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# Standard Deviation Calculator

The Standard Deviation is, by definition, the positive root mean square or root-mean-square error deviation of the data set.

Note: Please enter comma separated values e.g. 12, 20, 30, 15

## What is Standard Deviation?

Standard deviation, typically denoted by 𝛔, is a statistic that measures the dispersion or spread of a data distribution around the average, 𝛍. For a particular sample, a lower standard deviation means that the data values are closer to the mean or expected value, that is, they are less spread out. Conversely, a higher standard deviation for the same sample suggests that the data values are far apart from the mean, that is, they are spread out in a wider range of values.

## Formula of Standard Deviation

There are multiple ways in which the formula for standard deviation is amended to get unbiased results in different real-life situations. The definitional formula for standard deviation is as follows:

$$σ = \sqrt{\dfrac{ 1}{ N } \sum_{i=1}^{N}(x_i – 𝛍)^{2}}$$

Where xi = any value from the data set
μ = mean/expected value
N = total number of values in the data set/sample space

The above equation may seem intimidating for those looking at it for the first time.

However, it is not as complicated as it may seem. xi𝛍 is the deviation of a data value from the mean. For those unfamiliar with summation,

$$\text{Sum} = \sum_{i=1}^{n}x_i$$

, it merely implies the addition of all values starting from the first one ( i = 1 ) going through i = 2, i = 3, and so on till N (total number of values).

To make calculations less messy, the following is a computational formula:

$$σ = \sqrt{\dfrac{ 1}{ N } \sum_{i=1}^{N}(x_i)^{2} – (𝛍)^{2}}$$

In statistics, mathematical models are made, which are then used against real data to draw out crucial conclusions. In real life, the calculated mean rarely ever indicates that there is an equal number of data values less than and greater than the mean. Many factors affect the mean, including skewness and range of the data. Two data sets can have the same mean; however, it is only after calculating the standard deviation that statisticians can reach satisfactory conclusions. Here is a simple example for better understanding.

### Standard Deviation Calculation Example

The average rainfall in two cities, A and B, is 20mm. You are doing a monthly report on the water level of the two cities. You notice that the water level of city A has increased only by 0.5mm while, on the other hand, the water level of city B has risen abundantly by 10mm. If the amount of rainfall in both the cities was the same, what went wrong?

Considering the slope, terrain, weather conditions, and all other factors are kept constant or have a negligible effect, take a look at both the data sets:

 S.No. Date Rainfall in City A (mm) Rainfall in City B (mm) 1 May 5th 0 20 2 May 12th 0 20 3 May 13th 0 30 4 May 30th 0 5 5 May 17th 0 10 6 May 24th 70 0 7 May 25th 80 0 8 May 26th 100 25 9 May 27th 50 40 10 May 30th 0 50

After taking a look at the data sets, you will notice the following things:
The total rain in a span of 10 days for both cities is 200mm.
In City A, it rained heavily for four consecutive days, reaching a total of 200mm.
In City B, it rained less to moderate on ten days of the month, adding to a total of 200mm.

From the afore-mentioned observations, you can conclude that there was more dispersion in the data set for City B than City A. When you calculate the standard deviation, it indicates that the values in the data set for City B are closer to the mean than City A. As a result of consecutive heavy rainfall in City A, all the water from the rain drained to lower areas and did not get enough time to absorb. However, in City B, it rained consistently and moderately so that the water had time to get absorbed, so the water level was higher.

When using standard deviation for statistical inferences, two cases must be considered:
Population Standard Deviation
Sample Standard Deviation

## Population Standard Deviation

The population standard deviation is a parameter which is used when all the individual data points can be obtained from the entire population. It is the square root of the variance of a dataset in which all the values can be sampled from a population. The formula for population standard deviation is:

$$σ = \sqrt{\dfrac{ 1}{ N } \sum_{i=1}^{N}(x_i – 𝛍)^{2}}$$

Where xi = any value from the population
μ = mean/expected value
N = total number of values in the population

Calculation with example
A group of five students compare their scores on a math test. Calculate the standard deviation of their scores of 10.

7, 8, 6, 9, 5

Step 1:
Calculate the mean of the scores. This will be your 𝛍.

$$= \dfrac{7 + 8 + 6 + 9 + 5}{ 5 }$$

$$= 6$$

Step 2:
Subtract the mean from each of the individual scores. These differences are deviations. Scores below the mean will have negative deviations, and scores above the mean will have positive deviations. Scores equal to the mean will naturally have a deviation of 0.

 Score Deviation (xi – 𝛍) 7 7 – 6 = +1 8 8 – 6 = +2 6 6 – 6 = 0 9 9 – 6 = +3 5 5 – 6 = -1

Step 3:
Square each deviation so that it becomes positive.

