Consider the probability distribution shown here

1. Calculate ${\mathbf{\text{μ,σ}}}^{\mathbf{\text{2}}}\mathbf{\text{andσ}}$.
2. Graph $\mathbf{\text{p}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$. Locate$\mathbf{\text{μ,μ}}\mathbf{-}\mathbf{\text{2σandμ+2σ}}$ on the graph.
3. What is the probability that $\mathbf{x}$ is in the interval $\mathbf{\text{μ+2σ}}$ ?

The calculation $\mathbf{\mu }$ is shown below:

$$\begin{gathered} \mathbf{\text{μ=Summation\hspace{0.17em}of\hspace{0.17em}}}\mathbf{[}\mathbf{\text{Value\hspace{0.17em}of\hspace{0.17em}x}}\mathbf{\times }\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\mathbf{(}\mathbf{-}\mathbf{\text{4}}\mathbf{)}\mathbf{\text{0.2+}}\mathbf{(}\mathbf{-}\mathbf{\text{3}}\mathbf{)}\mathbf{\text{0.7+}}\mathbf{(}\mathbf{-}\mathbf{\text{2}}\mathbf{)}\mathbf{\text{0.1+}}\mathbf{(}\mathbf{-}\mathbf{\text{1}}\mathbf{)}\mathbf{\text{0.15+}}\mathbf{\left(}\mathbf{\text{0}}\mathbf{\right)}\mathbf{\text{0.3+}}\mathbf{\left(}\mathbf{\text{1}}\mathbf{\right)}\mathbf{\text{0.18+}}\mathbf{\left(}\mathbf{\text{2}}\mathbf{\right)}\mathbf{\text{0.1+}}\mathbf{\left(}\mathbf{\text{3}}\mathbf{\right)}\mathbf{\text{0.06+}}\mathbf{\left(}\mathbf{\text{4}}\mathbf{\right)}\mathbf{\text{0.02}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\mathbf{-}\mathbf{\text{2.61}} \end{gathered}$$

The calculation ${\mathbf{\text{σ}}}^{\mathbf{\text{2}}}\mathbf{\text{andσ}}$is shown below:

$$\begin{gathered} {\mathbf{\text{σ}}}^{\mathbf{\text{2}}}\mathbf{\text{=Summation\hspace{0.17em}of\hspace{0.17em}}}\mathbf{[}\mathbf{(}\mathbf{\text{Value\hspace{0.17em}of\hspace{0.17em}x}}\mathbf{-}\mathbf{\text{μ}}\mathbf{)}\mathbf{\times }\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=}}\mathbf{(}\mathbf{-}\mathbf{\text{4}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.2+}}\mathbf{(}\mathbf{-}\mathbf{\text{3}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.7+}}\mathbf{(}\mathbf{-}\mathbf{\text{2}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.1+}}\mathbf{(}\mathbf{-}\mathbf{\text{1}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.15+}}\mathbf{(}\mathbf{\text{0}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.3+}}\mathbf{(}\mathbf{\text{1}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.18+}}\mathbf{(}\mathbf{\text{2}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.1+}}\mathbf{(}\mathbf{\text{3}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.06+}}\mathbf{(}\mathbf{\text{4}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{)}\mathbf{\text{0.02}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=10.1957}} \end{gathered}$$

$\begin{array}{c}\mathbf{\text{σ=}}\sqrt{\mathbf{\text{10.1957}}}\\ \mathbf{\text{=3.193071}}\end{array}$

Therefore the ${\mathbf{\text{μ,σ}}}^{\mathbf{\text{2}}}\mathbf{\text{andσ}}$ are -2.61, 10.1957 and 3.193071.

The calculation $\mathbf{\text{μ}}\mathbf{-}\mathbf{\text{2σandμ+2σ}}$ is shown below:

$\begin{array}{c}\mathbf{\text{μ}}\mathbf{-}\mathbf{\text{2σ=}}\mathbf{-}\mathbf{\text{2.61}}\mathbf{-}\mathbf{\text{2}}\mathbf{\times }\mathbf{\text{3.193071}}\\ \mathbf{\text{=}}\mathbf{-}\mathbf{\text{8.99614}}\\ \mathbf{\text{μ+2σ=}}\mathbf{-}\mathbf{\text{2.61+2}}\mathbf{\times }\mathbf{\text{3.193071}}\\ \mathbf{\text{=3.78}}\end{array}$

Therefore, the values $\mathbf{\text{μ,μ}}\mathbf{-}\mathbf{\text{2σandμ+2σ}}$ are -2.61, -8.99614 and 3.78, respectively.

In the graphical representation shown above, the respective probabilities are given, and in the horizontal axis, the respective values of x are given. Within the horizontal axis, the respective variables$\mathbf{\text{μ,μ}}\mathbf{-}\mathbf{\text{2σandμ+2σ}}$ are labelled.

Chebycheb’s rule assumes that there are two standard deviations in the model. The rule says the maximum number of the values or observations must lie within the given standard deviations.

As the value $\mathbf{\text{μ+2σ}}$is 3.78, the summation of the probabilities from -4 to 3 will be the answer as shown below:

$\begin{array}{c}\mathbf{\text{Summation\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}probabilities=0.2+0.7+0.10+0.15+.30+0.18+0.10+0.06}}\\ \mathbf{\text{=1.79}}\end{array}$

$\begin{array}{c}\mathbf{\text{Summation\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}probabilities=0.2+0.7+0.10+0.15+.30+0.18+0.10+0.06}}\\ \mathbf{\text{=1.79}}\end{array}$

Multiplying 1.79 by 100, it becomes 179%, which means 179% of the probability distribution lies within the given distributions.