Standard Deviation

Standard Deviation is an essential statistical measure that quantifies the degree of variation or dispersion in a set of data values. Whether you're dealing with a small sample or a large population, understanding and calculating standard deviation is fundamental in statistics.

1.0What is Standard Deviation?

Standard deviation, often represented by the Greek letter sigma (σ) for a population or "s" for a sample, measures the spread of data points around the mean. In simpler terms, it tells you how much the data deviates from the average.

2.0Definition of Standard Deviation in Statistics

In statistical terms, Standard Deviation is defined as the square root of the variance. It gives insight into the reliability and variability of data. A low standard reveals that the data points are closely clustered around the mean, while a high standard deviation suggests a wide range of values.

3.0The Importance of Standard Deviation

Standard deviation plays a vital role in fields such as finance, economics, and the natural sciences. It helps in assessing risk, comparing datasets, and making informed decisions based on data variability.

4.0Formula for Finding Standard Deviation

The standard deviation can be determined using the following formulas:

For a Population:

$\sigma =\sqrt{\frac{\sum {\left(x_i-\mu \right)}^2}{N}}$

 For a Sample:

$s=\sqrt{\frac{\sum {\left(x_i-\bar{x}\right)}^2}{n-1}}$

Where:

• x_i = Each individual data point
• = Population mean 
• = Sample mean
• N = Total number of data points in the population
• n = Number of data points in the sample

Step-by-Step Example: Calculating Standard Deviation

Let’s walk through an example to find the standard deviation for a dataset.

Suppose we have the following data: 4, 8, 6, 5, 3.

Step 1: Calculate the mean:

$\text{ Mean }=\frac{4+8+6+5+3}{5}=5.2$

Step 2: Subtract the mean from each data point and then square the resulting difference:

(4 – 5.2)², (8 – 5.2)², (6 – 5.2)², (5 – 5.2)², (3 – 5.2)²

Step 3: Find the variance (mean of squared differences):

$\text{ Variance }=\frac{(1.44)+(7.84)+(0.64)+(0.04)+(4.84)}{5}=2.96$

Step 4: Compute square root of the variance:

$\text{ Standard Deviation }=\sqrt{2.96}\approx 1.72$

5.0Calculating Standard Deviation for Grouped Data

For grouped data, the process involves a weighted approach. Here’s a brief outline:

1. Find the mid-point (x_i) of each class interval.
2. Calculate the mean using these mid-points.
3. Compute the squared differences from the mean.
4. Multiply each squared difference by the frequency of the corresponding class.
5. Divide by the total number of data points to find the variance.
6. Obtain the standard deviation by finding the square root of the variance.

6.0Binomial Distribution and Standard Deviation

In binomial distribution, the standard deviation can be calculated using:

$\sigma =\sqrt{np(1-p)}$

Where:

• n = Number of trials
• p = Probability of success

This formula helps in understanding the spread of binomial outcomes.

7.0Applications of Standard Deviation

• Finance: Risk assessment and portfolio management.
• Quality Control: Monitoring process consistency.
• Research: Validating experimental results.

8.0Solved Examples of Standard Deviation

Example 1: Determine the standard deviation for the following data: 2, 4, 4, 6, 8, 10.

Solution: 

1. Mean = $\frac{2+4+4+6+8+10}{6}=5.67$
2. Variance = $\frac{(2-5.67{)}^2+(4-5.67{)}^2+(4-5.67{)}^2+(6-5.67{)}^2+(8-5.67{)}^2+(10-5.67{)}^2}{6}\approx 6.67$
3. Standard Deviation $=\sqrt{6.67}\approx 2.58$

Example 2: Consider the following set of data representing the number of hours studied by 5 students for an exam:  Data: 4, 8, 6, 5, 3

Solution: 

Step 1: Calculate the Mean

First, find the mean (average) of the data.

$\text{ Mean }=\frac{\sum x_i}{n}$

$=\frac{4+8+6+5+3}{5}$

$=\frac{26}{5}=5.2$

Step 2: Subtract the Mean and Square the Result

Next, subtract the mean from each data point, then square the result.

(4 – 5.2)² = (–1.2)² = 1.44 

(8 – 5.2)² = (2.8) ² = 7.84 

(6 – 5.2)² = (0.8) ² = 0.64 

(5 – 5.2)² = (–0.2) ² = 0.04 

(3 – 5.2)² = (–2.2) ² = 4.84 

Step 3: Find the Variance

Now, calculate the variance by finding the average of these squared differences.

$\text{ Variance }=\frac{\sum {\left(x_i-\mu \right)}^2}{n}$

$=\frac{1.44+7.84+0.64+0.04+4.84}{5}$

$=\frac{14.8}{5}$

= 2.96

Step 4: Calculate the Standard Deviation

Lastly, find the standard deviation by taking the square root of the variance.

$\text{ Standard Deviation }=\sqrt{\text{ Variance }}$

$=\sqrt{2.96}\approx 1.72$

Example 3: Consider a scenario where a researcher is studying the daily temperature fluctuations in a desert over a week. The recorded temperatures (in degrees Celsius) are as follows: 30, 35, 28, 32, 40, 29, 33.

