Standard deviation and variance both are important statistical terms and both terms are associated with each other. These terms play a key role in the measurements of dispersion for the statistical values. The degree to which values in a distribution deviate from the average of the distributions is known as its dispersion. There are several measures that can be used to measure the magnitude of the variation. The process of measuring the fluctuation of data points is used to calculate the degree of dispersion.

Variance and standard deviation are correlated terms because the square root of the standard deviation for the given data values is obtained by determining the square root of the variance. In this guide, we will address the concept of the standard deviation vs variance in detail.

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**What is Variance?**

The variance is defined as the mean of the square of deviation of all the observations from their mean. **R.A. Fisher** introduced the concept of variance in **1918**. Variance is a frequently used measure of dispersion due to its importance. The symbolic definition of variance is

**S****2****= ∑ (x – x̅)****2****/ n**where S2 is variance and ∑ is the summation sign.

Or

**S****2****= (∑ x****2****/ n) – (∑x / n)****2**

In the case of frequency distribution variance may be defined as

**S****2****= ∑f (x – x̅)****2****/ ∑ f**

Or

**S****2****= (∑ fx****2****/ ∑ f) – (∑ fx / ∑ f)****2**

The degree of dispersion in a set of data is measured by its variance. It can be concluded from the above relations that the variance is zero as the given values get close to each other so that they are equivalent. Positive variances are all of the values that are not zero.

A small variance shows that the data points are close to the mean and one another whereas the data points being far apart from the mean and one another indicate a large variance.

**What is Standard Deviation?**

The term standard deviation corresponds to the positive square root of variance. Three variables are used in the standard deviation calculation. Each point’s value in a data set is the initial variable, and a sum number indicates each successive variable (x, x 1, x 2, x 3, etc.)

The values of the variable x and the value of data provided to the variable **n** are subjected to the mean. Symbolically

**S = √ [∑ (x – x̅)****2****/ n]**where S is the standard deviation.

Or

**S = √ [(∑ x****2****/ n) – (∑x / n)****2****]**

In the case of frequency distribution standard deviation may be defined as:

**S = √ [∑f (x – x̅)****2****/ ∑ f]**

Or

**S = √ [∑ fx****2****/ ∑ f) – (∑ fx / ∑ f)****2****]**

The unit of expression for the standard deviation is the same as the unit of the observations. The standard deviation is a measure that indicates the degree of variation (dispersion or spread) from the mean. A common variation from the mean is illustrated by the standard deviation.

Because it reaches back to the initial units of data values of measurement, it is a preferred measure of variability. Similar to variance, there is a slight variation if the data points are close to the mean and a large variation if the data points are far dispersed from the mean.

The standard deviation determines how far the numbers deviate from the average. The simplest way to assess dispersion is using the standard deviation which is based on all data. As a result, the standard deviation is affected by a small change in one value.

Now we will elaborate on both terms with the help of examples.

**Examples:**

**Example 1:**

Calculate the variance and the standard deviation from the following marks obtained by the students.

45, 32, 37, 46, 39, 36, 41, 48, 36

**Solution:**

**Step 1:** We will determine the necessary values as

X | 45 | 32 | 37 | 46 | 39 | 36 | 41 | 48 | 36 | ∑X = 360 |

X2 | 2025 | 1024 | 1369 | 2116 | 1521 | 1296 | 1681 | 2304 | 1296 | ∑X2 = 14632 |

**Step 2:** Formula:

**For variance: **

S2 = (∑ x2 / n) – (∑x / n)2

Putting the relevant values:

S2 = (14632 / 9) – (360 / 9) ^ 2

S2 = 1625.78 – 1600

**S****2**** = 25.78 marks****^2**** Ans.**

**For standard deviation:**

S = √ [(∑ x2 / n) – (∑x / n) ^2]

S = √ (25.78)

**S = 5.08 marks**

You can take assistance from online websites like standarddevoiationcalcuator.io to find standard deviation and variance.

**Example 2:**

Determine the variance and the standard deviation from the following data:

Groups | 20 – 24 | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 |

f | 1 | 4 | 8 | 11 | 15 | 9 | 2 |

**Solution:**

**Step 1:** We will determine the necessary values as:

Classes | F | x | fx | fx2 |

20 – 24 | 1 | 22 | 22 | 484 |

25 – 29 | 4 | 27 | 108 | 2916 |

30 – 34 | 8 | 32 | 256 | 8192 |

35 – 39 | 11 | 37 | 407 | 15059 |

40 – 44 | 15 | 42 | 630 | 26460 |

45 – 49 | 9 | 47 | 423 | 19881 |

50 – 54 | 2 | 52 | 104 | 5408 |

Total | 50 | — | 1950 | 78400 |

**Step 2:** Formula:

For variance:

S2 = (∑ fx2 / ∑ f) – (∑ fx / ∑ f)2

S2 = (78400 / 50) – (1950 / 50) ^2

S2 = 1568 – 1521

**S****2**** = 47 Ans**.

For standard deviation:

S = √ [∑ fx2 / ∑ f) – (∑ f x / ∑ f) ^ 2]

S = √ 47

**S = 6.856 Ans**.

**Wrap Up:**

In this article, we have elaborated on the important concepts of variance and standard deviation in detail. We have addressed their definitions and their different mathematical relations that help us to calculate these important sorts of terms to analyze and observe the given information.

Hopefully, by reading this article you will be able to solve the problems about variance and standard deviation easily. Moreover, it will enable you to understand the basic difference between the variance and the standard deviation.

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