Consider the following probability distribution:

$\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\text{=}}\left(\begin{array}{c}5\\ x\end{array}\right){\mathbf{\left(}\mathbf{\text{0.7}}\mathbf{\right)}}^{\mathbf{\text{x}}}{\mathbf{\left(}\mathbf{\text{0.3}}\mathbf{\right)}}^{\mathbf{\text{5 - x}}}\mathbf{(}\mathbf{\text{x = 0,1,2,}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\text{,5}}\mathbf{)}$

a. Is x a discrete or a continuous random variable?

b. What is the name of this probability distribution?

c. Graph the probability distribution.

d. Find the mean and standard deviation of x.

e. Show the mean and the 2-standard-deviation interval on each side of the mean on the graph you drew in part c.

A researcher can regard a random variable to be discrete if the random variable shows a finite number of values. In contrast to it, a random variable is regarded as continuous by a researcher whenever there are infinite values.

The values of x are found to be ranging between 0 and 5 which indicates that there are finite numbers.This indicates as it is not reflecting infinite values, the random variable is discrete and not continuous.

With the given formula,$\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\left(\begin{array}{c}5\\ x\end{array}\right){\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{x}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{5}\mathbf{-}\mathbf{x}}$, the probability is found when value is 0 as shown below:

$\begin{array}{rcl}\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{=}& \left(\begin{array}{c}5\\ x\end{array}\right){\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{x}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{5}\mathbf{-}\mathbf{x}}\\ \mathit{W}\mathit{h}\mathit{e}\mathit{n}\mathit{x}& \mathbf{=}& \mathbf{0}\mathbf{,}\\ \mathit{p}\mathbf{\left(}\mathbf{0}\mathbf{\right)}& \mathbf{=}& \left(\begin{array}{c}5\\ 0\end{array}\right){\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{0}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{5}\mathbf{-}\mathbf{0}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{0}\mathbf{!}\mathbf{(}\mathbf{5}\mathbf{-}\mathbf{0}\mathbf{)}\mathbf{!}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{0}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{5}\mathbf{-}\mathbf{0}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{1}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{!}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{0}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{5}\mathbf{-}\mathbf{0}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{00243}\\ & & \end{array}$

With the given formula, $\mathit{p}\left(x\right)\mathbf{=}\left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}$, the probability is found when value is 1 as shown below:

role="math" localid="1668486774624" $\begin{array}{rcl}\mathit{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{=}& \mathbf{\left(}\begin{array}{c}\mathbf{5}\\ \mathbf{x}\end{array}\mathbf{\right)}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{x}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{5}\mathbf{-}\mathbf{x}}\\ \mathit{W}\mathit{h}\mathit{e}\mathit{n}\mathit{x}& \mathbf{=}& \mathbf{1}\mathbf{,}\\ \mathit{p}\mathbf{\left(}\mathbf{1}\mathbf{\right)}& \mathbf{=}& \mathbf{\left(}\begin{array}{c}\mathbf{5}\\ \mathbf{1}\end{array}\mathbf{\right)}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{1}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{5}\mathbf{-}\mathbf{1}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{1}\mathbf{!}\mathbf{(}\mathbf{5}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{!}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{1}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{5}\mathbf{-}\mathbf{1}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{1}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{!}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{7}\mathbf{)}}^{\mathbf{1}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{)}}^{\mathbf{4}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{02835}\\ & & \end{array}$

With the given formula, $\mathit{p}\left(x\right)\mathbf{=}\left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}$, the probability is found when value is 2 as shown below:

$\begin{array}{rcl}\mathit{p}\left(x\right)& \mathbf{=}& \left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}\\ \mathit{W}\mathit{h}\mathit{e}\mathit{n}\mathit{x}& \mathbf{=}& \mathbf{2}\mathbf{,}\\ \mathit{p}\left(2\right)& \mathbf{=}& \left(\begin{array}{c}5\\ 2\end{array}\right){\left(0.7\right)}^{\mathbf{2}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{2}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{2}\mathbf{!}\left(5-2\right)\mathbf{!}}{\left(0.7\right)}^{\mathbf{2}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{2}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{2}\mathbf{!}\left(3\right)\mathbf{!}}{\left(0.7\right)}^{\mathbf{2}}{\left(0.3\right)}^{\mathbf{3}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{1323}\end{array}$

