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# Standard Deviation

While working with statistical problems, we come across with standard deviation quite frequently. It is the measure of dispersion of the statistical data. Dispersion or variation is the property of the data to spread over a field. Dispersion measures the deviation of the data from its average or mean position. The degree of dispersion or variation is calculated by the means of measures of variation.

There are many different measures of variation most common of those are listed below :

1. Range
2. Mean deviation
3. Standard Deviation
4. Quartile deviation.

Among those, SD is the most frequently used measure of dispersion. It is denoted by $\sigma$, pronounced as sigma.

The measures of dispersion are the statistical tool which measures the variability or the dispersion of a data. The measures of central tendency, such as - mean, median and mode, are the central values of the values. Therefore, these are known as first order averages. On the other hand, the measures of variation or dispersion mentioned above are averages of deviations derived from the average values. Therefore, they are known as second order averages.

 Related Calculators Calculate Standard Deviation Calculator Relative Standard Deviation Calculator Confidence Interval Standard Deviation Calculator Normal Distribution Standard Deviation Calculator

## Definition

A measure of dispersion in statistics. It gives an idea of how the individual data in a data set is dispersed from the mean. Also known as the root mean square deviation. The symbol used is $\sigma$.

1. The minimum positive value is 0. i.e. it cannot be negative.
2. When the items in a series are more dispersed from the mean, then the standard deviation is also large.

### Purpose

The purpose of obtaining is to measure the standard distance from the mean.

Merits
1. It is based on all observations in a distribution.
2. It is capable of further algebraic treatment.
3. We can find out measures like coefficient of variation, combined standard deviation extra.

Demerits

1. It is difficult to calculate.
2. It gives more importance to bigger values.

## Calculate Standard Deviation

To find, we first find the arithmetic mean of the values. Then we find the deviation of each item from the mean. Find the squares of the deviations and add them. Then divide the sum by the number of items in the series and take the square root.

Steps to Calculate the Standard Deviation:

Step 1: Calculate the arithmetic mean.

Step 2: Find the deviation of each item from the mean.

Step 3: Square these deviations and add them.

Step 4: We get $\sum (x - \bar{x})^{2}$.

Step 5: Divide this sum by the total number of items.

Step 6:
Take the square root of the result of step 5.

Use below widget

### Solved Example

Question: Find the standard deviation of 1, 2, 3, 4, 5.
Solution:

Step 1: Calculate mean and deviation

 X M (X - M) $(X - M)^2$ 1 3 -2 4 2 3 -1 1 3 3 0 0 4 3 1 1 5 3 2 4

Step 2
: Find the sum of $(X - M)^2$

=> $(X - M)^2$ = 4 + 1 + 0 + 1 + 4 = 10

Step 3: N = 5, the total number of values.

=> N - 1 = 5 - 1 = 4.

Step 4: Now find the standard deviation by deviation formula.

=>   S = $\sqrt{\frac{\sum (X - M)^2}{N - 1}}$

= $\sqrt{\frac{10}{4}}$

=> S = 1.58113.

## Mean and Standard Deviation

The mean and standard deviation are the most useful measures. Mean is the measure of central tendency and the standard deviation is the measure of dispersion, which is using the mean. We can say that the mean is the first order average and the standard deviation is the second order average.

## Graph

The graph of Normal distribution shows the area between the mean and standard deviation. The area between the Mean $\pm$ standard deviation is 68.27 % of the total area. The area between Mean $\pm$ 2 standard deviation is 95.45 % of the total area. Area between Mean $\pm$ 3 standard deviation is 98.73 % of the total area. The normal distribution graph depends on two factors, the mean and the standard deviation. The mean of the distribution determines the location of the centre of the graph, and the standard deviation determines the height and width of the graph.

Case 1:
When the standard deviation is large, the curve is short and wide.

Case 2:
When the standard deviation is small, the curve is tall and narrow. ## Standard Deviation Table

The Standard normal distribution is used in various hypothesis tests including tests on single means, the difference between two means, and tests on proportions. Has a mean of 0 and a standard deviation of 1. This table also called as Z-score table. ### Solved Example

Question: By using Z-score table, find the percent of Population Between 1 and 1.11.
Solution:

Step 1:
Start at the row for 1.1, and read along until 1.11.

=> There is the value 0.3665

Step 2:

0.3665 of the population are between 1 and 1.11.

=> Standard deviations from the mean and 0.3665 is 36.65%

So 36.65% of the population are between 1 and 1.11.

