Question

Consider the probability distributions shown here:

1. Use your intuition to find the mean for each distribution. How did you arrive at your choice?
2. Which distribution appears to be more variable? Why?
3. Calculate${\mathbf{\text{μ\hspace{0.17em}and\hspace{0.17em}σ}}}^{\mathbf{\text{2}}}$ for each distribution. Compare these answers with your answers in parts a and b.

Short answer

1. 0.33 for both
2. 0.33 for both
3. $\begin{array}{c}{\mathbf{\text{μ}}}_{\mathbf{\text{X}}}\mathbf{\text{=1}}\\ {\mathbf{\text{μ}}}_{\mathbf{\text{y}}}\mathbf{\text{=1}}\\ {\mathbf{\text{σ}}}_{\mathbf{\text{X}}}^{\mathbf{\text{2}}}\mathbf{\text{=0.9}}\\ {\mathbf{\text{σ}}}_{\mathbf{\text{y}}}^{\mathbf{\text{2}}}\mathbf{\text{=0.2}}\end{array}$
   

Step-by-step solution

(a)  Definition of a probability distribution

A probability distribution indicates the likelihood of a random variable to take a particular value within a given range.In a probability distribution, the percentages of probabilities are mainly represented as decimals.

Elucidation on the mean

With the help of intuition, the mean can be found by finding the average of the probabilities as shown below:

$$\begin{gathered} \mathbf{\text{Mean\hspace{0.17em}for\hspace{0.17em}x=}}\frac{\mathbf{\text{0.3+0.4+0.3}}}{\mathbf{\text{3}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.33}} \\ \mathbf{\text{Mean\hspace{0.17em}for\hspace{0.17em}}}\mathbf{y}\mathbf{\text{=}}\frac{\mathbf{\text{0.1+0.8+0.1}}}{\mathbf{\text{3}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.33}} \end{gathered}$$

Therefore, the mean of the distributions is 0.33 for both.

(b) Definition of variance

Finding a variance determines the variation in the values of a dataset around the mean of the values. The researcher often uses it to determine the consistency of a particular dataset.

Elucidation of the variance

The second distribution appears to be more variable compared to the first one. It is because

$\begin{array}{c}\mathbf{\text{Variance\hspace{0.17em}of\hspace{0.17em}x\hspace{0.17em}distribution=}}{\mathbf{(}\mathbf{\text{0.3}}\mathbf{-}\mathbf{\text{0.33}}\mathbf{)}}^{\mathbf{\text{2}}}\mathbf{\text{+}}{\mathbf{(}\mathbf{\text{0.4}}\mathbf{-}\mathbf{\text{0.33}}\mathbf{)}}^{\mathbf{\text{2}}}\mathbf{\text{+}}{\mathbf{(}\mathbf{\text{0.3}}\mathbf{-}\mathbf{\text{0.33}}\mathbf{)}}^{\mathbf{\text{2}}}\\ \mathbf{\text{=0.0067}}\\ \mathbf{\text{Variance\hspace{0.17em}of\hspace{0.17em}}}\mathbf{y}\mathbf{\text{\hspace{0.17em}distribution=}}{\mathbf{(}\mathbf{\text{0.1}}\mathbf{-}\mathbf{\text{0.33}}\mathbf{)}}^{\mathbf{\text{2}}}\mathbf{\text{+}}{\mathbf{(}\mathbf{\text{0.8}}\mathbf{-}\mathbf{\text{0.33}}\mathbf{)}}^{\mathbf{\text{2}}}\mathbf{\text{+}}{\mathbf{(}\mathbf{\text{0.1}}\mathbf{-}\mathbf{\text{0.33}}\mathbf{)}}^{\mathbf{\text{2}}}\\ \mathbf{\text{=0.3267}}\end{array}$

From the variances in the probabilities, it can be found that the y distribution has a greater variance.

(c) Elucidation of the mean

The means of x and y distributions are shown below:

$\begin{array}{c}\mathbf{}\mathbf{}\mathbf{}{\mathbf{\mu }}_{\mathbf{X}}\mathbf{\text{=0}}\mathbf{\times }\mathbf{\text{0.3+1}}\mathbf{\times }\mathbf{\text{0.4+2}}\mathbf{\times }\mathbf{\text{0.3}}\\ \mathbf{\text{=1}}\\ {\mathbf{\mu }}_{\mathbf{y}}\mathbf{\text{=0}}\mathbf{\times }\mathbf{\text{0.1+1}}\mathbf{\times }\mathbf{\text{0.8+2}}\mathbf{\times }\mathbf{\text{0.1}}\\ \mathbf{\text{=1}}\end{array}$

In comparing part a, it can be found that in part c, the means of y and x distribution are equal, as observed in part a.

Elucidation of the variance

The variances of x and y distributions are shown below:

$$\begin{gathered} \mathbf{Variance}\mathbf{of}\mathbf{x}\mathbf{distribution}\mathbf{=}{(\mathbf{0}-\mathbf{1})}^2\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{3}\mathbf{+}{(\mathbf{1}-\mathbf{1})}^2\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{4}\mathbf{+}{(\mathbf{2}-\mathbf{1})}^2\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{3} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{9} \\ \mathbf{Variance}\mathbf{of}\mathbf{y}\mathbf{distribution}\mathbf{=}{\mathbf{(}\mathbf{0}\mathbf{-}\mathbf{1}\mathbf{)}}^2\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{+}{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{1}\mathbf{)}}^2\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{+}{\mathbf{(}\mathbf{2}\mathbf{-}\mathbf{1}\mathbf{)}}^2\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{1} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{2} \end{gathered}$$

In comparing to part b, it can be found that in part b, the variance of y distribution appeared to be greater than x, but in part c, it is the opposite.