With the given formula, $\mathit{p}\left(x\right)\mathbf{=}\left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}$, the probability is found when value is 3 as shown below:

$\begin{array}{rcl}\mathit{p}\left(x\right)& \mathbf{=}& \left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}\\ \mathit{W}\mathit{h}\mathit{e}\mathit{n}\mathit{x}& \mathbf{=}& \mathbf{3}\mathbf{,}\\ \mathit{p}\left(3\right)& \mathbf{=}& \left(\begin{array}{c}5\\ 3\end{array}\right){\left(0.7\right)}^{\mathbf{3}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{3}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{3}\mathbf{!}\left(5-3\right)\mathbf{!}}{\left(0.7\right)}^{\mathbf{3}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{3}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{3}\mathbf{!}\left(2\right)\mathbf{!}}{\left(0.7\right)}^{\mathbf{3}}{\left(0.3\right)}^{\mathbf{2}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{3087}\\ & & \end{array}$

With the given formula,$\mathit{p}\left(x\right)\mathbf{=}\left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}$, the probability is found when value is 4 as shown below:

$\begin{array}{rcl}\mathit{p}\left(x\right)& \mathbf{=}& \left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}\\ \mathit{W}\mathit{h}\mathit{e}\mathit{n}\mathit{x}& \mathbf{=}& \mathbf{4}\mathbf{,}\\ \mathit{p}\left(4\right)& \mathbf{=}& \left(\begin{array}{c}5\\ 4\end{array}\right){\left(0.7\right)}^{\mathbf{4}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{4}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{4}\mathbf{!}\left(5-4\right)\mathbf{!}}{\left(0.7\right)}^{\mathbf{4}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{4}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{4}\mathbf{!}\left(1\right)\mathbf{!}}{\left(0.7\right)}^{\mathbf{4}}{\left(0.3\right)}^{\mathbf{1}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{36015}\\ & & \end{array}$

With the given formula,$\mathit{p}\left(x\right)\mathbf{=}\left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}$ , the probability is found when value is 5 as shown below:

$\begin{array}{rcl}\mathit{p}\left(x\right)& \mathbf{=}& \left(\begin{array}{c}5\\ x\end{array}\right){\left(0.7\right)}^{\mathbf{x}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{x}}\\ \mathit{W}\mathit{h}\mathit{e}\mathit{n}\mathit{x}& \mathbf{=}& \mathbf{5}\mathbf{,}\\ \mathit{p}\left(5\right)& \mathbf{=}& \left(\begin{array}{c}5\\ 5\end{array}\right){\left(0.7\right)}^{\mathbf{5}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{5}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{5}\mathbf{!}\left(5-5\right)\mathbf{!}}{\left(0.7\right)}^{\mathbf{5}}{\left(0.3\right)}^{\mathbf{5}\mathbf{-}\mathbf{5}}\\ & \mathbf{=}& \frac{\mathbf{5}\mathbf{!}}{\mathbf{5}\mathbf{!}\left(0\right)\mathbf{!}}{\left(0.7\right)}^{\mathbf{5}}{\left(0.3\right)}^{\mathbf{0}}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{16807}\\ & & \end{array}$

As the probabilities are not zero and range between 0 and 1, the probability distribution cannot be continuous. For it to be discrete the summation has to be 1 as shown below:

$\left(0.00243+0.02835+0.1323+0.3087+0.36015+0.16807\right)\mathbf{=}\mathbf{1}$

As the summation of $\mathbf{\text{0.00243,0.02835,0.1323,0.3087,0.36015 and 0.16807}}$is 1, the probability distribution can be deduced to be discrete and not continuous.

c. The list shown below shows the probabilities of the values of x ranging from 0 to 5 which have been calculated in part b of the question:

The graph shown above has two axes consisting of the probabilities and the values of x.The vertical axis consists of the probabilities ranging between 0 and 1 and the horizontal axis shows the values of x ranging from 0 to 5.

d.

