Statistical-Measure-of-Dispersion -

Statistical Measure of Central Tendency

Statistical Measures of Dispersion

In this part, we discuss about Dispersion, computation, problems and solutions related to Dispersion

Dispersion
Range
Quartile Deviation
Mean Deviation
Standard Deviation

Statistical Dispersion

Statistical Dispersion is the degree of the scatter or the variation of the variables about a central value.

Dispersion shows mean deviation or scatteredness. A measure of central tendency alone is not sufficient to give complete idea of the distribution. The study of dispersion helps to know whether the distribution is homogenous (evenly spread) or non-homogenous (non-evenly spread). It explains the spread or scattering of the individual values around the central value.

Characteristics of Dispersion

Dispersion measurement is done through different methods through which variations can be measured in quantitative manner. Dispersion deals with a statistical series. Dispersion indicates the degree or extent to which the various items of a series deviate from its central value. Dispersion indicates more reliability or otherwise of the average value of a series.

Measures of Dispersion

Absolute measures of Dispersion : Range, Quartile Deviation (or Semi-interquartile range), Mean Deviation (or Average Deviation), Standard Deviation
Relative measures of Dispersion : Coefficient of Quartile Deviation, Coefficient of Dispersion, Coefficient of Variation

Range

For a set of observations, Range is the difference between the extremes.

So, Range (R)= Maximum (or largest) Value (L_(L)) – Minimum (or Smallest) value (L_(S).), i.e L_(L)– L_(S).

Co-efficient of Range = (Maximum Value – Minimum value) / (Maximum Value + Minimum value), ie. [(L_(L)– L_(S).) / (L_(L)+ L_(S).)]

Range is normally used in the field of quality control, stock-market fluctuations etc

Advantages and Disadvantages of Range

– Advantages: Range is easy to understand and is simple to compute.

– Disadvantages: It is very much effected by the extreme values. It does not depend on all the observations, but only on the extreme values. Range cannot be computed in case of open-end distribution.

Range – Problems

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Ex. The marks obtained by 6 students were 22, 10, 14, 9, 38, 40. Find the Range. If the highest mark is omitted, find the percentage change in range.

Here maximum mark = 40, minimum marks = 9. Range = 40 – 9 = 31 marks.

If the highest mark 40 is omitted, then amongst the remaining, maximum marks is 38.

So, range (revised) = 38 – 11 = 27 marks. Change in range = 31 – 27 = 4 marks.

Percentage change = (4/31) x 100 = 12.90 %

Ex. Compute the Range and co-efficient from the following data.

Marks :	10-20	20-30	30-40	40-50	50-60	60-80
No. of students	3	3	5	6	6	8

Range = L_(L)– L_(S). = 80-10 = 70

Co-efficient of Range = [(L_(L)– L_(S).) / (L_(L)+ L_(S).)] = [(80-10)/ (80+10)] = 70/90 = .77

Quartile Deviation

The Quartile Deviation is half of the difference between the upper and lower quartiles

Quartile Deviation = = (Q₃ – Q₁) /2

Inter-quartile range, indicates the difference between two quartiles and its half indicates Semi-interquartile range (semi stands for half).

Since 50% of the observation lie between two quartiles, as such Inter-quartile range gives a fair measure of variability. Interquartile range also does not depend on all observations, and it is effected by fluctuations.

Coefficient of Quartile Deviation

Quartile Deviation (Q.D.) is an absolute measure of dispersion. If it is divided by average value of two quartiles, we will find Coefficient of Quartile Deviation (a relative measure of dispersion).

Coefficient of quartile deviation = [{(Q₃ – Q₁) /2} / {(Q₃ + Q₁) /2}] = [{(Q₃ – Q₁)} / {(Q₃ + Q₁)}]

Merits and Demerits of Quartile Deviation

– Merits: Quartile Deviation is superior to Range as measures of dispersion. In case of open-end distributions, it can be computed. It is not effected by the presence of extreme values.

