Absolute deviation
The absolute deviation gives information about how close or far a datum is form the other data of the set. Intuitively, we can see one can calculate it as the difference of a value and the average of the data: $$$D_i=x_i-\overline{x}$$$
We can see that to calculate this deviation, if we know the average, we only need the value which deviation is to be calculated.
Also, we can say that if we have a datum and its absolute deviation, it is possible to isolate the average by applying a simple subtraction:$$$\overline{x}=x_i-D_i$$$and later we can use it to calculate other deviations.
In a mathematics exam Pedro got $$9$$, while the average of the class is $$6.7$$. Calculate the absolute deviation of Pedro's mark.
Applying the formula $$$D_i=x_i-\overline{x}=9-6.7=2.3$$$
The sign of the absolute deviation shows whether or not the value is above average (positive sign), or below average (negative sign).
The absolute value of the absolute deviation shows how far the value is from the average. A value equal to zero shows that the value is the same as the average, while a high value with regard to other deviations shows that the value is far from the other data.
In a basketball match, we have the following points of the players of a team: $$0, 2, 4, 5, 8, 10, 10, 15, 38$$. Calculate the absolute deviation of the scoring of the players of the team.
Applying the formula$$$\displaystyle \overline{x}=\frac{0+2+4+5+8+10+10+15+38}{9}=\frac{92}{9}=10.22$$$the average is obtained. The deviations can be represented in a table:
| Punctuation | $$D_i=x_i-\overline{x}=x_i-10.22$$ |
| $$0$$ | $$-10.22$$ |
| $$2$$ | $$-8.22$$ |
| $$4$$ | $$-6.22$$ |
| $$5$$ | $$-5.22$$ |
| $$8$$ | $$-2.22$$ |
| $$10$$ | $$-0.22$$ |
| $$10$$ | $$-0.22$$ |
| $$15$$ | $$4.78$$ |
| $$38$$ | $$27.78$$ |
The following table shows John's grades in the mathematics exams throughout the year. Calculate the different absolute deviations.
| Note | $$f_i$$ |
| $$3$$ | $$1$$ |
| $$4$$ | $$3$$ |
| $$5$$ | $$4$$ |
| $$6$$ | $$2$$ |
| $$7$$ | $$3$$ |
| $$9$$ | $$1$$ |
First the average is calculated $$$\displaystyle \overline{x}=\frac{3\cdot 1+4\cdot 3+5\cdot 4+6\cdot 2+7\cdot 3+9\cdot 1}{1+3+4+2+3+1}=\frac{77}{14}=5.5$$$ Next, it is possible to calculate the absolute deviation, including it in the table:
| Note | $$f_i$$ | $$D_i=x_i-\overline{x}=x_i-5.5$$ |
| $$3$$ | $$1$$ | $$-2,5$$ |
| $$4$$ | $$3$$ | $$-1,5$$ |
| $$5$$ | $$4$$ | $$-0,5$$ |
| $$6$$ | $$2$$ | $$0,5$$ |
| $$7$$ | $$3$$ | $$1,5$$ |
| $$9$$ | $$1$$ | $$3,5$$ |
Standard deviation
Standard deviation is the arithmetical average of the absolute values of the absolute deviations. It is symbolized by $$D_{\overline{x}}$$ and can be calculated applying the formula$$$\displaystyle D_{\overline{x}}=\frac{\displaystyle \sum_{i=1}^{N} |x_i-\overline{x}|}{N}=\frac{|x_1-\overline{x}|+|x_2-\overline{x}|+\ldots+|x_N-\overline{x}|}{N}$$$ It shows whether the information is dispersed (or not). High standard deviation involves a lot of changeability in the information, while a standard deviation equal to zero implies that all the values are equal and, therefore, they have the same value as the average.
George's grades in mathematics throughout the course are the following: $$8, 7, 9, 8, 8, 10, 9, 7, 4, 9$$. Calculate standard deviation.
