Standard deviation is an important concept in statistics that is used to measure the spread of data around the mean value. This term is also called standard deviation. In data analysis, we often want to know how a variable or data spreads out from its environment.

In this guide we will discuss standard deviation in full, including the definition and meaning of standard deviation, the formula for calculating it, and examples of its use in data analysis.

You will learn how to calculate the standard deviation for individual and group data, as well as understand the meaning of the standard deviation value in interpreting data.

## What is standard deviation?

Standard deviation is defined as a statistical value used to measure the distribution of data in a sample and how close the location of individual data points is to the mean or median value of the sample. This concept is used to provide information about the spread of data from the average value.

The higher the standard deviation, the greater the variance or difference between individual data points and your average. The standard deviation in a data set can be zero or less than zero.

If the value is zero, all values â€‹â€‹in the data set have the same value. However, if it is valuable **standard deviation** **Or standard deviation** Smaller or larger than zero, it indicates that individual data points lie far from the mean value.

To calculate the standard deviation value for a data set, you must follow several steps. First, calculate the average value (mean) of all data points. Next, calculate the variance of the data by taking the difference between each data point from the mean value.

Then square the deviation value at each data point, and differentiate it by the square of the mean value. This value is called variance. After obtaining the variance value, the next step is to take the square root of the variance value to get the standard deviation value.

This way, you will get information about how close or far individual data points are from the average value in the data set.

## Is the standard deviation the same as the average deviation?

In the world of statistics, there are two important concepts that are often used to measure the distribution of data, which are the standard deviation and the mean deviation. There are differences between these two concepts that must be understood well.

### 1. Average deviation

Mean deviation is a concept describing how far individual data points are from the average value in a set of data. The smaller the mean deviation, the closer the data points are to the mean value, which indicates that the data tends to be more homogeneous.

This concept is useful for knowing the extent to which data deviates from the central value.

### 2. Standard deviation

On the other hand, standard deviation is used to measure the spread of data in a sample and the extent of variation or difference between individual data points and the average. The higher the standard deviation, the greater the variance in the data, and the further away the individual data points are from the mean.

These two concepts are important in analyzing data and providing information about data variations in a sample. However, they have different formulas and uses, so it is important to understand the difference between standard deviation and mean deviation well.

With the right understanding, you can use these two concepts effectively in your statistical analysis.

## Standard deviation function

Some standard deviation functions are:

### 1. Measure the distribution of data

The standard deviation is used to evaluate the spread of data from the mean value. The higher the standard deviation, the greater the variance or difference between individual data points and the average value.

### 2. Determine data homogeneity

You can use standard deviation to evaluate the homogeneity or uniformity of data in a group. If the standard deviation is close to zero, the data tends to be homogeneous or have little variation.

Conversely, if the standard deviation is large, the data tends to be heterogeneous or have greater variance.

### 3. Identify outliers

Standard deviation functions to help you identify outliers or data points that are far from the mean value. If there are data points with a large standard deviation, these points can be considered outliers that need to be paid special attention to.

### 4. Risk measurement

Standard deviation is also useful for measuring risk or changes in investment results. In the context of investor money management for startups, standard deviation helps you understand how much investment results differ from the expected average.

## Standard deviation formula

Below is the formula for standard deviation.

### 1. Sample

Information:

- xi is the value of each data point in the sample.
- xÌ„ is the average value of all data in the sample.
- n is the number of data in the sample.
- âˆš is the square root symbol.
- Î£ is the symbol for the sum of all values â€‹â€‹in parentheses.
- (n – 1) is the correction factor used because we are using sample data, not the entire population. This correction factor is used to obtain a more accurate estimate of the population standard deviation.

### 2. Population

Information:

- xi is every value in the data society.
- Î¼ is the mean (median) value of the data set.
- N is the total amount of data in the population.
- âˆš is the symbol for the square root sign.
- Î£ is the symbol for adding all the values â€‹â€‹in parentheses.

The main difference between the formula for population standard deviation and sample standard deviation lies in the denominator. In the population standard deviation formula, the denominator is the total amount of data in the population (N).

Meanwhile, in the sample standard deviation formula, the denominator is the total amount of data in the sample minus one (n-1). A correction factor (n-1) is used in the standard deviation formula model to provide a more accurate estimate.