– Demerits: Quartile deviation is neither based on all the observations, nor it is capable of further algebraic treatment. Its value is much effected by sampling fluctuations. It is not a measure of dispersion, particularly for series having considerable variation.

Ex. Find the quartile deviation and coefficient of quartile deviation of observations of Marks :

10, 11, 13, 16, 18, 20, 26, 27, 29, 31, 32

Here, n = 11, and observations are arranged in order.

Q₁ ₌ size of [(n+1) /4] th Item = size of [11 +1) /4] th or 3rd Item = 13 marks

Q₃ ₌ size of [3 (n+1) /4] th Item = size of 3x[11 +1) /4]th or 9th Item = 29 marks

Quartile Deviation (Q.D.) = [(29 – 13) /2] = 16/2 = 8 marks

Coefficient of quartile Deviation = (29-13) / (29+13) = 16/42 = .3809

Quartile Deviation – Problems

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Ex. Compute quartile deviation and its co-efficient.

Quartile Frequency Table

Daily Wages	Freq	Cum Freq
12.50 – 17.50 17.50 – 22.50 22.50 – 27.50 27.50 – 32.50 32.50 – 37.50	4 6 4 5 9	4 10 14 19 28
	N=28

Q₁= Value of (N/4) th term = Value of (28/4) th term = Value of 7th term, which lies between 17.50- 22.50 class interval. By interpolation, Q₁ = L₁+ {(L₂– L₁ ) / f₁} x ((q₁– c) = 17.50 + [{(22.50 – 17.50 )} / 6 x (7-4)] = 17.50 + (5/6) x 3 = 17.50 + 5/2 = 20

Q3 = Value of 3 x (N/4) the item = value of 3 x (28/4) th item, or value of 21st term, which lies in 32.50- 37.50 class interval. By interpolation, Q₃ = L₁+ {(L₂– L₁ ) / f₁} x ((q₃– c) = 32.50 + {(37.50 – 32.50 ) / 9}x (28-19) = 32.50 + (5/ 9)x 9 = 32.5 + 5 = 37.5

Quartile Deviation (QD) = (Q₃– Q₁) /2 = (37.50 – 20) / 2 = 17.5 /2 = 8.75

Co-efficient of QD = (Q₃– Q₁) / (Q₃+ Q₁) = (37.50 – 20) / (37.50 + 20) = 17.50 / 57.50 = .3043

Mean Deviation

Mean Deviation of a series is the arithmetic average of the deviations of various items from the median or mean of that series.

Median is preferred, since the sum of the deviations from the median is less than that from the mean. So the values of mean deviation calculated from median is usually less than that calculated from mean. Mode is not considered, as its values is indeterminate.

Mean deviation is known as First Moment of dispersion.

Calculation of Mean Deviation.

For individual observation: M.D. = $\displaystyle \frac{{\sum{{\left| D \right|}}}}{n}$ , where $\displaystyle \left| D \right|$ is absolute value of D (D within two vertical lines) denotes deviations from mean (or median), ignoring algebraic signs (i.e., + and -).

For Discrete Series: M.D. = $\displaystyle \frac{{\sum{{f\left| D \right|}}}}{{\sum{f}}}$ , where $\displaystyle \left| D \right|$ = deviations from mean (or median) ignoring ± sign.

Steps to find M.D. : 1.Find mean or median; 2. Take deviations ignoring + signs; 3. Get total of deviations; 4. Divide the above total by the number of items.

Coefficient of Mean Deviation

About Mean, Coefficient of M.D = MD / Mean

About Median, Coefficient of M.D = MD / Median

Merits and Demerits of Mean Deviation

Merits: Mean Deviation is based on all the observations. Any change in any item would change the value of mean deviation. Mean Deviation is readily understood. Mean Deviation is the average of the deviations from a measure of central tendency. Mean deviation is less affected by the extreme items than the standard deviation.

Demerits: Mean deviation ignores the algebraic signs of the deviations, and as such it is not capable of further algebraic treatment. Mean Deviation is not an accurate measure, particularly when it is calculated from mode. Mean Deviation is not as popular as standard deviations.

