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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:

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

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$\mathit{渭}$, and the respective intervals are determined by$\mathit{渭}\mathbf{-}\mathbf{2}\mathit{蟽}$and$\mathit{渭}\mathbf{+}\mathbf{2}\mathit{蟽}$.

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

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

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