Measures of Dispersion
Dr Anshul Singh Thapa
MEASURES OF DISPERSION FROM�AVERAGE
Mean Deviation
MEAN DEVIATION
UNGROUPED DATA
DISCRETE SERIES
DIRECT METHOD
ASSUMED MEAN METHOD
FROM MEDIAN
GROUPED DATA
FROM ARITHMETIC MEAN
CONTINUOUS SERIES
DIRECT METHOD
SHORT CUT METHOD
FROM ARITHMETIC MEAN
SHORT CUT METHOD
SHORT CUT METHOD
FROM MEDIAN
FROM MEDIAN
FROM ARITHMETIC MEAN
DIRECT METHOD
DIRECT METHOD
MEAN DEVIATION
UNGROUPED DATA
DISCRETE SERIES
FROM MEDIAN
GROUPED DATA
FROM ARITHMETIC MEAN
CONTINUOUS SERIES
FROM ARITHMETIC MEAN
FROM MEDIAN
FROM MEDIAN
FROM ARITHMETIC MEAN
ΣI dx I
n
ΣI dx I
n
ΣI dx I
n
ΣI dx I
n
ΣI dx I
n
ΣI dx I
n
CALCULATION OF MEAN DEVIATION FOR UNGROUPED DATA
STEPS (FROM MEDIAN) | STEPS (FROM MEAN) |
I. Arrange the data in ascending order. II. Calculate the median Me = Size of (N + 1/ 2)th item III. Find out the deviation of the items from median. Ignore (-) and (+) signs. Express it by I dm I sign. IV. Sum up the deviations, and express the same by using the relevant formula to find out the required answer. MDm = ΣI dm I N V. After that solve the question with the help of formula. |
MDx = ΣI dx I N |
Calculation of Mean Deviation from Arithmetic Mean for ungrouped data – Direct Method
Steps:
Example
X | |
2 | |
4 | |
7 | |
8 | |
9 | |
| |
Example
X | |
2 | |
4 | |
7 | |
8 | |
9 | |
30 | |
Example
X | |dx| = X – X = 6 |
2 | |
4 | |
7 | |
8 | |
9 | |
| |
Example
X | |dx| = X – X = 6 |
2 | 4 |
4 | 2 |
7 | 1 |
8 | 2 |
9 | 3 |
| |
Example
X | |dx| = X – X = 6 |
2 | 4 |
4 | 2 |
7 | 1 |
8 | 2 |
9 | 3 |
| Σ|dx| = 12 |
Example
X | |dx| = X – X = 6 |
2 | 4 |
4 | 2 |
7 | 1 |
8 | 2 |
9 | 3 |
| Σ|dx| = 12 |
N 5
Calculation of Mean Deviation from Arithmetic Mean for ungrouped data – Assumed Mean Method
Example
MDx = Σ|dx| + (X – Ax) (ΣfB – ΣfA)
n
X | |dx| = Ax – X |
2 | 5 |
4 | 3 |
7 | 0 |
8 | 1 |
9 | 2 |
| Σ|dx| = 11 |
MDx = 11 + (6 – 7) (3 – 2) = 12 = 2.4
5 5
Mean Deviation from median for ungrouped data
Direct Method
Mean Deviation from median for ungrouped data
Median = Size of N + 1 th item = 5+ 1 = 3rd item
2 2
Size of the third item is 4400
M.D.median = |dx| = 1200 = 240
n 5
Income (X) | |
4000 | |
4200 | |
4400 | |
4600 | |
4800 | |
| |
Mean Deviation from median for ungrouped data
Median = Size of N + 1 th item = 5+ 1 = 3rd item
2 2
Size of the third item is 4400
Income (X) | |
4000 | |
4200 | |
4400 | |
4600 | |
4800 | |
| |
Mean Deviation from median for ungrouped data
Median = Size of N + 1 th item = 5+ 1 = 3rd item
2 2
Size of the third item is 4400
Income (X) | Deviation from Median 4400 |dx| |
4000 | 400 |
4200 | 200 |
4400 | 0 |
4600 | 200 |
4800 | 400 |
| |
Mean Deviation from median for ungrouped data
