4.4 Mean and Standard Deviation of a Discrete Variable

Given the probability distribution of a discrete random variable X, we are able to calculate the mean, variance, and standard deviation.

The mean of a discrete variable X can be calculated as

[latex]\mu = \sum xP(X=x),[/latex]

which is a weighted average over all possible values of X. Each possible value x is weighted by its probability [latex]P(X=x)[/latex]. The mean is also called the expected value or the expectation of X. Note that a probability distribution can be viewed as the relative frequency distribution of some population. In this regard, the mean of a discrete random variable is equivalent to the population mean.

Some students might find it easier to find the mean by constructing a working table (see the following example).

Even though the random variable X = # of siblings can only take integer values, we should keep the decimal place for the mean of X. That is, do not round the mean 1.6 to 2. To demonstrate why we keep the mean at 1.6, let us suppose that this probability distribution describes a much larger population of students. Although it is counterintuitive to say that we expect a student to have 1.6 siblings, it is quite natural to say that we expect 10 students to have a total of 16 siblings, 100 students to have a total of 160 siblings, and so on. If we sample the entire population of students, then the combined number of siblings is 1.6 times greater than the number of students. Hence, the average number of siblings per student is 1.6.

Here we explain why the population mean is given by [latex]\mu = \sum xP(X=x)[/latex]. Suppose there are [latex]N=5[/latex] students, one has no siblings [latex](x_1 = 0)[/latex], one has one sibling [latex](x_2 = 1)[/latex], two have two siblings [latex](x_3 = x_4 =2)[/latex], and one has three siblings [latex](x_5 =3)[/latex]. Recall that the population mean [latex]\mu[/latex] is calculated as:

[latex]\begin{align*}\mu&=\frac{\sum x_i}{N}=\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{0+1+2+2+3}{5}=\frac{{\color{blue} 0}\times {\color{red} 1}+{\color{blue} 1}\times {\color{red} 1}+{\color{blue} 2}\times{\color{red} 2}+{\color{blue} 3}\times {\color{red} 1}}{5}\\&={\color{blue} 0} \times {\color{red} \frac{1}{5}}+{\color{blue} 1}\times {\color{red} \frac{1}{5}}+{\color{blue} 2}\times {\color{red} \frac{2}{5}}+{\color{blue} 3}\times {\color{red} \frac{1}{5}}\\&={\color{blue} 0}\times {\color{red} P(X=0)}+{\color{blue} 1}\times {\color{ red} P(X=1)}+{\color{blue} 2}\times{\color{red} P(X=2)}+{\color{blue} 3}\times{\color{red} P(X=3)}\\&=\sum {\color{blue} x} {\color{red} P(X=x)}.\end{align*}[/latex]

Similarly, the variance of a discrete variable X can be calculated as

[latex]\sigma^2=\underbrace{\sum (x-\mu)^2 P(X=x)}\limits_{\mbox{defining formula}}.[/latex]

which is a weighted average of the squared distance from each value [latex]x[/latex] to the population mean [latex]\mu[/latex], weighted by its probability [latex]P(X=x)[/latex] (relative frequency). It can be shown that

[latex]\sigma^2=\sum (x-\mu)^2 P(X=x)=\underbrace{\sum x^2P(X=x)-\mu^2}\limits_{\mbox{computing formula}}.[/latex]

Taking the square root of the variance [latex]\sigma^2[/latex] gives the standard deviation [latex]\sigma[/latex]

[latex]\sigma=\sqrt{\sigma^2}=\underbrace{\sqrt{\sum (x-\mu)^2 P(X=x)}}\limits_{\mbox{defining formula}}=\underbrace{\sqrt{\sum x^2 P(X=x)-\mu^2}}\limits_{\mbox{computing formula}}.[/latex]