## Fractional and Binomial

The fractional deviation, $\frac{\sigma}{r}$, may be used in place of the standard deviation to describe the precision of data. If r represents the mean then $\sigma$ = $\sqrt{r}$ and the fractional deviation is $\sqrt{r}$. The percent standard deviation is the fractional standard deviation multiplied by $100$.

The binomial standard deviation applies to events with two outcomes. Standard deviation equal to the square root of the number of trials, $N$, multiplied by the probability, $p$, multiplied by the opposite probability.

Standard Deviation = $\sqrt{Np\ (1\ -\ p)}$

Where,
$N$ = Number of trials
$p$ = Probability of trials.

### Solved Example

Question: For a count rate of 10000 cpm, what counting time is necessary to achieve a percent standard deviation of 2%.
Solution:

Here, $\sigma$% = 2%, c = 10000

Step 1:

$\sigma$% = 100$\sqrt{\frac{1}{ct}}$

=> 2 % = 100 $\sqrt{\frac{1}{10000 * t}}$

=> 2 = 100 * $\frac{1}{10}$$\frac{1}{\sqrt{t}}$

=> 2 = 10 $\frac{1}{\sqrt{t}}$

=> $\frac{2}{10}$ = $\frac{1}{\sqrt{t}}$

= $\frac{1}{5}$ = $\frac{1}{\sqrt{t}}$

Step 2:
On squaring both sides, we have

=> $\frac{1}{25}$ = $\frac{1}{t}$

or t = 25 min. answer

## Normal Standard Deviation

A standard normal distribution is a normal distribution with mean $0$ and standard deviation $1$. Normal distributions are symmetrical with a single central peak at the mean of the data. The shape of the curve is described as bell-shaped with the graph falling off evenly on either side of the mean.
Normal distributions can be transformed to standard normal distributions by the formula:

$Z$ = $\frac{x - \mu}{\sigma}$

where, $x$ = values from the original normal distribution, $\mu$ = the mean and $\sigma$ = the standard deviation .

## Standard Deviation Example

### Solved Examples

Question 1: Find the standard deviation of the following values 2, 3, 4, 5, 6.
Solution:

We first find the arithmetic mean of the values. Then we find the deviation of each item from the mean. Find the squares of the deviations and add them. Then divide the sum by the number of items in the series and take the square root.
The calculations are shown in the table below

 X $(x_{1} - \bar{x})$ $(x_{1} - \bar{x})^{2}$ 2 2 - 4 = - 2 4 3 3 - 4 = - 1 1 4 4 - 4 = 0 0 5 5 - 4 = 1 1 6 6 - 4 = 2 4 10

Mean $\bar{x}$ = $\frac{2 + 3 + 4 + 5 + 6}{5}$ = 4

Standard deviation $\sigma$ = $\sqrt{\frac{\sum (x_{1} - \bar{x})^{2}}{n}}$

= $\sqrt{\frac{10}{4}}$

= $\sqrt{2.5}$

=1.58

=> $\sigma$ = 1.58

Here you can see that all items are taken into consideration. Also we can see that it give more importance to bigger values since we are squaring the values.

Question 2: Find the standard deviation of
 X 1 2 3 4 f 10 4 3 3

Solution:

We first find the arithmetic mean of the values. Then we find the deviation of each item from the mean. Find the squares of the deviations, multiply by the corresponding frequencies and add them. Then divide the sum by the number of items in the series and take the square root.

Step 1: The calculations are shown in the table below

 x f $(x_{1}-\bar{x})$ $(x_{1}-\bar{x})^{2}$ $f(x_{1}-\bar{x})^{2}$ 1 10 1-1.95 = -0.95 0.9025 9.025 2 4 2-1.95 = 0.05 0.0025 0.01 3 3 3-1.95 = 1.05 1.1025 3.3075 4 3 4-1.95 = 2.05 4.2025 12.6075 20 Total 24.95

Step 2:
Mean, $\bar{x}$ = $\frac{\sum fx}{\sum f}$

= $\frac{1 * 10 + 2 * 4 + 3 * 3 + 4 * 3}{20}$

= $\frac{10 + 8 + 9 + 12}{20}$

= $\frac{39}{20}$

= 1.95

=> Mean = 1.95

Step 3:
Now,
Standard deviation, $\sigma$ = $\sqrt{\frac{\sum f(x_{1}-\bar{x})^{2}}{N}}$

= $\sqrt{\frac{24.95}{20}}$

= 1.12

=> $\sigma$ = 1.12

Here also we can see that its based on all the observation and it gives more importance to bigger values.