O standard deviation is a measure of dispersion, as are variance and coefficient of variation. When determining the standard deviation, we can establish a range around the arithmetic mean (division between the sum of numbers in a list and the number of numbers added) where most of the data is concentrated. The greater the value of the standard deviation, the greater the variability of the data, that is, the greater the deviation from the arithmetic mean.
Read too: Mode, mean, and median — the main measures of central tendencies
Standard deviation summary
- Standard deviation is a measure of variability.
- Standard deviation notation is the lowercase Greek letter sigma (σ) or the letter s.
- The standard deviation is used to verify the variability of the data around the mean.
- The standard deviation determines a range \(\left[\mu-\sigma,\mu+\sigma\right]\), where most of the data is located.
- To calculate the standard deviation, we must find the square root of the variance:
\(\sigma=\sqrt{\frac{\sum_{i=1}^{N}\left (x_i-\mu\right)^2}{N}}\)
What is standard deviation?
The standard deviation is a dispersion measure adopted in Statistics. Its use is linked to variance interpretation, which is also a measure of dispersion.
In practice, the standard deviation determines an interval, centered on the arithmetic mean, in which most of the data is concentrated. Thus, the greater the value of the standard deviation, the greater the irregularity of the data (more information heterogeneous), and the smaller the value of the standard deviation, the smaller the irregularity of the data (more information homogeneous).
How to calculate the standard deviation?
To calculate the standard deviation of a data set, we must find the square root of the variance. So, the formula for calculating the standard deviation is
\(\sigma=\sqrt{\frac{\sum_{i=1}^{N}\left (x_i-\mu\right)^2}{N}}\)
- \(x_1,x_2,x_3,\ldots, x_N\) → data involved.
- μ → arithmetic mean of the data.
- N → amount of data.
- \( \sum_{i=1}^{N}\left (x_i-\mu\right)^2\ =\ \left (x_1-\mu\right)^2+\left (x_2-\mu\right )^2+\left (x_3-\mu\right)^2+...+\left (x_N-\mu\right)^2 \)
The last item, which refers to the numerator of the radicand, indicates the sum of squares of the difference between each data point and the arithmetic mean. please note that the unit of measure for the standard deviation is the same unit of measure as the data x1,x2,x3,…,xNo.
Although the writing of this formula is a bit complex, its application is simpler and more direct. Below is an example of how to use this expression to calculate the standard deviation.
- Example:
For two weeks, the following temperatures were recorded in a city:
Week/Day |
Sunday |
Second |
Third |
Fourth |
Fifth |
Friday |
Saturday |
week 1 |
29°C |
30°C |
31°C |
31.5°C |
28°C |
28.5°C |
29°C |
week 2 |
28.5°C |
27°C |
28°C |
29°C |
30°C |
28°C |
29°C |
In which of the two weeks did the temperature remain more regular in this city?
Resolution:
To analyze temperature regularity, we must compare the standard deviations of the temperatures recorded in weeks 1 and 2.
- Let's first look at the standard deviation for week 1:
Note that the average μ1 It is No1 they are
\(\mu_1=\frac{29+30+31+31.5+28+28.5+29}{7}\approx29.57\)
\(N_1=7 \) (7 days a week)
Also, we need to calculate the square of the difference between each temperature and the average temperature.
\(\left (29-29.57\right)^2=0.3249\)
\(\left (30-29.57\right)^2=0.1849\)
\(\left (31-29.57\right)^2=2.0449\)
\(\left (31.5-29.57\right)^2=3.7249\)
\(\left (28-29.57\right)^2=2.4649\)
\(\left (28.5-29.57\right)^2=1.1449\)
\(\left (29-29.57\right)^2=0.3249\)
Adding the results, we have that the numerator of the radicand in the standard deviation formula is
\(0,3249\ +\ 0,1849\ +2,0449+3,7249+2,4649+1,1449+0,3249\ =\ 10,2143\)
So the week 1 standard deviation is
\(\sigma_1=\sqrt{\frac{\sum_{i=1}^{7}\left (x_i-\mu_1\right)^2}{N_1}}=\sqrt{\frac{10,2143} {7}}\ \approx1.208\ °C\)
Note: This result means that most of the week 1 temperatures are in the interval [28.36 °C, 30.77 °C], that is, the interval \(\left[\mu_1-\sigma_1,\mu_1+\sigma_1\right]\).
