91Ó°ÊÓ

A discrete probability distribution refers to a probability distribution where the count is finite. The outcomes mainly do not fall anywhere in the distribution on a continuum as they are finite.

The calculation $\mathbf{μ}$for the given discrete probability distribution is shown below:

role="math" localid="1653629636380" $$\begin{gathered} \mathbf{\text{μ=·¡}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=E}}\mathbf{\left[}\mathbf{\text{xp}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{\right]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=1}}\mathbf{\left(}\mathbf{\text{0.1}}\mathbf{\right)}\mathbf{\text{+2}}\mathbf{\left(}\mathbf{\text{0.2}}\mathbf{\right)}\mathbf{\text{+3}}\mathbf{\left(}\mathbf{\text{0.2}}\mathbf{\right)}\mathbf{\text{+4}}\mathbf{\left(}\mathbf{\text{0.3}}\mathbf{\right)}\mathbf{\text{+5}}\mathbf{\left(}\mathbf{\text{0.2}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.1+0.4+0.6+1.2+1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{3} \end{gathered}$$

Discrete probability distributions include a finite number of observations. A typical example of this can be tossing a coin where everyone knows that there will be either of the two outcomes, head or tail.

The calculation $\mathbf{σ}\mathbf{=}\sqrt{\mathbf{E}\mathbf{\left[}{\mathbf{(}\mathbf{x}\mathbf{−}\mathbf{μ}\mathbf{)}}^{\mathbf{2}}\mathbf{\right]}}$is shown below:

role="math" localid="1653630450721" $$\begin{gathered} \mathbf{\text{σ=}}\sqrt{\mathbf{\text{E}}\mathbf{\left[}{\mathbf{(}\mathbf{\text{x}}\mathbf{−}\mathbf{\text{μ}}\mathbf{)}}^{\mathbf{\text{2}}}\mathbf{\right]}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\sqrt{\mathbf{\text{E}}\mathbf{\left[}{\mathbf{(}\mathbf{\text{x}}\mathbf{−}\mathbf{\text{μ}}\mathbf{)}}^{\mathbf{\text{2}}}\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\right]}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\sqrt{\mathbf{\left(}\mathbf{\text{1-3.3}}\mathbf{\right)}\mathbf{\text{×0.1+}}\mathbf{(}\mathbf{\text{2}}\mathbf{−}\mathbf{\text{3.3}}\mathbf{)}\mathbf{\text{×0.2+}}\mathbf{(}\mathbf{\text{3}}\mathbf{−}\mathbf{\text{3.3}}\mathbf{)}\mathbf{\text{×0.2+}}\mathbf{(}\mathbf{\text{4}}\mathbf{−}\mathbf{\text{3.3}}\mathbf{)}\mathbf{\text{×0.3+}}\mathbf{(}\mathbf{\text{5}}\mathbf{−}\mathbf{\text{3.3}}\mathbf{)}\mathbf{\text{×0.2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\sqrt{\mathbf{\text{1.61}}} \\ \mathbf{}\mathbf{}\mathbf{} \end{gathered}$$

The calculation of the interval is shown below:

$$\begin{gathered} \mathbf{\text{μ+2σ=3.3+2×1.27}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=5.84}} \\ \mathbf{\text{μ}}\mathbf{−}\mathbf{\text{2σ=3.3}}\mathbf{−}\mathbf{\text{2×1.27}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.76}} \end{gathered}$$

As 0.76 is less than 1 and 5.84 is greater than 5, the calculation of probability via empirical rule must consider all the associated probabilities as shown below:

$$\begin{gathered} \mathbf{\text{Probability=E}}\mathbf{\left[}\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\right]}\mathbf{\text{×100}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\mathbf{\left[}\mathbf{\left(}\mathbf{\text{0.1}}\mathbf{\right)}\mathbf{\text{+}}\mathbf{\left(}\mathbf{\text{0.2}}\mathbf{\right)}\mathbf{\text{+}}\mathbf{\left(}\mathbf{\text{0.2}}\mathbf{\right)}\mathbf{\text{+}}\mathbf{\left(}\mathbf{\text{0.3}}\mathbf{\right)}\mathbf{\text{+}}\mathbf{\left(}\mathbf{\text{0.2}}\mathbf{\right)}\mathbf{\right]}\mathbf{\text{×100}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=1×100}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=100\%}} \end{gathered}$$

It shows that as it is 100 percent, all the values lie within the interval.