Median = Size of N + 1 th item = 5+ 1 = 3rd item
2 2
Size of the third item is 4400
Income (X) | Deviation from Median 4400 |dx| |
4000 | 400 |
4200 | 200 |
4400 | 0 |
4600 | 200 |
4800 | 400 |
| Σ|dx| = 1200 |
Mean Deviation from median for ungrouped data
Median = Size of N + 1 th item = 5+ 1 = 3rd item
2 2
Size of the third item is 4400
M.D.median = |dx| = 1200 = 240
n 5
Income (X) | Deviation from Median 4400 |dx| |
4000 | 400 |
4200 | 200 |
4400 | 0 |
4600 | 200 |
4800 | 400 |
| Σ|dx| = 1200 |
Mean Deviation from median for ungrouped data
Short-cut method
MDmedian = Σ|dx| + (Median – A) (ΣfB – ΣfA)
n
Mean Deviation from mean for discrete series
X | 10 | 11 | 12 | 13 | 14 |
f | 3 | 12 | 18 | 12 | 3 |
x | f | | | |
10 | 3 | | | |
11 | 12 | | | |
12 | 18 | | | |
13 | 12 | | | |
14 | 3 | | | |
| | | | |
x | f | | | |
10 | 3 | | | |
11 | 12 | | | |
12 | 18 | | | |
13 | 12 | | | |
14 | 3 | | | |
| Σf = 48 | | | |
x | f | fx | | |
10 | 3 | 30 | | |
11 | 12 | 132 | | |
12 | 18 | 216 | | |
13 | 12 | 156 | | |
14 | 3 | 42 | | |
| Σf = 48 | | | |
x | f | fx | | |
10 | 3 | 30 | | |
11 | 12 | 132 | | |
12 | 18 | 216 | | |
13 | 12 | 156 | | |
14 | 3 | 42 | | |
| Σf = 48 | Σfx = 576 | | |
Σf 48
x | f | fx | | |
10 | 3 | 30 | | |
11 | 12 | 132 | | |
12 | 18 | 216 | | |
13 | 12 | 156 | | |
14 | 3 | 42 | | |
| Σf = 48 | Σfx = 576 | | |
Σf 48
x | f | fx | |d| | |
10 | 3 | 30 | 2 | |
11 | 12 | 132 | 1 | |
12 | 18 | 216 | 0 | |
13 | 12 | 156 | 1 | |
14 | 3 | 42 | 2 | |
| Σf = 48 | Σfx = 576 | | |
Σf 48
x | f | fx | |d| | f|d| |
10 | 3 | 30 | 2 | 6 |
11 | 12 | 132 | 1 | 12 |
12 | 18 | 216 | 0 | 0 |
13 | 12 | 156 | 1 | 12 |
14 | 3 | 42 | 2 | 6 |
| Σf = 48 | Σfx = 576 | | |
Σf 48
x | f | fx | |d| | f|d| |
10 | 3 | 30 | 2 | 6 |
11 | 12 | 132 | 1 | 12 |
12 | 18 | 216 | 0 | 0 |
13 | 12 | 156 | 1 | 12 |
14 | 3 | 42 | 2 | 6 |
| Σf = 48 | Σfx = 576 | | 36 |
Σf 48
Σf 48
x | f | fx | |d| | f|d| |
10 | 3 | 30 | 2 | 6 |
11 | 12 | 132 | 1 | 12 |
12 | 18 | 216 | 0 | 0 |
13 | 12 | 156 | 1 | 12 |
14 | 3 | 42 | 2 | 6 |
| Σf = 48 | Σfx = 576 | | 36 |
Mean Deviation from mean for discrete series
Height | 158 | 159 | 160 | 161 | 162 | 163 | 164 | 165 | 166 |
Students | 15 | 20 | 32 | 35 | 33 | 22 | 20 | 10 | 8 |
Mean Deviation Mean = Σf|d| = 338.59 = 1.74
Σf 195
Mean Deviation from median for discrete series
X | 10 | 11 | 12 | 13 | 14 |
f | 3 | 12 | 18 | 12 | 3 |
x | f | | | |
10 | 3 | | | |
11 | 12 | | | |
12 | 18 | | | |
13 | 12 | | | |
14 | 3 | | | |
| Σf = 48 | | | |
x | f | | | c.f |
10 | 3 | | | 3 |
11 | 12 | | | 15 |
12 | 18 | | | 33 |
13 | 12 | | | 45 |
14 | 3 | | | 48 |
| Σf = 48 | | | |
2
Size of 5th item lies in 33 cumulative frequecy therefore the median is 12