Solution: 

Step 1: Calculate the Mean

$\text{ Mean }=\frac{30+35+28+32+40+29+33}{7}$

$=\frac{227}{7}\approx 32.43$

Step 2: Subtract the Mean and Square the Result

(30 – 32.43)² = (-2.43)² = 5.90 

(35 – 32.43)² = (2.57)² = 6.60 

(28 – 32.43)² = (–4.43)² = 19.62 

(32 – 32.43)² = (–0.43)² = 0.18 

(40 – 32.43)² = (7.57)² = 57.29 

(29 – 32.43)² = (–3.43)² = 11.76 

(33 – 32.43)² = (0.57)² = 0.32 

Step 3: Find the Variance

$\text{ Variance }=\frac{5.90+6.60+19.62+0.18+57.29+11.76+0.32}{7}$

$=\frac{101.67}{7}\approx 14.52$

Step 4: Calculate the Standard Deviation

$\text{ Standard Deviation }=\sqrt{14.52}\approx 3.81$

Example 4: In a physics experiment, the time taken for 10 oscillations of a simple pendulum was recorded five times as follows (in seconds): 20.2, 19.8, 20.4, 20.0, 19.6. Determine the standard deviation of the recorded times.

Solution: 

Step 1: Calculate the Mean

$\text{ Mean }=\frac{20.2+19.8+20.4+20.0+19.6}{5}$

$=\frac{100.0}{5}=20.0$

Step 2: Subtract the Mean and Square the Result

(20.2 – 20.0)² = (0.2)² = 0.04 

(19.8 – 20.0) ² = (–0.2)² = 0.04 

(20.4 – 20.0) ² = (0.4)² = 0.16 

(20.0 – 20.0)² = (0.0)² = 0.00 

(19.6 – 20.0)² = (–0.4)² = 0.16 

Step 3: Find the Variance

$\text{ Variance }=\frac{0.04+0.04+0.16+0.00+0.16}{5}$

$=\frac{0.40}{5}=0.08$

Step 4: Calculate the Standard Deviation

$\text{ Standard Deviation }=\sqrt{0.08}\approx 0.28$

Example 5: A financial analyst is studying the weekly returns of a particular stock over a month (4 weeks). The returns (in percentage) are as follows: 2%, -1%, 3%, -2%.

Solution: 

Step 1: Calculate the Mean**

$\text{ Mean }=\frac{2+(-1)+3+(-2)}{4}$

Step 2: Subtract the Mean and Square the Result

(2 – 0.5)² = (1.5)² = 2.25 

(–1 – 0.5)² = (-1.5)² = 2.25 

(3 – 0.5)² = (2.5)² = 6.25 

(–2 – 0.5)² = (–2.5)² = 6.25 

Step 3: Find the Variance

$\text{ Variance }=\frac{2.25+2.25+6.25+6.25}{4}$

$=\frac{17.0}{4}=4.25$

Step 4: Calculate the Standard Deviation

$\text{ Standard Deviation }=\sqrt{4.25}\approx 2.06\%$

Example 6: The heights of 6 students in a class (in cm) are as follows: 150, 160, 155, 165, 158, 162. Calculate the standard deviation of the heights.

Solution: 

Step 1: Calculate the Mean

$\text{ Mean }=\frac{150+160+155+165+158+162}{6}$

$=\frac{950}{6}\approx 158.33$

Step 2: Subtract the Mean and Square the Result

(150 - 158.33)² = (-8.33)² = 69.39 

(160 - 158.33)² = (1.67)² = 2.79 

(155 - 158.33)² = (-3.33)² = 11.09 

(165 - 158.33)² = (6.67)² = 44.49 

(158 - 158.33)² = (-0.33)² = 0.11 

(162 - 158.33)² = (3.67)² = 13.47 

Step 3: Find the Variance

$\text{ Variance }=\frac{69.39+2.79+11.09+44.49+0.11+13.47}{6}$

$=\frac{141.34}{6}\approx 23.56$

Step 4: Calculate the Standard Deviation

$\text{ Standard Deviation }=\sqrt{23.56}\approx 4.85$

9.0Practice Questions on Standard Deviation

1. The following data represents the number of books read by five students in a month: 2, 4, 3, 5, 6. Calculate the standard deviation.
2. Calculate the standard deviation for the following data set: 50, 60, 70, 80, 90, 100. Assume the data represents the scores of students in an exam.
3. In a classroom of 10 students, their test scores out of 100 are as follows: 75, 80, 85, 70, 65, 90, 95, 88, 72, 78. Determine the standard deviation of the test scores.
4. The daily closing values of a stock over 7 days are given as follows: ₹150, ₹152, ₹155, ₹148, ₹150, ₹153, ₹151. Calculate the standard deviation.
5. In an experiment, the following data points are recorded: 12, 14, 16, 18, 20, 22, 24, 26. Compute the standard deviation.

10.0Sample Question on Standard Deviation

1. What is the Formula for Standard Deviation?

Ans: For a sample of data, the standard deviation (s) is calculated as: $s=\sqrt{\frac{\sum {\left(x_i-\bar{x}\right)}^2}{n-1}}$

where x_i is each individual data point, \bar{X} is the mean of the data, and n is the number of data points. For a population, the denominator is n instead of n – 1.