In statistics, the term standard deviation is widely used for various purposes. It is commonly used in different fields of education and daily life dealing such as weather forecast, finance, science, engineering, etc.
This statistical measurement is the square root of another statistical measurement known as a variance. In this post, we’ll learn sample and population standard deviations along with their formulas and solved examples.
What is the standard deviation?
In statistics, the measure of the variability of a set of sample and population data is said to be the standard deviation. It is a well-known way to measure the spread of data such as whether it lies closer or far from the expected value.
The square root of the other statistical measure known as variance will give you the result of standard deviation. The term variance is the average of squared differences from the expected value of the set of population and sample data.
There are two types of data values in this measurement that is sample standard deviation and population standard deviation.
Sample Standard Deviation
In statistics, the sample standard deviation is a type of STD that is used to measure the variability of a sample set of data. A sample is the smaller estimated values from a larger population of objects, individuals, or measurements.
For example, measuring the height of each student in the school is rather tough so we have to take the height of some students from the whole to measure estimate heights. The formula for measuring the sample standard deviation is:
| Formula | Components |
| s = √(Σ(xi– x̄)² / (n – 1)) |
|
Population Standard Deviation
In statistics, the population standard deviation is a type of STD that is used to measure the variability of a population set of data. A population is the entire group of the larger population of objects, individuals, or measurements.
For example, measuring the height of each student in the school is the population estimation of heights. The formula for measuring the population standard deviation is:
| Formula | Components |
| σ = √(Σ(xi – μ)² / N) |
|
Note
In the formula for sample standard deviation, we divide by (n-1) instead of n. This is because we are estimating the population standard deviation based on a sample, and dividing by n would underestimate the variability of the population.
How to calculate the sample and population standard deviations?
Here are a few examples to calculate standard deviation of sample and population data values.
Example I: for population data
Calculate the given data values to measure the variability of the given entire experiments of an object.
2, 9, 12, 14, 16, 18, 22, 23, 25, 27, 30
Solution
Step 1: Find the sum of given experimental values and divide it by total numbers to calculate the expected value.
| Experiment values | xi = 2, 9, 12, 14, 16, 18, 22, 23, 25, 27, 30 |
| Addition of values | Σ xi = 2 + 9 + 12 + 14 + 16 + 18 + 22 + 23 + 25 + 27 + 30 Σ xi = 198 |
| Population Average | μ = (Σ xi ) ÷ n μ = 198 ÷ 11 = 18 |
Step 2: Now minus each experimental value from the expected value and take the square of subtracted results.
| Experimental values | xi – μ | (xi – μ)² |
| 2 | 2 – 18 = -16 | (-16)² = 256 |
| 9 | 9 – 18 = -9 | (-9)² = 81 |
| 12 | 12 – 18 = -6 | (-6)² = 36 |
| 14 | 14 – 18 = -4 | (-4)² = 16 |
| 16 | 16 – 18 = -2 | (-2)² = 4 |
| 18 | 18 – 18 = 0 | (-0)² = 0 |
| 22 | 22 – 18 = 4 | (4)² = 16 |
| 23 | 23 – 18 = 5 | (5)² = 256 |
| 25 | 25 – 18 = 7 | (7)² = 49 |
| 27 | 27 – 18 = 9 | (9)² = 81 |
| 30 | 30 – 18 = 12 | (12)² = 144 |
Step 3: Now take the sum of squared differences of data observations from the expected value.
Σ(xi– x̄)² = 256 + 81 + 36 + 16 + 4 + 0 + 16 + 25 + 49 + 81 + 144
Σ(xi- x̄)² = 708
Step 4: Now divide the above summed result by the total number of experimental values.
Σ(xi- x̄)² ÷ n = 708 ÷ 11
Σ(xi- x̄)² ÷ n = 64.364
Step 5: Now take the above result and find the square root of it.
√[Σ(xi- x̄)² ÷ (n)] = √64.364
√[Σ(xi- x̄)² ÷ (n)] = 8.023
Example-2: for sample data
Calculate the given data values to measure the variability of the given sample observations from the entire experiments of an object.
1, 9, 19, 23, 29, 31, 37, 39, 43, 49
Solution
Step 1: Find the sum of given experimental values and divide it by total numbers to calculate the expected value.
| Sample Observations | xi = 1, 9, 19, 23, 29, 31, 37, 39, 43, 49 |
| Sum of Sample | Σ xi = 1 + 9 + 19 + 23 + 29 + 31 + 37 + 39 + 43 + 49 Σ xi = 280 |
| Sample Average | x̄ = (Σ xi ) ÷ n x̄ = 280 ÷ 10 = 28 |
Step 2: Now minus each experimental value from the expected value and take the square of subtracted results.
| Experimental values | xi – x̄ | (xi – x̄)² |
| 2 | 1 – 28 = -27 | (-27)² = 729 |
| 9 | 9 – 28 = -19 | (-19)² = 361 |
| 12 | 19 – 28 = -9 | (-9)² = 81 |
| 14 | 23 – 28 = -5 | (-5)² = 25 |
| 16 | 29 – 28 = 1 | (1)² = 1 |
| 18 | 31 – 28 = 3 | (3)² = 9 |
| 22 | 37 – 28 = 9 | (9)² = 81 |
| 23 | 39 – 28 = 11 | (11)² = 121 |
| 25 | 43 – 28 = 15 | (15)² = 225 |
| 27 | 49 – 28 = 21 | (21)² = 441 |
Step 3: Now take the sum of squared differences of data observations from the expected value.
∑ (xi – x̄)² = 729 + 361 + 81 + 25 + 1 + 9 + 81 + 121 + 225 + 441
∑ (xi – x̄)² = 2074
Step 4: Now divide the above sum by n – 1.
∑ (xi – x̄)² ÷ (N – 1) = 2074 ÷ 10 – 1
∑ (xi – x̄)² ÷ (N – 1) = 2074 ÷ 9
∑ (xi – x̄)² ÷ (N – 1) = 230.44
Step 5: Now take the above result and find the square root of it.
√ [∑ (xi – x̄)² ÷ (N – 1)] = √230.44
√ [∑ (xi – x̄)² ÷ (N – 1)] = 15.18
Conclusion
The term standard deviation is used to calculate the measure of the variability of sample and population data values. The formulas for calculating sample and population standard deviation are used to find the results easily and quickly.