This is because samples tend to have greater variance than the total population. Meanwhile, in the total population, no correction factors are needed because all data are included in the calculation.

### 3. Varian

The formula for the variance of the data is:

This formula can be used as a way to find the standard deviation, i.e. by s = âˆšvariance. This may be an alternative you can do in solving problems related to standard deviation.

## Example of standard deviation questions

Below is an example of a question that can increase your understanding of standard deviation. You can try to complete it first before watching the discussion.

**1. Question: Calculate the standard deviation of the data for the following group:**

Season | repetition |

10 – 20 | 5 |

20 – 30 | 10 |

30 – 40 | 15 |

40 – 50 | 20 |

Answer:

- Calculating the midpoint of the semester:

Midpoint for class 10-20: (10 + 20) / 2 = 15 Midpoint for class 20-30: (20 + 30) / 2 = 25 Midpoint for class 30-40: (30 + 40) / 2 = 35 midpoint 40-50: (40 + 50) / 2 = 45

Rate rate (Î¼) = ((15 x 5) + (25 x 10) + (35 x 15) + (45 x 20)) / (5 + 10 + 15 + 20) Rate rate (Î¼) = (75 + 250 + 525 + 900) / 50 rate ratio (Î¼) = 1750 / 50 rate ratio (Î¼) = 35

Variables = (((15-35)^2 * 5) + ((25-35)^2 * 10) + ((35-35)^2 * 15) + ((45-35)^2 * 20) ) / 50 variables = ((400 * 5) + (100 * 10) + (0 * 15) + (100 * 20)) / 50 variables = (2000 + 1000 + 0 + 2000) / 50 variables = 5000 / 50 Varian = 100

- Calculate the standard deviation:

S = âˆšVarians

x = âˆš100

S = 10.52 (two digits taken after the comma)

So the standard deviation of the group data is about 10.52.

**2. Question: A teacher wants to know how much his students’ scores on a math test differ from the class average. The average class score is 75 and the students’ scores are as follows: 80, 70, 85, 90, 60. You are asked to calculate the standard deviation of the data.**

Solution:

- Calculate the average: (80 + 70 + 85 + 90 + 60) / 5 = 77
- Calculate the difference between each value and the average: (80 â€“ 77), (70 â€“ 77), (85 â€“ 77), (90 â€“ 77), (60 â€“ 77) = 3, -7, 8, 13, – 17
- The square of each difference: 3^2, (-7)^2, 8^2, 13^2, (-17)^2 = 9, 49, 64, 169, 289
- Add all the squared results: 9 + 49 + 64 + 169 + 289 = 580
- Divide the sum of the result by the amount of data: 580 / 5 = 116
- Take the square root of the division result: root (116) â‰ˆ 10.77

So, the standard deviation of the data is about 10.77.

- Problem: The company wants to know how well its employees’ salaries are distributed. The following is the monthly salary data for employees: 3 million, 4 million, 5 million, 6 million, 7 million, 8 million, 9 million, 10 million. You are asked to calculate the standard deviation of the data.

Solution:

- Calculate the average: (3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) / 8 = 6.125 million
- Calculate the difference between each value and the average: 3 – 6.125, 4 – 6.125, 5 – 6.125, 6 – 6.125, 7 – 6.125, 8 – 6.125, 9 – 6.125, 10 – 6.125 = -3.125, -2.125, -1.125, -0.125 , 0.875, 1.875, 2.875, 3.875
- The square of each difference: (-3.125)^2, (-2.125)^2, (-1.125)^2, (-0.125)^2, 0.875^2, 1.875^2, 2.875^2, 3.875^2 = 9.766, 4,516, 1,266, 0.016, 0.766, 3,516, 8,016, 15,016
- Add all the squared results: 9.766 + 4.516 + 1.266 + 0.016 + 0.766 + 3.516 + 8.016 + 15.016 = 43.92
- Divide the total result by the total data: 43.92 / 8 = 5.49
- Take the square root of the division: root (5.49) â‰ˆ 2.34

So, the standard deviation of the data is about 2.34 million.

## Close

The sample questions above can help you understand standard deviation and implement it in everyday life. You can repeat the discussion if you are still confused and try to understand it again.

Standard deviation is a mathematical concept that must be well understood. Understanding the concept and knowing the correct formula is the key to solving the problem. You can continue to practice sample questions to understand the material better.