Standard Deviation for Ungrouped Data: A Complete Guide

The standard deviation is one of the important terms in the study of the subject of statistics. The standard deviation is a statistical term used for determining to what extent fluctuations or dispersion there is in a set of data values. It is measured as the difference between the data points and the mean (average) value.

It is frequently used to observe and understand the range of data. A greater standard deviation signifies the dispersion or spread of the data points, whereas a smaller standard deviation shows the degree of variation in the data set or given data information precisely.

In this comprehensive detail, we will describe the concept of the standard deviation for ungrouped data with a complete guide on how to solve the problems of the standard deviation for ungrouped data. We will elaborate on this comprehensive concept with the help of solved examples.

Standard Deviation (SD)

An assessment of the variation or dispersion in a collection of data values is made using a statistical metric known as the standard deviation. The mean (average) value is often used to understand the distribution of data points around it.

The formula used to find the standard deviation can differ depending on whether you are using a sample or the entire population. There are two variations of the standard deviation formula: one for the entire population and another for a particular portion or sample.

The sample standard deviation (SSD) is used when we have the data on a subset of the population data values and the population standard deviation (PSD) is used when we have data on the entire population.

How to Calculate SD for Ungrouped Data

A list of values that may or may not be categorized makes up the raw ungrouped data. The standard deviation is then determined using the relevant formula either there is to compute sample standard deviation or population standard deviation.

Explanation

The degree to which each observation deviates from the mean is expressed as a standard deviation. The positive would exactly balance the negative if the disparities were to be combined, giving a sum of zero and as a result, we sum up all the squares of the deviations.

The sum of the squares is then divided by the number of observations minus one to determine the mean of the squares. The square root is then applied for converting the data to its original units. (To get the mean square, use “degrees of freedom” and divide by the number of observations minus one instead of the total number of observations.) In this case, they are one less than the total.

Mathematically,

For population Standard deviation:

σ = √ [Σ (x_k – μ)² / N]

Where Σ is the sigma or summation sign and u is the mean for population data. Moreover, N is the size of the population data.

For sample standard deviation:

S_k = √ [Σ (x_k – x̅)² / (N – 1)]

Example Section:

Example 1: Determine what the standard deviation for the following ungrouped data given in the table that represents the number of passed candidates in the CSS exam from different districts:

City	A	B	C	D	E	F
X	1	2	4	7	9	12

Solution:

Step 1: Determine the mean for the given data in the above table.

x̅ = μ= (1 + 2 + 4 + 7 + 9 + 12) / 6

x̅ = μ= 35 / 6

x̅ = μ = 5.8333

Step 2: Now we will do the following necessary computations to get the desired result.

x	xk – x̅ = xk – μ	(xk – x̅)2 = (xk – μ)2
1	-4.83	23.3289
2	-3.83	14.6689
4	-1.83	3.3489
7	1.17	1.3489
9	3.17	10.0489
12	6.17	38.0689
Σ		90.8334

Step 3: Formula for population SD:

σ = √ [Σ (x_k – μ)² / N]

σ = √ [(90.8334) / 6]

σ = √ (15.1389)

σ = 3.8909 Ans.

Step 4: Formula for sample SD:
S_k = √ [Σ (x_k – x̅)² / (N – 1)]

S_k = √ [(90.8334) / (6 – 1)]

S_k = √ (90.8334) / 5

S_k = √ (18.1667)

S_k = 4.26

Example 2: Determine the standard deviation for the set of numbers given in the table that shows the marks of a candidate in his annual exam.

Subject	Eng	Ur	IS	PS	Math	Phy	Chem	Bio
Marks (xk)	29	31	28	29	31	46	39	31

Solution:

Step 1: Mean of the marks:

x̅ = μ= (29 + 31+ 28 + 29 + 31 + 46 + 39 + 31) / 8

x̅ = μ= 264 / 8

x̅ = μ = 33

Step 2: Now we will do the following necessary computations to get the desired result.

x	xk – x̅ = xk – μ	(xk – x̅)2 = (xk – μ)2
29	-4	16
31	-2	4
28	-5	25
29	-4	16
31	-2	4
46	13	169
39	6	36
31	-2	4
Σ		274

Step 3: Formula for population SD:

σ = √ [Σ (x_k – μ)² / N]

σ = √ [(274) / 8]

σ = √ (34.25)

σ = 5.8524 Ans.

Step 4: Formula for sample SD:
S_k = √ [Σ (x_k – x̅)² / (N – 1)]

S_k = √ [(274) / (8 – 1)]

S_k = √ (274) / 7

S_k = √ (39.1429)

S_k = 6.2564

A mean standard deviation calculator could be used as an alternative to find sample or population STD to get rid of lengthy manual solutions.

Wrap Up:

In this comprehensive guide, we have addressed the topic of the standard deviation for in details. We have elaborated on how to compute the standard deviation for ungrouped data. We have explained a complete guide with solved examples to determine the population standard deviation and sample standard deviation if there is given ungrouped data. We hope that by reading this article you can solve problems about standard deviation easily.