x | f | | | c.f |
10 | 3 | | | 3 |
11 | 12 | | | 15 |
12 | 18 | | | 33 |
13 | 12 | | | 45 |
14 | 3 | | | 48 |
| Σf = 48 | | | |
2
Size of 5th item lies in 33cumulative frequecy therefore the median is 12
x | f | |d| | | c.f |
10 | 3 | 2 | | 3 |
11 | 12 | 1 | | 15 |
12 | 18 | 0 | | 33 |
13 | 12 | 1 | | 45 |
14 | 3 | 2 | | 48 |
| Σf = 48 | | | |
2
Size of 5th item lies in 33cmlativefrequecy therefore the median is 12
x | f | |d| | f|d| | c.f |
10 | 3 | 2 | 6 | 3 |
11 | 12 | 1 | 12 | 15 |
12 | 18 | 0 | 0 | 33 |
13 | 12 | 1 | 12 | 45 |
14 | 3 | 2 | 6 | 48 |
| Σf = 48 | | Σfd = 36 | |
2
Size of 5th item lies in 33cmlativefrequecy therefore the median is 12
Σf 48
x | f | |d| | f|d| | c.f |
10 | 3 | 2 | 6 | 3 |
11 | 12 | 1 | 12 | 15 |
12 | 18 | 0 | 0 | 33 |
13 | 12 | 1 | 12 | 45 |
14 | 3 | 2 | 6 | 48 |
| Σf = 48 | | Σfd = 36 | |
Mean Deviation from Mean for�Continuous distribution
Steps:
Example
Class – interval | Frequency |
10 – 20 | 5 |
20 – 30 | 8 |
30 – 50 | 16 |
50 – 70 | 8 |
70 – 80 | 3 |
| 40 |
Class Interval | f | | | | |
10 – 20 | 5 | | | | |
20 – 30 | 8 | | | | |
30 – 50 | 16 | | | | |
50 – 70 | 8 | | | | |
70 – 80 | 3 | | | | |
| | | | | |
Class Interval | f | | | | |
10 – 20 | 5 | | | | |
20 – 30 | 8 | | | | |
30 – 50 | 16 | | | | |
50 – 70 | 8 | | | | |
70 – 80 | 3 | | | | |
| Σf = 40 | | | | |
Class Interval | f | Mid point m | | | |
10 – 20 | 5 | 15 | | | |
20 – 30 | 8 | 25 | | | |
30 – 50 | 16 | 40 | | | |
50 – 70 | 8 | 60 | | | |
70 – 80 | 3 | 75 | | | |
| Σf = 40 | | | | |
Class Interval | f | Mid point m | fm | | |
10 – 20 | 5 | 15 | 75 | | |
20 – 30 | 8 | 25 | 200 | | |
30 – 50 | 16 | 40 | 640 | | |
50 – 70 | 8 | 60 | 480 | | |
70 – 80 | 3 | 75 | 225 | | |
| Σf = 40 | | | | |
Class Interval | f | Mid point m | fm | | |
10 – 20 | 5 | 15 | 75 | | |
20 – 30 | 8 | 25 | 200 | | |
30 – 50 | 16 | 40 | 640 | | |
50 – 70 | 8 | 60 | 480 | | |
70 – 80 | 3 | 75 | 225 | | |
| Σf = 40 | | Σfm = 1620 | | |
Class Interval | f | Mid point m | fm | | |
10 – 20 | 5 | 15 | 75 | | |
20 – 30 | 8 | 25 | 200 | | |
30 – 50 | 16 | 40 | 640 | | |
50 – 70 | 8 | 60 | 480 | | |
70 – 80 | 3 | 75 | 225 | | |
| Σf = 40 | | Σfm = 1620 | | |
Σf 40
Σf 40
Class Interval | f | Mid point m | fm | |d| | |
10 – 20 | 5 | 15 | 75 | 25.5 | |
20 – 30 | 8 | 25 | 200 | 15.5 | |
30 – 50 | 16 | 40 | 640 | 0.5 | |
50 – 70 | 8 | 60 | 480 | 19.5 | |
70 – 80 | 3 | 75 | 225 | 34.5 | |
| Σf = 40 | | Σfm = 1620 | | |
Σf 40
Σf 40
Class Interval | f | Mid point m | fm | |d| | f|d| |
10 – 20 | 5 | 15 | 75 | 25.5 | 127.5 |
20 – 30 | 8 | 25 | 200 | 15.5 | 124.0 |
30 – 50 | 16 | 40 | 640 | 0.5 | 8.0 |
50 – 70 | 8 | 60 | 480 | 19.5 | 156.0 |
