To know the main difference between Variance and Standard Deviation is very important to understand. Both of these mathematical terms are usually used in many mathematical equations to solve problems. Initially, variance and standard deviation are used as metrics or methods to solve statistical problems.
Both of these terminologies are used in probability theory as well. To know the extent of the spread of data from its mean, these are used in many numerical measurements.In terms of how it is calculated, the square root of the variance can be used to calculate the standard deviation value once the variance of the provided data has been determined.
In this article, we will discuss the main difference between variance and standard deviation, their formulas, measuring steps, and examples from different directions.
What is variance?
By definition, a variance is the average of the squared differences from the mean. find the variance by calculating the difference between each data point in the set and the average. Variance is written in squared units. It is mathematically symbolized as (σ^{2}) and is computed as follows:
σ^{2} = ∑ (x_{i} – x)^{2}/n
Variance is a great way to show how dispersed individuals are in the group. We can say that it is a numerical value, which explains how much variability exists in the observations.
What is Standard deviation?
Standard deviation is the level of dispersion of data points relative to its mean. It also tells about how the values are spread across the data sample and the variation of the data points from the mean. Standard Deviation is stated in the same units of the data provided. It is mathematically symbolized as (σ) and computed as follows
σ = √[∑ (x_{i} – x)^{2}/n]
Understanding the connection between Variance and Standard deviation
By taking the mean of the data points, subtracting the mean from each data point in the set, squaring each of these outputs, and then taking another mean of these squares, one can estimate variance. Standard deviation is just the square root of the output of variance.
When compared to the data point mean value, the variance helps in determining the extent of the data spread. Data values become more variable as variance increases, and there may be a significant difference between values. The variance will be minimal if all of the data values are near to one another.
However, because variances indicate a squared result that might not be expressively depicted on the same graph as the real dataset, they are more difficult to understand than standard deviations.
Standard deviation is usually easier to estimate. While finding standard deviation, in the second last step, the value is the output of variance because only by taking the square root of variance we can find the value of standard deviation.
Statisticians can quickly determine whether a data set has a normal curve or some other mathematical combination by using the standard deviation. About 68% of data points will lie within one standard deviation of the average and data points if the data appears to be a normal curve. More data points fall beyond the standard deviation when variances are larger. More data are close to the data set average when variances are smaller.
Examples Section
Example 1:
Find the standard deviation and variance for the following data 7,4,6,3,8,4,7,9,4,8.
Solution:
To evaluate ∑ (x_{i} – x̄)^{2}, we have to find the mean first
x̄ = ∑x/n
x̄ = (7+4+6+3+8+4+7+9+4+8)/10 = 60/10 = 6
x̄ = 6
x_{i} | [x_{i} – x̄] | [x_{i} – x̄]^{2} |
7 | 7-6 = 1 | 1 |
4 | 4-6 = -2 | 4 |
6 | 6-6 = 0 | 0 |
3 | 3-6 = -3 | 9 |
8 | 8-6 = 2 | 4 |
4 | 4-6 = -2 | 4 |
7 | 7-6 = 1 | 1 |
9 | 9-6 = 3 | 9 |
3 | 3-6 = -3 | 9 |
8 | 8-6 = 2 | 4 |
∑ (x_{i} – x̄)^{2} = 45 |
σ^{2} = (45)/10
σ^{2} = 4.5
Now we’ll calculate the standard deviation
S.D. = √Variance
σ = √4.5
σ = 2.12
An online tool, Standarddeviationcalculator.io could be used for dealing with the problems of standard deviation and variance as it will give the results with steps in a fraction of a second to get rid of manual calculations.
Example 2:Find variance and standard deviation for the following data
Weights | 10-12 | 13-15 | 16-18 | 19-21 | 22-24 |
Frequency | 4 | 8 | 7 | 6 | 5 |
Solution:
Weights | Frequency | x | Fx | x- x̄ | (x- x̄)^{2} | F(x- x̄)^{2} |
10-12 | 4 | 11 | 44 | 11-17=-6 | 36 | 144 |
13-15 | 8 | 14 | 112 | 14-17=-3 | 9 | 72 |
16-18 | 7 | 17 | 119 | 17-17=0 | 0 | 0 |
19-21 | 6 | 20 | 120 | 20-17=3 | 9 | 54 |
22-24 | 5 | 23 | 115 | 23-17=6 | 36 | 180 |
∑f =30 | ∑fx = 510 | ∑ F(x- x̄)^{2} = 450 |
x̄ = ∑fx/ ∑f
x̄ = 510/30 = 17
Now,
σ^{2} = [∑F(x- x̄)^{2}/ ∑ (F-1)]
σ^{2} = 450/29
σ^{2}= 15.72
And,
σ = √Variance
σ = √15.72
σ = 3.94
Conclusion
In this section, we have discussed the concept of Standard deviation and Variance in detail. The difference between both metrics has also been discussed thoroughly. By covering its relationship, one can easily defend this topic precisely.
By taking into mind its practical applications, we easily use these terminologies in many of the statistical complications. Examples of each data type e.g. group and ungroup data have also been solved above to understand these metrics efficiently.