 Score Deviation (xi – 𝛍) Squared Deviation (xi – 𝛍)2 7 7 – 6 = +1 1 8 8 – 6 = +2 4 6 6 – 6 = 0 0 9 9 – 6 = +3 9 5 5 – 6 = -1 1

Step 4:
Sum up the squared deviations together.

$$= 1 + 4 + 0 + 9 + 1$$

$$= 15$$

Step 5:
Divide the sum of the squared deviations by the total number of data values in the population. The calculated result is called the variance.

$$= \dfrac{ 15 }{ 5 }$$

$$= 3$$

Step 6:
Calculate the square root of the variance to get the standard deviation of the scores.

$$= 3$$

$$= 1.732$$

## Sample Standard Deviation

Sample standard deviation is a statistic that is most commonly used as an estimator for population standard deviation, σ. In cases where it is not possible to sample every member of the population, a random sample is taken for convenience purposes and the sample standard deviation, s, is calculated using a modified formula. Here is why:
While calculating the sample standard deviation, s, the μ is unknown. ( μ is the mean of the population, not sample). Therefore, we use x as the mean of the sample which introduces a slight bias in the calculation. Since x is calculated using the data values from the sample, the data values (xi)will be closer to x than μ on an average. As a result, the square of the deviations will also be smaller for the sample which will result in a greater standard deviation.
This bias can be corrected by dividing by N-1 instead of N. N-1 is called the degrees of freedom.
⇒ This corrected version has a significant amount of error for samples N>10.

$$s = \sqrt{\dfrac{ 1}{ N – 1 } \sum_{i=1}^{n}(x_i – \overline{x})^{2}}$$

Where xi= any value from the sample
x= sample mean
N = sample space

Calculation with example
→A group of five students compare their scores on a math test. Calculate the standard deviation of their scores of 10.

7, 8, 6, 9, 5

Step 1:
Calculate the mean of the scores. This will be your 𝛍.

$$= \dfrac{ 7 + 8 + 6 + 9 + 5 }{ 5 }$$

$$= 6$$

Step 2: Subtract the mean from each of the individual scores. These differences are deviations. Scores below the mean will have negative deviations, and scores above the mean will have positive deviations. Scores equal to the mean will naturally have a deviation of 0.

 Score Deviation (xi – 𝛍) 7 7 – 6 = +1 8 8 – 6 = +2 6 6 – 6 = 0 9 9 – 6 = +3 5 5 – 6 = -1

Step 3:
Square each deviation so that it becomes positive.

 Score Deviation (xi – 𝛍) Squared Deviation (xi – 𝛍)2 7 7 – 6 = +1 1 8 8 – 6 = +2 4 6 6 – 6 = 0 0 9 9 – 6 = +3 9 5 5 – 6 = -1 1

Step 4:
Sum up the squared deviations together.

$$= 1 + 4 + 0 + 9 + 1$$

$$= 15$$

Step 5:
Divide the sum of the squared deviations by one less than the total number of data values in the sample.

$$= \dfrac{ 15 }{ 4 }$$

$$= 3.75$$

Step 6:
Calculate the square root of the variance to get the standard deviation of the scores.

$$=\sqrt{3.75}$$

$$= 1.936$$

Notice here that the sample standard deviation is greater (1.936) than the population standard deviation (1.732).

## Why do we need to calculate Standard Deviation?

Statistical parameters like the standard deviation are widely used in business, finance, and the manufacturing industry. It is an important tool financial analysts and business-owners use to manage risk and make decisions. Measuring market volatility and performance trends is also done by calculating standard deviation. For example, in investment, the index fund is likely to have a low standard deviation when compared to its benchmark index. On the contrary, aggressive growth funds are expected to have a high standard deviation from relative stock indices due to aggressive bets by portfolio managers to generate higher-than-average returns.IN this scenario, alow standard deviation will not be ideal.

In the case of slumping sales or a rise in bad customer reviews, potent risk management maneuvers can be devised using standard deviation. It can also be used to calculate the volatility of stock prices and margins of error in surveys taken by the company.

Quality control measures employ standard deviation to monitor and maintain the caliber of manufactured products. For example, a company manufactures potato chips. Each bag of chips is monitored for its weight. If the average weight is 25grams, and the standard deviation is 士2 grams, the minimum amount of chips in a bag can be 23 grams and the maximum can be 27 grams. Anything greater or lesser than that cannot be distributed by the company.

In scientific research and reasoning, standard deviation plays a crucial role when it comes to analyzing the obtained results from an experiment.
Customer reviews, employee surveys, presidential election polls are all conducted with the aid of standard deviation and calculating margin of error. Even teachers make use of standard deviation when setting exam papers and scoring them.

These are only a handful of practical applications of standard deviation out of many others. It is an essential statistical tool that is easy to employ and gives crucial information about a data set.

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