70 – 80 | 3 | 75 | 225 | 34.5 | 103.5 |
| Σf = 40 | | Σfm = 1620 | | |
Σf 40
Σf 40
Class Interval | f | Mid point m | fm | |d| | f|d| |
10 – 20 | 5 | 15 | 75 | 25.5 | 127.5 |
20 – 30 | 8 | 25 | 200 | 15.5 | 124.0 |
30 – 50 | 16 | 40 | 640 | 0.5 | 8.0 |
50 – 70 | 8 | 60 | 480 | 19.5 | 156.0 |
70 – 80 | 3 | 75 | 225 | 34.5 | 103.5 |
| Σf = 40 | | Σfm = 1620 | | Σf|d| = 519.0 |
Σf 40
Σf 40
Class Interval | f | Mid point m | fm | |d| | f|d| |
10 – 20 | 5 | 15 | 75 | 25.5 | 127.5 |
20 – 30 | 8 | 25 | 200 | 15.5 | 124.0 |
30 – 50 | 16 | 40 | 640 | 0.5 | 8.0 |
50 – 70 | 8 | 60 | 480 | 19.5 | 156.0 |
70 – 80 | 3 | 75 | 225 | 34.5 | 103.5 |
| Σf = 40 | | Σfm = 1620 | | Σf|d| = 519.0 |
Σf 40
Σf 40
Mean Deviation from median for�Continuous distribution
Class Interval | f | | | |
20 – 30 | 5 | | | |
30 – 40 | 10 | | | |
40 – 60 | 20 | | | |
60 – 80 | 9 | | | |
80 – 90 | 6 | | | |
| Σf = 50 | | | |
Mean Deviation from median for�Continuous distribution
Class Interval | Mid Value | f | | | |
20 – 30 | 25 | 5 | | | |
30 – 40 | 35 | 10 | | | |
40 – 60 | 50 | 20 | | | |
60 – 80 | 70 | 9 | | | |
80 – 90 | 85 | 6 | | | |
| | Σf = 50 | | | |
Mean Deviation from median for�Continuous distribution
Class Interval | Mid Value | f | c.f | | |
20 – 30 | 25 | 5 | 5 | | |
30 – 40 | 35 | 10 | 15 | | |
40 – 60 | 50 | 20 | 35 | | |
60 – 80 | 70 | 9 | 44 | | |
80 – 90 | 85 | 6 | 50 | | |
| | Σf = 50 | | | |
Median Class: N/2 = 50/2 = 25
Median = L + (N/2 - c.f.) × i
f
Median = 40 + (25 - 15) x 20 = 50
20
Mean Deviation from median for�Continuous distribution
Class Interval | Mid Value | f | c.f | | |
20 – 30 | 25 | 5 | 5 | | |
30 – 40 | 35 | 10 | 15 | | |
40 – 60 | 50 | 20 | 35 | | |
60 – 80 | 70 | 9 | 44 | | |
80 – 90 | 85 | 6 | 50 | | |
| | Σf = 50 | | | |
Mean Deviation from median for�Continuous distribution
Class Interval | Mid Value | f | c.f | |d| | |
20 – 30 | 25 | 5 | 5 | 25 | |
30 – 40 | 35 | 10 | 15 | 15 | |
40 – 60 | 50 | 20 | 35 | 0 | |
60 – 80 | 70 | 9 | 44 | 20 | |
80 – 90 | 85 | 6 | 50 | 35 | |
| | Σf = 50 | | | |
Mean Deviation from median for�Continuous distribution
Class Interval | f | c.f | |d| | f|d| |
20 – 30 | 5 | 5 | 25 | 125 |
30 – 40 | 10 | 15 | 15 | 150 |
40 – 60 | 20 | 35 | 0 | 0 |
60 – 80 | 9 | 44 | 20 | 180 |
80 – 90 | 6 | 50 | 35 | 210 |
| Σf = 50 | | | |
Mean Deviation from median for�Continuous distribution
Class Interval | f | c.f | |d| | f|d| |
20 – 30 | 5 | 5 | 25 | 125 |
30 – 40 | 10 | 15 | 15 | 150 |
40 – 60 | 20 | 35 | 0 | 0 |
60 – 80 | 9 | 44 | 20 | 180 |
80 – 90 | 6 | 50 | 35 | 210 |
| Σf = 50 | | | Σf|d| = 665 |
Median = L + (N/2 - c.f.) × i
f
Median = 40 + (25 - 15) x 20 = 50
20
Mean Deviation median = Σf|d| = 665
Σf 50