Mean Deviation & its co-efficient – Problems

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Ex. Compute the mean deviation of following data about mean and median

2, 6, 11, 14, 16, 19, 23.

Deviation about Mean and Median
About Mean			About Median
Serial no.	x (Rs.)	Dev. from A.M.	Serial no.	x (Rs.)	Dev. from Med.
1 2 3 4 5 6 7	2 6 11 14 16 19 23	11 7 2 1 3 6 10	1 2 3 4 5 6 7	2 6 11 14 16 19 23	12 8 3 0 2 5 9
Total	–	40	Total	–	39

A.M = 1/7(2 + 6 + 11 + 14 + 16 + 19 + 23) = 1/7 x91 = 13

Median = size of (7+1) /2 th term = size of 4th term = 14

Median deviation (about Mean) = ( $\displaystyle {\sum{{\left| D \right|}}}$ ) / n = 40/7 = 5.71

Median deviation (about Median) = ( $\displaystyle {\sum{{\left| D \right|}}}$ ) / n = 39/7 = 5.57

Note: The sum of deviation (å çDç) about median is 39, less than çDç about mean (i.e 40.). M.D. about Median (i.e., 5.57) is less than about Mean (i.e., 5.71).

Coefficient of M.D (about Mean) = M.D / Mean = 5.71 / 13 = .44

Coefficient of M.D (about Median) = M.D / Median = 5.57 / 14 = .40

Mean Deviation & its co-efficient – Problems

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Ex. Compute mean deviation and coefficient of dispersion of the following series:

(Marks) x :	5	10	15	20	25	Total
(Student) f :	6	7	8	11	8	40

Computation of Mean Deviation

A.M = A + { $\displaystyle \frac{{\sum{{fd'}}}}{{\sum{f}}}$ } x I = 15 + [( $\displaystyle \frac{8}{{40}}$ ) x 5 ]= 15+1 =16

MD = $\displaystyle {\frac{{\sum{{f\left| D \right|}}}}{{\sum{f}}}}$ = $\displaystyle \frac{{232}}{{40}}$ = 5.8

Coefficient of dispersion (about mean) = M.D / Mean = $\displaystyle \frac{{5.8}}{{16}}$ = .363

Computation of Mean Deviation

Median =size of the [latex]\displaystyle \frac{{\left( {40+1} \right)}}{2}[/latex] th item = size of 20.5th item₌15

MD = ( $\displaystyle \frac{{\sum{{f\left| D \right|}}}}{{\sum{f}}}$ ) = $\displaystyle \frac{{230}}{{40}}$ = 5.75

Coefficient of dispersion (about median) = M.D / Median = $\displaystyle \frac{{5.75}}{{15}}$ = .383

Mean Deviation & its co-efficient

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Ex. Compute mean deviation from mean from the following data (Continuous series)

Class Interval :	0 -10	10 -20	20 -30	30 -40	40 -50	50 -60	60 -70
Frequency :	6	12	10	10	5	4	5

Computation of Mean

Class Interval	Mid Pt (X)	f	d=x-35 (A)	fd
0-10	5	6	-30	-180
10-20	15	12	-20	-240
20-30	25	10	-10	-100
30-40	35	10	0	0
40-50	45	5	10	50
50-60	55	4	20	80
60-70	65	5	30	150

$\displaystyle \overline{x}=\frac{{\left( {\sum{{fd}}} \right)}}{N}$ + A = 35 + ( $\displaystyle \frac{{-240}}{{52}}$ ) = 35-4.6= 30.4

Computation of Mean Deviation

Mean deviation from the mean = 1 $\displaystyle \frac{{\sum{{f\left( {x-\overline{x}} \right)}}}}{n}$ = 1 X ( $\displaystyle \frac{{780.80}}{{52}}$ ) =16

Co-efficient of Mean deviation from the mean = M.D / Mean = $\displaystyle \frac{{16}}{{29}}$ = 0.5517

Standard Deviation

Standard Deviation is the square root of the arithmetic average of the squares of all the deviations from the mean. It may be defined as the root-means-square deviation from the mean. Standard deviation is denoted by “s” (read as sigma).

