Dispersion is also known as scatter, spread and variation.
Sr.No 
Sneh 
Amit 
Minna 
1 
15000 
7000 
5000 
2 
15000 
10000 
2000 
3 
15000 
14000 
8000 
4 
15000 
17000 
10000 
5 
 
20000 
50000 
6 
 
22000 
 
Total income 
60000 
90000 
75000 
Average income 
15000 
15000 
15000 
Arithmetic Mean
Solution:
Range =LS
=207
=13
Coeff. of Range =LS/L+S*100
=207/20+7*100
 Discrete series
x  f 
5  3 
10  7 
15  5 
20  12 
25  9 
Calculate range and coeff. of range?
Solution:
255
=20
Coeff. of Range
LS/L+S*100
=255/25+5*100
=66.67%
 Continuous Series
class  f 
2025  2 
2530  5 
3035  9 
3540  4 
4045  7 
Range= LS
4520
=25
Coeff. of range=LS/L+S*100
=4520/45+20*100
=38.46%
Exclusive
class  f 
2024  2 
2529  5 
3034  9 
3539  4 
4044  7 
class  f 
19.524.5  2 
24.529.5  5 
29.534.5  9 
34.539.5  4 
39.544.5  7 
Calculate range and Coeff. of Range.?
=44.519.5
=25
Coeff. of Range
=LS/L+S*100
=44.519.5/44.5+19.5*100
=39.06%
2. Quartile Deviation
In such a situation, if the entire data is divided into four equal parts, each containing 25% of the values, we get the values of quartiles and median. The upper and lower quartiles (Q3 and Q1 , respectively) are used to calculate interquartile range which is Q3 – Q1 .
Inter quartile range is based upon middle 50% of the values in a distribution and is, therefore, not affected by extreme values. Half of the interquartile range is called quartile deviation (Q.D.).
Calculate range and Q.D. of the following observations: 20, 25, 30, 41, 29, 35, 39, 48, 51, 60 and 70
Sol. Firstly arrange value in ascending or descending order.
For Q.D., we need to calculate values of Q3 and Q1
Q1 is the size of n +1 th/4th value.
(11+1)/4=3
n being 11, Q1 is the size of 3rd value. As the values are already arranged in ascending order, it can be seen that Q1 , the 3rd value is 29.
Similarly,
Q3 is size of 3( n +1)/4)value;
3(11+1)/4
36/4=9
i.e. 9th value which is 51. Hence Q3 = 51 =
Q.D=Q3Q1/2
=(5129)/2
Month  Income 
Jan  139 
Feb  140 
Mar  140 
Apr  141 
May  141 
Jun  142 
July  142 
Aug  143 
Sep  143 
Oct  144 
Nov  144 
Dec  145 
x  f 
5  2 
2  2 
9  1 
10  5 
15  3 
x  f  c.f 
2  1  1 
5  2  3 
9  2  5 
10  5  10 
15  3  13 
Q3=3(n+1)/4
=3(13/4)
=10.5 th term
10th term+0.5(11th 10th)
=10+.5(5)
=12.5
Q.D= (Q3Q1)/2
=5.5/2
=2.75
Coeff of Q.D= (Q3Q1/Q3+Q1)*100
=(12.57/12.5+7.5)100
=28.2%
3)Continuous series
Class interval C I  No of student (f) 
010  5 
1020  8 
2040  16 
4060  7 
6090  4 
40 
Class interval C I  No of student (f)  CF 
010  5  5 
1020  8  13 
2040  16  29 
4060  7  36 
6090  4  40 
40 
Q1 is the size of n th/ 4 value in a continuous series.
Thus, N=40 , 40/4=10, it is the size of the 10th value.
The class containing the 10th value is 10–20. Hence, Q1 lies in class 10–20. Now, to calculate the exact value of Q1 , the following formula is used:
Where L = 10 (lower limit of the relevant Quartile class)
and f = 8 (frequency of the quartile class)
Q1= 10+(105)/8*10=16.25
Similarly, Q3 is the size of 3n/4 th value
value; i.e., 30th value, which lies in class 40–60. Now using the formula for Q3 , its value can be calculated as follows:
Q3=40+(3029)/7*20=42.87
QD=Q3Q1/2
(42.8716.25)/2=13.31
Standard Deviation
The standard deviation is the average amount of variability in your dataset. It tells you, on average, how far each value lies from the mean.
A high standard deviation means that values are generally far from the mean, while a low standard deviation indicates that values are clustered close to the mean.
Standard deviation is a useful measure of spread for normal distributions.
In normal distributions, data is symmetrically distributed with no skew. Most values cluster around a central region, with values tapering off as they go further away from the center. The standard deviation tells you how spread out from the center of the distribution your data is on average.
Many scientific variables follow normal distributions, including height, standardized test scores, or job satisfaction ratings. When you have the standard deviations of different samples, you can compare their distributions using statistical tests to make inferences about the larger populations they came from.
Standard deviation formulas
Formula  Explanation 


