Arithmetical mean

The arithmetical mean is the average value of the samples. It is independent of the width of the intervals. It is symbolized as $\overline{x}$ and it is only used for quantitative variables. We find it by adding up all the values and dividing by the total number of data.

The general formula for $N$ elements is: $$\displaystyle \overline{x}=\frac{x_1+x_2+x_3+\ldots+x_n}{n}$$

In a basketball match, we have the following points for the players of a team: $$0, 2, 4, 5, 8, 8, 10, 15, 38$$ Calculate the mean of points of the team.

Applying the formula $$\displaystyle \overline{x}=\frac{0+2+4+5+8+9+10+15+38}{9}=\frac{90}{9}=10$$

Calculation of the mean for grouped information

The average in the case of $N$ data grouped in $n$ intervals is given by the formula $$\displaystyle \overline{x}=\frac{x_1\cdot f_1+x_2\cdot f_2+x_3\cdot f_3+\ldots+x_n\cdot f_n}{f_1+f_2+f_3+\ldots+f_n}$$

where $f_i$ represents the times that the value $x_i$ is repeated. The grouping can also be done by intervals, using then the intermediate value of the interval to calculate the mean.

The height in $cm$ of the players of a basketball team is in the following table. Calculate the mean.

Interval	$x_i$	$f_i$	$x_i\cdot f_i$
$[160,170)$	$165$	$1$	$165$
$[170,180)$	$175$	$2$	$350$
$[180,190)$	$185$	$4$	$740$
$[190,200)$	$195$	$3$	$585$
$[200,210)$	$205$	$2$	$410$
		$12$	$2250$

We calculate the mean for grouped data: $$\displaystyle \overline{x}=\frac{165 \cdot 1+175 \cdot 2+185\cdot 4+195\cdot 3+205\cdot 2}{1+2+4+3+2}=$$ $$=\frac{2250}{12}=187.5$$

If there is an interval with a non determinated width it is not possible to calculate the mean:

$[160,170)$	$165$	$1$	$16$
$[170,180)$	$175$	$2$	$350$
$[180,190)$	$185$	$4$	$740$
$[190,200)$	$195$	$3$	$585$
$[200,)$	$2$
		$12$	$2250$

It is also important to mention that the arithmetical mean is very sensitive to extreme punctuations.

In a basketball match, we have the following points for the players of a team: $$0, 1, 3, 4, 5, 6, 7, 8, 47$$ Calculate the mean of points of the team.

$$\displaystyle \overline{x}=\frac{0+1+3+4+5+6+7+8+47}{9}=\frac{81}{9}=9$$

In this case the mean does not illustrate well the information, since all the values except one are below the mean.