Absolute deviation and standard deviation
We throw $10$ times a dice, obtaining the following results: $1, 1, 1, 3, 3, 4, 4, 5, 6, 6$. Calculate the absolute deviation of each thrown.
We must know the average before applying the formula $D_i=x_i-\overline{x}$.
To find the average we add up all the terms and divide by the number of throws: $$\overline{x}=\dfrac{1+1+1+3+3+4+4+5+6+6}{10}=\dfrac{34}{10}=3,4$$ Again the deviations are classified in a table, regardless of the frequencies:
| Result of the throw | $D_i=x_i-\overline{x}=x_i-3,4$ |
| $1$ | $-2,4$ |
| $3$ | $-0,4$ |
| $4$ | $0,6$ |
| $5$ | $1,6$ |
| $6$ | $2,6$ |
Taking $1, 1, 1, 3, 3, 4, 4, 5, 6, 6$ as samples , we have:
| Result of the throw | $D_i$ |
| $1$ | $-2,4$ |
| $3$ | $-0,4$ |
| $4$ | $0,6$ |
| $5$ | $1,6$ |
| $6$ | $2,6$ |
Albert scored $14$ points in the basketball match, this supposes an absolute deviation from his team of $3,2$ points. His friend Marc scored only $3$ points. Calculate:
a) The average points of the team.
b) The absolute deviation of Marc's scoring with respect to the team average.
a) Applying the formula $$\overline{x}=x_i-D_i=14-3,2=10,8$$ the average is found.
b) Next, we apply the equation $$D_i=x_i-\overline{x}=3-10,8 = -7,8$$ to find the deviation of Marc's scoring with regard to the average.
a) $\overline{x}=10,8$
b) $D_i=-7,8$
Next the results of the handball league are shown. Calculate the total goals of every match. With the obtained result, calculate the absolute deviations of every match.
| Team | Team | Final Score |
| TEAM A | TEAM B | $30-28$ |
| TEAM C | TEAM D | $30-28$ |
| TEAM E | TEAM F | $34-23$ |
| TEAM G | TEAM H | $37-32$ |
| TEAM I | TEAM J | $26-27$ |
| TEAM K | TEAM L | $33-27$ |
| TEAM M | TEAM N | $29-30$ |
The goals are: $58, 58, 57, 69, 53, 60, 59$ and $57$. The average is $$\overline{x}=\dfrac{58+58+57+69+53+60+59+57}{8}\simeq58,9$$ Next we calculate the deviations indicating the result in a table:
| Goals | $D_i=x_i-\overline{x}=x_i-58,9$ |
| $53$ | $-5,9$ |
| $57$ | $-1,9$ |
| $57$ | $-1,9$ |
| $58$ | $-0,9$ |
| $58$ | $-0,9$ |
| $59$ | $0,1$ |
| $60$ | $1,1$ |
| $69$ | $10,1$ |
| Note | $D_i$ |
| $53$ | $-5,9$ |
| $57$ | $-1,9$ |
| $57$ | $-1,9$ |
| $58$ | $-0,9$ |
| $58$ | $-0,9$ |
| $59$ | $0,1$ |
| $60$ | $1,1$ |
| $69$ | $10,1$ |
In the following table the results of throwing $20$ times a dice are shown. Find the standard deviation.
| $x_i$ | $f_i$ |
| $1$ | $4$ |
| $2$ | $2$ |
| $3$ | $4$ |
| $4$ | $3$ |
| $5$ | $4$ |
| $6$ | $3$ |
We add a column to help with the calculation of the average:
| $x_i$ | $f_i$ | $x_i f_i$ |
| $1$ | $4$ | $4$ |
| $2$ | $2$ | $4$ |
| $3$ | $4$ | $12$ |
| $4$ | $3$ | $12$ |
| $5$ | $4$ | $20$ |
| $6$ | $3$ | $18$ |
| $70$ |
Find then the average $\overline{x}=\dfrac{70}{20}=3,5$, and find the two columns that simplify the calculation of the standard deviation:
| $x_i$ | $f_i$ | $x_i f_i$ | $|x_i-\overline{x}|$ | $|x_i-\overline{x}|\cdot f_i$ |
| $1$ | $4$ | $4$ | $2,5$ | $10$ |
| $2$ | $2$ | $4$ | $1,5$ | $3$ |
| $3$ | $4$ | $12$ | $0,5$ | $2$ |
| $4$ | $3$ | $12$ | $0,5$ | $1,5$ |
| $5$ | $4$ | $20$ | $1,5$ | $6$ |
| $6$ | $3$ | $18$ | $2,5$ | $7,5$ |
| $70$ | $30$ |
The standard deviation is then: $D_{\overline{x}}=\dfrac{30}{20}=1,5$.
$$D_{\overline{x}}=1,5$$
The height of the students of a class is measured, grouping the results in the following table. Calculate the standard deviation.
| $x_i$ | $f_i$ | |
| $[140, 155)$ | $147,5$ | $3$ |
| $[155,165)$ | $160$ | $6$ |
| $[165,175)$ | $170$ | $17$ |
| $[175,190)$ | $182,5$ | $5$ |
The table is filled in to make easier to calculate the average and the standard deviation:
| $x_i$ | $f_i$ | $x_i f_i$ | $|x_i-\overline{x}|$ | $|x_i-\overline{x}|\cdot f_i$ | |
| $[140, 155)$ | $147,5$ | $3$ | $442,5$ | ||
| $[155,165)$ | $160$ | $6$ | $960$ | ||
| $[165,175)$ | $170$ | $17$ | $2890$ | ||
| $[175,190)$ | $182,5$ | $5$ | $912,5$ | ||
| $5205$ |
To be able to fill in the last 2 columns the average is calculated $\overline{x}=\dfrac{5205}{31}=167,9$
| $x_i$ | $f_i$ | $x_i f_i$ | $|x_i-\overline{x}|$ | $|x_i-\overline{x}|\cdot f_i$ | |
| $[140, 155)$ | $147,5$ | $3$ | $442,5$ | $20,4$ | $61,2$ |
| $[155,165)$ | $160$ | $6$ | $960$ | $7,9$ | $47,4$ |
| $[165,175)$ | $170$ | $17$ | $2890$ | $2,1$ | $35,7$ |
| $[175,190)$ | $182,5$ | $5$ | $912,5$ | $14,6$ | $73$ |
| $5205$ | $217,3$ |
The standard deviation is then $D_{\overline{x}}=\dfrac{217,3}{31}=7,01$.
With these values, we have $D_{\overline{x}}=7,01$
The following table shows the final classification of the East Conference of the NBA. Find the standard deviation of the number of victories achieved in the period.
The average is calculated first $$\overline{x}=\dfrac{65+54+54+54+53+50+49+48+46+29+24+24+23+19+17}{15}=$$ $$=\dfrac{609}{15}=40,6$$
The standard deviation is then: $$D_{\overline{x}}=\dfrac{|65-40,6|+|54-40,6|+|54-40,6|+|54-40,6|+|53-40,6|+|50-40,6|}{15}+$$ $$+\dfrac{|49-40,6|+|48-40,6|+|46-40,6|+|29-40,6|+|24-40,6|+|24-40,6|}{15}+$$ $$+\dfrac{|23-40,6|+|19-40,6|+|17-40,6|}{15}=$$ $$=\dfrac{24,4+13,4+13,4+13,4+12,4+9,4+8,4+7,4+5,4+11,6+16,6+16,6}{15} $$ $$+\dfrac{17,6+21,6+23,6}{15}=\dfrac{215,2}{15}\simeq14,35$$
$$D_{\overline{x}}\simeq 14,35$$