- Now let's look at the week 2 standard deviation:
Following the same reasoning, we have
\(\mu_2=\frac{28.5+27+28+29+30+28+29}{7}=28.5\)
\(N_2=7\)
\(\left (28.5-28.5\right)^2=0\)
\(\left (27-28.5\right)^2=2.25\)
\(\left (28-28.5\right)^2=0.25\)
\(\left (29-28.5\right)^2=0.25\)
\(\left (30-28.5\right)^2=2.25\)
\(\left (28-28.5\right)^2=0.25\)
\(\left (29-28.5\right)^2=0.25\)
\(0\ +\ 2,25\ +\ 0,25\ +\ 0,25+2,25+0,25+0,25\ =\ 5,5\)
So the week 2 standard deviation is
\(\sigma_2=\sqrt{\frac{\sum_{i=1}^{7}\left (x_i-\mu_1\right)^2}{N_2}}=\sqrt{\frac{5,5} {7}}\ \approx0.89\ °C\)
This result means that most week 2 temperatures are in the range \(\left[\mu_2-\sigma_2,\mu_2+\sigma_2\right]\), that is, the range \(\left[\mu_2-\sigma_2,\mu_2+\sigma_2\right]\).
realize that \(\sigma_2, that is, the week 2 standard deviation is less than the week 1 standard deviation. Therefore, week 2 presented more regular temperatures than week 1.
What are the types of standard deviation?
The types of standard deviation are related to the type of data organization. In the previous example, we worked with the standard deviation of ungrouped data. To calculate the standard deviation of a set of otherwise organized data (grouped data, for example), you would need to adjust the formula.
What are the differences between standard deviation and variance?
the standard deviation is the square root of the variance:
\(\sigma=\sqrt{\frac{\sum_{i=1}^{N}\left (x_i-\mu\right)^2}{N}}\)
\(V=\frac{\sum_{i=1}^{N}\left (x_i-\mu\right)^2}{N}\)
When using variance to determine the variability of a data set, the result has the data unit squared, which makes its analysis difficult. Thus, the standard deviation, which has the same unit as the data, is a possible tool to interpret the variance result.
Know more:Absolute frequency — the number of times the same response appeared during data collection
Solved exercises on standard deviation
question 1
(FGV) In a class of 10 students, the students' grades in an assessment were:
6 |
7 |
7 |
8 |
8 |
8 |
8 |
9 |
9 |
10 |
The standard deviation of this list is approximately
A) 0.8.
B) 0.9.
C) 1.1.
D) 1.3.
E) 1.5.
Resolution:
Alternative C.
According to the statement, N = 10. The average of this list is
\( \mu=\frac{6+7+7+8+8+8+8+9+9+10}{10}=8 \)
Furthermore,
\(\left (6-8\right)^2=4\)
\(\left (7-8\right)^2=1\)
\(\left (8-8\right)^2=0\)
\(\left (9-8\right)^2=1\)
\(\left (10-8\right)^2=4\)
\(4+1+1+0+0+0+0+1+1+4=12\)
So the standard deviation of this list is
\(\sigma=\sqrt{\frac{\sum_{i=1}^{10}\left (x_i-8\right)^2}{10}}=\sqrt{\frac{12}{10} }\approx1.1\)
question 2
Consider the statements below and rate each one as T (True) or F (False).
i. The square root of the variance is the standard deviation.
II. The standard deviation has no relationship with the arithmetic mean.
III. Variance and standard deviation are examples of measures of dispersion.
The correct order, from top to bottom, is
A) V-V-F
B) F-F-V
C) F-V-F
D) F-F-F
E) V-F-V
Resolution:
E alternative.
i. The square root of the variance is the standard deviation. (true)
II. The standard deviation has no relationship with the arithmetic mean. (false)
The standard deviation indicates an interval around the arithmetic mean in which most of the data fall.
III. Variance and standard deviation are examples of measures of dispersion. (true)
By Maria Luiza Alves Rizzo
Math teacher
Source: Brazil School - https://brasilescola.uol.com.br/matematica/desvio-padrao.htm