Computation of Standard Deviation.

Computation may be done in two ways : (a) by taking deviations from actual mean. (b) by taking deviation from assumed mean

Steps to find Deviations from actual mean : 1.Find the actual mean 2. Find the deviations from the mean 3. Make squares of the deviations and add up . 4. Divide the addition by total number of items and then find square root.

Steps to find Deviation from assumed mean : 1.Find the deviations of the items from an assumed mean and denote it by d. 2. Find also Sd. 3 Square the deviations find Sd^2. 4. -Apply the

Standard Deviation:

S.D $\displaystyle \left( \sigma \right)$ = $\displaystyle \sqrt{{}}$ [{( $\displaystyle \sum{{}}$ d² / n)} – {( $\displaystyle \sum{{}}$ d / n)²}], or [{( $\displaystyle \sum{{}}$ d² / n)} – {( $\displaystyle \sum{{}}$ d / n)²}]^1/2

Co–efficient of Standard Deviation

Co–efficient of standard deviation is expressed as the ratio of the absolute standard deviation to the arithmetic average of the series.

_{Co-efficient of = / ()}

Co–efficient of Variation

Co–efficient of variation is the percentage of standard deviation of the arithmetic average of the series.

_{C.V. = {/ ()} x 100}

The co-efficient of variation guides us in determining the relative degree of variability or stability of a series. A series with more co-efficient of variation is regarded as less stable or less consistent than a series with less co-efficient of variation.

Standard Deviation – Problem

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Ex. Calculate the Standard Deviation from Actual Mean and Deviation from Assumed Mean of the marks obtained by 10 students in a certain test.

Serial No. :	1	2	3	4	5	6	7	8	9	10
Marks :	43	48	65	57	31	60	37	48	78	58

Calculation of Standard Deviation

For Method (a), A.M = (1/10) x 525 = 52.5 marks

S.D $\displaystyle \left( \sigma \right)$ = $\displaystyle \sqrt{{}}$ (1746.50 / 10) = $\displaystyle \sqrt{{}}$ (174.65) = 13.21 marks

Here the average marks are 52.5, and they deviate on an average from the average by

13.21 marks.

For Method (b) : S.D. $\displaystyle \left( \sigma \right)$ ₌[{(Sd² / n)} – {(Sd / n)²}] = $\displaystyle \sqrt{{}}$ [ (1809 / 10) – (25/10)]²

_{(180.9 – 6.25) = 174.65 = 13.21 marks}

Note: If the actual mean is in fraction, then it is better to take deviations from an assumed mean, to avoid complicated calculations.

Standard Deviation – Discrete Series

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Ex. Compute standard deviation for following data using 1. Direct method. 2. Actual mean method. 3. Step deviation method.

X :	5	10	15	20	25
F :	3	12	18	20	7

direct method:

X	F	X²	FX²	FX
5	3	25	75	15
10	12	100	1200	120
15	18	225	4050	270
20	20	400	8000	400
25	7	625	4375	175
Total	N = 60	–	17700	980

$\displaystyle \sigma$ =[{( $\displaystyle \sum{{}}$ FX²) / N} – {( $\displaystyle \sum{{}}$ FX / N)²] ^1/2 = [(17700 / 60) – (980 / 60) ²] ^1/2

= $\displaystyle \sqrt{{}}$ [(295- (16.33)²] = $\displaystyle \sqrt{{}}$ (295- 266.67) = $\displaystyle \sqrt{{}}$ 28.33 = 5.32 Appx

Actual Mean Method :

X	F	FX	x=(X -X)	x²	Fx²
5	3	15	-10.6	112.36	337.08
10	12	120	-5.6	31.36	376.32
15	18	270	-.6	.36	6.48
20	20	400	4.4	19.36	387.20
25	7	175	9.4	88.36	618.52
Total	N = 60	980	–	–	1725.60