we have given data set below:
Data Set(score)  
46  69  32  60  52  41 
1.Find the mean.
x̅ = (46 + 69 + 32 + 60 + 52 + 41) ÷ 6 = 50
2.Find each score’s deviation from the mean.
Score  Deviation from the mean 
46  46 – 50 = 4 
69  69 – 50 = 19 
32  32 – 50 = 18 
60  60 – 50 = 10 
52  52 – 50 = 2 
41  41 – 50 = 9 
3. Square each deviation from the mean and find the sum of square.
Score  Deviation from the mean  (Xx̅)2 
46  46 – 50 = 4  16 
69  69 – 50 = 19  361 
32  32 – 50 = 18  324 
60  60 – 50 = 10  100 
52  52 – 50 = 2  4 
41  41 – 50 = 9  81 
886 
4.Put the values in the formula
√ Σ(Xx̅)2/n
Coeff. of S.D= σ/ x̅
=.2428
Variances= (σ)2=147.6
Calculate Standard Deviation by Assumed Mean Method
10,12,13,15,20
Sol
Firstly we will find the some of d and sum of square of d according to formula,
Here we assume d= 13(middle value) ,
x  d=x13  d2 
10  3  9 
12  1  1 
13  0  0 
15  4  16 
20  7  49 
total  Σd=5  Σd2=63 
Now we will put the values in formula
=√ 63/5(5/5)2
=√12.61
=√11.6
=3.4
Coeff. of S.D= σ/ x̅
=3.4/14
=.2428
Variance=(σ)2
11.6
Coeff. of Variance= (σ/ x̅)*100
=.2428*100
=24.28
2.Continuous Series
σ= (√ Σfd2/N(Σfd/N)2 ) x h
∑ fd
i(xifd
x¯)2N
Class  f 
010  15 
1020  15 
2030  23 
3040  22 
4050  25 
5060  10 
6070  5 
7080  10 
Sol
Class  f  Mid Value  d  d2  fd  fd2 
010  15  5  3  9  45  135 
Oct20  15  15  2  4  30  60 
2030  23  25  1  1  23  23 
3040  22  35  0  0  0  0 
4050  25  45  1  1  25  25 
5060  10  55  2  4  20  40 
6070  5  65  3  9  15  45 
7080  10  75  4  16  40  160 
N=125  2  488 
σ=(√ 488/125(2/125)2 ) *10
Mean=A+(Σfd/Σf) x h
35+(2/125)*10
35+0.16
35.16
σ/ x̅
=19.16/35.16
=0.5620
Variance=(σ)2
=(19.76)2
=390.45
Coeff. of Varience= σ/ x̅*100
=56.2%
Mean Deviation
MEASURES OF DISPERSION FROM AVERAGE
Since the average is a central value, some deviations are positive and some are negative. If these are added as they are, the sum will not reveal anything. In fact, the sum of deviations from Arithmetic Mean is always zero. Look at the following two sets of values.
Suppose a college is proposed for students of five towns A, B, C, D and E which lie in that order along a road. Distances of towns in kilometers from town A and number of students in these towns are given below:
Town  Distance from town  Students 
a  0  90 
b  2  150 
c  6  100 
d  14  200 
e  18  80 
620 
Now, if the college is situated in town A, 150 students from town B will have to travel 2 kilometers each (a total of 300 kilometers) to reach the college. The objective is to find a location so that the average distance travelled by students is minimum.
You may observe that the students will have to travel more, on an average, if the college is situated at town A or E. If on the other hand, it is somewhere in the middle, they are likely to travel less. Mean deviation is the appropriate statistical tool to estimate the average distance travelled by students. Mean deviation is the average. The average used is either the arithmetic mean or median .
Calculation of Mean Deviation from Arithmetic Mean for ungrouped data
Direct Method Steps:
(i) The A.M. of the values is calculated
(ii) Difference between each value and the A.M. is calculated. All differences are considered positive. These are denoted as d
(iii) The A.M. of these differences (called deviations) is the Mean Deviation
A.M=SUM of X/N
30/5=6
X  I XX bar=d I 
2  4 
4  2 
7  1 
8  2 
9  3 
12 
M.D(x) =12/5=2.4
Mean Deviation from median for ungrouped data.
Method Using the values in Example 3, M.D. from the Median can be calculated as follows
X  Id I (Xmedian) 
2  5 
4  3 
7  0 
8  1 
9  2 
11 
Steps:
(i) Calculate the median which is 7.
(ii) Calculate the absolute deviations from median, denote them as d.
(iii) Find the average of these absolute deviations.
M.D (median)=11/5=2.21
Mean Deviation from Mean for Continuous Distribution
Example:
Profits of Companies (Rs in lakh) Class intervals  Number of companies 
1020  5 
2030  8 
3050  16 
5070  8 
7080  3 
40 
Steps:
(i) Calculate the mean of the distribution.
(ii) Calculate the absolute deviations d of the class midpoints from the mean.
(iii) Multiply each d value with its corresponding frequency to get fd values. Sum them up to get Σ fd.
(iv) Apply the following formula,
M.D( x ) = Σ f  d  / Σ f
Profits of Companies (Rs in lakh) Class intervals  Number of companies(F)  Midpoint  I d I  fI d I 
1020  5  15  25  125 
2030  8  25  15  120 
3050  16  40  0  0 
5070  8  60  20  152 
7080  3  75  35  102 
40  499 
M.D( x ) = Σ f  d  / Σ f
=499/40
=12.47
Means deviation : comment
Mean deviation is based on all values. A change in even one value will affect it. Mean deviation is the least when calculated from the median i.e., it will be higher if calculated from the mean. However it ignores the signs of deviations and cannot be calculated for open ended distributions.
As you can see in above example ,Mean is same but dispersion is very at every point .If you have another value which reflects the quantum of variation in values, your understanding of a distribution improves considerably. For example, per capita income gives only the average income. A measure of dispersion can tell you about income inequalities, thereby improving the understanding of the relative standards of living enjoyed by different strata of society.
Thank you
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