The calculation of the mean of the discrete probability distribution is shown below:

$\begin{array}{rcl}\mathit{M}\mathit{e}\mathit{a}\mathit{n}& \mathbf{=}& \mathbf{\sum }_{\mathbf{i}\mathbf{=}\mathbf{1}}^{\mathbf{n}}\mathit{x}\mathit{p}\left(x\right)\\ & \mathbf{=}& \mathbf{0}\left(0.00243\right)\mathbf{+}\mathbf{1}\left(0.02835\right)\mathbf{+}\mathbf{2}\left(0.1323\right)\mathbf{+}\mathbf{3}\left(0.3087\right)\mathbf{+}\mathbf{4}\left(0.36015\right)\mathbf{+}\mathbf{5}\left(0.16807\right)\\ & \mathbf{=}& \mathbf{0}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{02835}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{02835}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{2646}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{9261}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{4406}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{84035}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{5}\\ & & \end{array}$

The mean of the discrete probability distribution is 3.5.

The calculation of the standard deviation of the discrete probability distribution is shown below:

$$\begin{gathered} \begin{array}{rcl}\mathbf{\text{Standard deviation}}& \mathbf{=}& \sqrt{{\mathbf{(}\mathbf{0}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{00243}\mathbf{)}\mathbf{+}{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{02835}\mathbf{)}\mathbf{+}{\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1323}\mathbf{)} \\ \mathbf{+}{\mathbf{(}\mathbf{3}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{3087}\mathbf{)}\mathbf{+}{\mathbf{(}\mathbf{4}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{36015}\mathbf{)}\mathbf{+}{\mathbf{(}\mathbf{5}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{5}\mathbf{)}}^{\mathbf{2}}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{16807}\mathbf{)}}\\ & \mathbf{=}& \sqrt{(0.029768+0.177188+0.297675+0.077175+0.090038+0.378158)}\\ & \mathbf{=}& \sqrt{\mathbf{1}\mathbf{.}\mathbf{05}}\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{024695}\\ & & \end{array} \end{gathered}$$

The standard deviation of the discrete probability distribution is 1.024695.

The mean is determined by$\mathit{\mu }$, and the respective intervals are determined by$\mathit{\mu }\mathbf{-}\mathbf{2}\mathit{\sigma }$and$\mathit{\mu }\mathbf{+}\mathbf{2}\mathit{\sigma }$.

$\begin{array}{rcl}\mathit{\mu }\mathbf{+}\mathbf{2}\mathit{\sigma }& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{2}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{024695}\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{5}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{04939}\\ & \mathbf{=}& \mathbf{5}\mathbf{.}\mathbf{54939}\\ \mathit{\mu }\mathbf{-}\mathbf{2}\mathit{\sigma }& \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{5}\mathbf{-}\left(2\times 1.024695\right)\\ & \mathbf{=}& \mathbf{3}\mathbf{.}\mathbf{5}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{04939}\\ & \mathbf{=}& \mathbf{1}\mathbf{.}\mathbf{45061}\\ & & \end{array}$

The mean is 3.5, and the intervals are 1.45061 and 5.54939.

In the diagram, as 5.54939 is greater than 5,$\mathit{\mu }\mathbf{+}\mathbf{2}\mathit{\sigma }$ is situated towards the right of 5, and as 1.45061 lies between 1 and 2,$\mathit{\mu }\mathbf{-}\mathbf{2}\mathit{\sigma }$ is situated between 1 and 2. As 3.5 lies between 3 and 4,$\mathit{\mu }$ is situated between 3 and 4.