The first step is to look for the average: $$$\displaystyle \overline{x}=\frac{8+7+9+8+8+10+9+7+4+9}{10}=\frac{79}{10}=7.9$$$ Next, we apply the definition: $$$\displaystyle D_{\overline{x}}=\frac{|8-7.9|+|7-7.9|+|9-7.9|+|8-7.9|+|8-7.9|+}{10}=\frac{+|10-7.9|+|9-7.9|+|7-7.9|+|9-7.9|}{10}=\\=\frac{0.1+0.9+1.1+0.1+0.1+2.1+1.1+0.9+3.9+1.1}{10}=\frac{11.4}{10}=1.14$$$
In a basketball match, we have the following points of the players of a team: $$0, 2, 4, 5, 8, 10, 10, 15, 38$$. Calculate standard deviation of the scorings for the players of the team.
Applying the formula $$$\displaystyle \overline{x}=\frac{0+2+4+5+8+10+10+15+38}{9}=\frac{92}{9}=10.22$$$ the average is obtained. The deviations can be represented in a table:
| Punctuation | $$D_i=x_i-\overline{x}-10,22$$ |
| $$0$$ | $$-10.22$$ |
| $$2$$ | $$-8.22$$ |
| $$4$$ | $$-6.22$$ |
| $$5$$ | $$-5.22$$ |
| $$8$$ | $$-2.22$$ |
| $$10$$ | $$-0.22$$ |
| $$10$$ | $$-0.22$$ |
| $$15$$ | $$4.78$$ |
| $$38$$ | $$27.78$$ |
Applying the formula $$$\displaystyle D_{\overline{x}}=\frac{10.22+8.22+6.22+5.22+2.22+0.22+0.22+4.78+27.78}{9}=\frac{65.1}{9}=7.23$$$ we obtain the standard deviation.
Calculation of the standard deviation for grouped information
If $$N$$ information are grouped in $$n$$ classes, the formula is $$$\displaystyle D_{\overline{x}}=\frac{\displaystyle \sum_{i=1}^n |x_i-\overline{x}| f_i}{N}=\frac{|x_1- \overline{x}|f_1+|x_2- \overline{x}|f_2+\ldots+|x_n- \overline{x}|f_n}{N}$$$
The height in cm of the players of a basketball team are in the following table. Calculate the standard deviation.
| $$x_i$$ | $$f_i$$ | |
| $$[160,170)$$ | $$165$$ | $$1$$ |
| $$[170,180)$$ | $$175$$ | $$2$$ |
| $$[180,190)$$ | $$185$$ | $$4$$ |
| $$[190,200)$$ | $$195$$ | $$3$$ |
| $$[200,210)$$ | $$205$$ | $$2$$ |
First, fill in the following table
| $$x_i$$ | $$f_i$$ | $$x_if_i$$ | $$|x_i-\overline{x}|$$ | $$|x_i-\overline{x}|f_i$$ | |
| $$[160,170)$$ | $$165$$ | $$1$$ | $$165$$ | $$22.5$$ | $$22.5$$ |
| $$[170,180)$$ | $$175$$ | $$2$$ | $$350$$ | $$12.5$$ | $$25$$ |
| $$[180,190)$$ | $$185$$ | $$4$$ | $$740$$ | $$2.5$$ | $$10$$ |
| $$[190,200)$$ | $$195$$ | $$3$$ | $$585$$ | $$7.5$$ | $$22.5$$ |
| $$[200,210)$$ | $$205$$ | $$2$$ | $$410$$ | $$17.5$$ | $$35$$ |
| $$12$$ | $$2250$$ | $$115$$ |
The average is calculated $$\overline{x}=\displaystyle\frac{2250}{12}=187.5$$ to be able to fill in the last two columns.
The standard deviation is calculated: $$\displaystyle D_{\overline{x}}=\frac{115}{12}=9.58$$.