We have $\displaystyle {\overline{x}}$ = ( $\displaystyle \sum{{}}$ FX) / N = $\displaystyle \frac{{980}}{{60}}$ = 15.6

$\displaystyle \sigma$ =[{( $\displaystyle \sum{{}}$ FX²) / N ] = $\displaystyle \sqrt{{}}$ (1725.60 / 60) = $\displaystyle \sqrt{{}}$ ( 28.76) = 5.36 appx

Assumed Mean Method: (at A = 15)

X	F	d=(X – A)	d²	Fd²	Fd
5	3	-10	100	300	-30
10	12	-5	25	300	-60
15	18	0	0	0	0
20	20	5	25	500	100
25	7	10	100	700	70
Total	N = 60	–	–	1800	80

$\displaystyle \sigma$ =[{( $\displaystyle \sum{{}}$ Fd²) / N} – {( $\displaystyle \sum{{}}$ Fd / N)²] ^1/2 = $\displaystyle \sqrt{{}}$ [ ( 1800/60) – (80 /60)²] = $\displaystyle \sqrt{{}}$ (30 – 1.78) = $\displaystyle \sqrt{{}}$ 28.22 = 5.31 appx

Step deviation method (at A = 15)

X	F	(X – A) d	d‘ (c = 5)	d’²	Fd’²	Fd‘
5	3	-10	-2	4	12	-6
10	12	-5	-1	1	12	-12
15	18	0	0	0	0	0
20	20	5	1	1	20	20
25	7	10	2	4	28	14
Total	N = 60	–	–	–	72	16

$\displaystyle \sigma$ =[{( $\displaystyle \sum{{}}$ Fd²) / N} – {( $\displaystyle \sum{{}}$ Fd’ / N)²] ^1/2 x c = [ $\displaystyle \sqrt{{}}$ (72/60) – (16/60) ² ]x5 = [ $\displaystyle \sqrt{{}}$ 1.2 – .071] x5 = ( $\displaystyle \sqrt{{}}$ 1.129) x 5 = 1.062 x 5 = 5.31 appx

Standard Deviation – Continuous Series

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Find the standard deviation from the following frequency distribution:

Height (cm)	131-140	141-150	151-160	161-170	171-180	181-190	191-200	211-240
No. of	2	5	4	9	7	5	3	1
persons:

S.D. $\displaystyle \left( \sigma \right)$ = [{( $\displaystyle \sum{{}}$ fd²) / $\displaystyle \sum{{}}$ f } – {( $\displaystyle \sum{{}}$ fd’ / $\displaystyle \sum{{}}$ f)²] ^1/2 x i = { $\displaystyle \sqrt{{}}$ ( 567/36) – (27/36)² } x 5

= { $\displaystyle \sqrt{{}}$ (15.75- .5626)} x 5 = ( $\displaystyle \sqrt{{}}$ 15.1874) x 5 = 3.897 x 5 = 19.485 appx

Note: If we are to find the mean, then

AM = A + {( $\displaystyle \sum{{}}$ fd) / ( $\displaystyle \sum{{}}$ f)} = 165.5 + (27/36) x 5 = 165.5 +( 0.75 x 5) = 165.5 + 3.75 = 169.25 lbs.

Co-efficient of Standard Deviation – Problems

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Ex. Compute the co-efficient of standard deviation if standard deviation = 12 and mean = 4.

Here, s = 12, X = 4. Co-efficient of $\displaystyle \sigma$ = ( $\displaystyle \sigma$ / X) = $\displaystyle \frac{{12}}{4}$ = 3

Ex. Find the Co-efficient of variation if the sum of squares of the deviations of 10 observation taken from the mean 60 is 360.

