>2\right)=0.2+0.1\\ P\left(x>2\right)=0.3\end{array}$

Hence, theprobability that $x$ is greater than 2is $P\left(x>2\right)=0.3$ .

d.

To determine

To find: the probability that $x$ is less than 3.

Answer to Problem 4.83E

  $P\left(x\le 3\right)=0.9$

Explanation of Solution

Given:

A random variable can assume five values. i.e. $0,1,2,3,4,5$ .

A portion of the probability distribution is shown below:

  

Calculation:

According to question,

  $\begin{array}{l}P\left(x\le 3\right)=p\left(0\right)+p\left(1\right)+p\left(2\right)+p\left(3\right)\\ P\left(x\le 3\right)=0.1+0.3+0.3+0.2\\ P\left(x\le 3\right)=0.9\end{array}$

Hence, the probability that $x$ is less than 3is $P\left(x\le 3\right)=0.9$ .