STANDARD DEVIATION
The best measure of dispersion is, usually, standard deviation which does not possess the demerits of range and mean deviation. And also standard deviation can be treated mathematically.
Standard deviation for a given set of observations is defined as the root mean square deviation when the deviations are taken from the AM of the observations.
If a variable x assumes n values x1, x2, x3,........xn then its standard deviation is given by
For a grouped frequency distribution, the standard deviation is given by
The above two formulas can be simplified to the following forms.
Sometimes the square of standard deviation, known as variance, is regarded as a measure of
dispersion. We have, then,
A relative measure of dispersion using standard deviation is given by coefficient of variation
(cv) which is defined as the ratio of standard deviation to the corresponding arithmetic mean,
expressed as a percentage.
Properties of Standard Devation
1. If all the observations assumed by a variable are constant i.e. equal, then the SD is zero. This means that if all the values taken by a variable x is k, say, then s = 0. This result applies to range as well as mean deviation.
2. Standard deviation remains unaffected due to a change of origin but is affected in the same ratio due to a change of scale. That is, if there are two variables x and y related as y = a + bx for any two constants a and b, then standard deviation of y is given by
Sy = |b|Sx
Example 1 :
Find the standard deviation and the coefficient of variation for the following numbers: 5, 8, 9, 2, 6.
Solution :
|
xi |
xi2 |
|
5 8 9 2 6 |
25 65 81 4 36 |
|
∑xi = 30 |
∑xi2 = 210 |
Formula for standard deviation for an ungrouped data :
s = √(∑xi2/n) - (∑xi/n)2
s = √(210/5) - (30/5)2
s = √42 - 62
s = √(42 - 36)
s = √6
s = 2.45
The coefficient of variation
= (Standard deviation/Arithmetic mean) x 100
= (2.45/6) x 100
= 40.83
Example 2 :
Find the standard deviation for the following distribution :
x
0 - 2
2 - 4
4 - 6
6 - 8
8 - 10
f
17
35
28
15
5
Solution :
x
fi
Mid-value (xi)
fixi
fixi2
0 - 2
2 - 4
4 - 6
6 - 8
8 - 10
17
35
28
15
5
1
3
5
7
9
17
105
140
105
45
17
315
700
735
405
Total
100
-
412
2172
Formula for standard deviation for a grouped frequency distribution :
s = √(∑xi2/n) - (∑xi/n)2
s = √(2172/100) - (412/100)2
s = √21.72 - (4.12)2
s = √21.72 - 16.97
s = √4.7456
s = 2.18
So, the required standard deviation is 2.18.
Example 3 :
Find the standard deviation of first n natural numbers.
Solution :
Let xi = 1, 2, 3, ..........n.
Arithmetic mean of first n natural numbers :
Standard deviation of first n natural numbers :
Example 4 :
Find the standard deviation of first 25 natural numbers.
Solution :
Standard deviation of first n natural numbers :
Substitute n = 25.
Example 5 :
If arithmetic mean and coefficient of variation of x are 10 and 40 respectively, what is the variance of (15 – 2x) ?
Solution :
Let y = 15 - 2x.
When x and y are related as y = a + bx, then
Sy = |b|Sx
Sy = |-2|Sx
Sy = 2Sx ----(1)
Coefficient of variation of x = 40 and mean of x = 10.
Coefficient of variation of x = 40
100 ⋅ (Sx/AM) = 40
100 ⋅ (Sx/10) = 40
10 ⋅ Sx = 40
Sx = 4
Sy = 2 ⋅ 4
Sy = 8
Variance of y = 82
Variance of (15 - 2x) = 64
Example 6 :
Answer each question using the list below
123, 100, 111, 124, 132, 154, 132, 160
- Mean =
- Median =
- Standard deviation =
- Range =
What does the standard deviation mean in this case?
Solution :
Mean :
Mean = (123 + 100 + 111 + 124 + 132 + 154 + 132 + 160)/8
= 1036/8
= 129.5
So, the required mean is 129.5
Median :
Total number of terms = 8 (even)
Arranging from least to greatest.
100, 111, 123, 124, 132, 132, 154, 160
median = (124 + 132)/2
= 128
so, the median is 128.
Standard deviation :
|
x 100 111 123 124 132 132 154 160 |
d 100 - 129.5 = -29.5 111 - 129.5 = -18.5 123 - 129.5 = -6.5 124 - 129.5 = -5.5 132 - 129.5 = 2.5 132 - 129.5 = 2.5 154 - 129.5 = 24.5 160 - 129.5 = 30.5 |
d2 870.25 342.25 42.25 30.25 6.25 6.25 600.25 930.25 |
∑d2 = 2828
s = √(∑d2/n) - (∑d/n)2
s = √(2828/8) - 02
s = √353.5
s = 18.8
So, the required stadard deviation is 18.8
Review
We may now have a review of the different measures of dispersion on the basis of their relative merits and demerits.
Standard deviation, like arithmetic mean, is the best measure of dispersion. It is rigidly defined, based on all the observations, not too difficult to compute, not much affected by sampling fluctuations and moreover it has some desirable mathematical properties.
All these merits of standard deviation make standard deviation as the most widely and commonly used measure of dispersion.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Recent Articles
-
Digital SAT Math Problems and Solutions (Part - 200)
Jul 03, 25 07:35 AM
Digital SAT Math Problems and Solutions (Part - 200) -
Digital SAT Math Problems and Solutions (Part - 199)
Jul 03, 25 07:24 AM
Digital SAT Math Problems and Solutions (Part - 199) -
Logarithm Questions and Answers Class 11
Jul 01, 25 10:27 AM
Logarithm Questions and Answers Class 11