_{Coefficient of Variation =}(S.D/ Mean) x 100 . Here S.D = $\displaystyle \sqrt{{}}$ ( $\displaystyle \frac{{360}}{{10}}$ ) = 6

So, Coefficient of Variation = ( $\displaystyle \frac{6}{{60}}$ ) x 100 = 10%

_{Click here to read PDF}

About Mean			About Median
Marks x	f	Dev. from assumed mean(15) d	Step deviation d’ = d/5	fd’	Deviation from actual mean(16) $\displaystyle \left\| D \right\|$	f $\displaystyle \left\| D \right\|$
(1)	(2)	(3)	(4)	(5) = (2) x (4)	(6)	(7) = (2) x (6)
5 10 15 20 25	6 7 8 11 8	-10 – 5 0 5 10	– 2 – 1 0 1 2	– 12 – 7 0 11 16	11 6 1 4 9	66 42 8 44 72
Total	40	–	–	8	–	232

_Class _interval	Mid point (x)	f	_{d = x}_–_{35 (A)}	fd
0-10 10-20 20-30 30-40 40-50 50-60 60-70	5 15 25 35 45 55 65	6 12 10 10 5 4 5	-30 -20 -10 0 10 20 30	-180 -240 -100 0 50 80 150
A = 35		N = $\displaystyle \sum{{}}$ f = 52		-240

x	f	x – $\displaystyle \overline{x}$	\|x – $\displaystyle \overline{x}$ \|	f\|x – $\displaystyle \overline{x}$ \|
5 15 25 35 45 55 65	6 12 10 10 5 4 5	– 25.2 – 15.20 – 5.2 + 4.8 + 14.8 + 24.8 + 34.8	25.2 15.2 5.2 4.8 14.8 24.8 34.8	151.2 182.4 52 48 74 99.20 174
	52			780.80

Statistical Measures of Dispersion

Statistical Dispersion

Range

Range – Problems

Quartile Deviation

Quartile Deviation – Problems

Mean Deviation

Mean Deviation & its co-efficient – Problems

Mean Deviation & its co-efficient – Problems

Mean Deviation & its co-efficient

Standard Deviation

Standard Deviation – Problem

Standard Deviation – Discrete Series

Standard Deviation – Continuous Series

Co-efficient of Standard Deviation – Problems

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About Mean			About Median
Marks (x)	f	Cum. freq	Dev. From median (15) \|D\|	f $\displaystyle \left\| D \right\|$

5 10 15 20 25	6 7 8 11 8	6 13 21 32 40(=N)	10 5 0 5 10	60 35 0 55 80
Total	40	–	–	230

	Method (a)			Method (b)
Serial		Dev. from	Square of		Dev. from	Square of
No.	Marks	actual	dev.	Marks	assumed	dev.
		mean			mean
		52.5 (d)	(d)²		50 (d)	(d)²
1	43	– 9.5	90.25	43	– 7	49
2	48	– 4.5	20.25	48	– 2	4
3	65	12.5	156.25	65	15	225
4	57	4.5	20.25	57	7	49
5	31	– 21.5	462.25	31	– 19	361
6	60	7.5	56.25	60	10	100
7	37	– 15.5	240.25	37	– 13	169
8	48	– 4.5	20.25	48	– 2	4
9	78	25.5	650.25	78	28	784
10	58	5.5	30.25	58	8	64
Total	525		1746.50	525	$\displaystyle \sum{{}}$ d = 25	$\displaystyle \sum{{}}$ d² = 1809

Calculation of Standard Deviation
Height		Mid-value
(cms)	f	x	D = X-A (A=165.5)	d’ = $\displaystyle \frac{d}{5}$	fd’	fd’²
131- 140	2	135.5	– 30	– 6	– 12	72
141- 150	5	145.5	– 20	– 4	– 20	80
151-160	4	155.5	– 10	– 2	– 8	16
161- 170	9	165.5	0	0	0	0
171- 180	7	175.5	10	2	14	28
181- 190	5	185.5	20	4	20	80
191- 200	3	200.5	35	7	21	147
211- 240	1	225.5	60	12	12	144
Total	N = 36	–	–	–	27	567