Measures of Dispersion | Business Statistics MGT 202 | TU BBS

This unit covers absolute and relative measures of dispersion: range, quartile deviation, mean deviation, standard deviation, coefficient of variation, combined standard deviation, and Lorenz’s curve as a graphical method.

Meaning of Dispersion

Dispersion means scatter, spread, or variation. It is defined as the measure of variation in the items from the central value.

Methods of measuring dispersion

Range
Semi-interquartile range or quartile deviation
Mean deviation or average deviation
Standard deviation
Lorenz’s curve

Requisites of a good measure of dispersion

It should be rigidly defined.
It should be simple to understand and easy to calculate.
It should be based on all observations.
It should be suitable for further mathematical treatment.
It should be least affected by fluctuation of sampling.
It should not be unduly affected by extreme values.

Absolute and Relative Measures

Absolute measures are expressed in the same unit as the data. Relative measures (coefficients) are pure numbers used to compare variability across series with different units or means.

Measure	Absolute	Relative
Range	$R = L - S$	Coefficient of range $= \dfrac{L-S}{L+S}$
Quartile deviation	$Q.D. = \dfrac{Q_3 - Q_1}{2}$	Coefficient of Q.D. $= \dfrac{Q_3 - Q_1}{Q_3 + Q_1}$
Mean deviation	M.D. from mean/median/mode	Coefficient of M.D. $= \dfrac{\text{M.D.}}{\text{average used}}$
Standard deviation	$\sigma$	Coefficient of S.D. $= \dfrac{\sigma}{\bar{X}}$; C.V. $= \dfrac{\sigma}{\bar{X}}\times 100\%$

Where $S$ = smallest value and $L$ = largest value. Inter-quartile range $= Q_3 - Q_1$.

Range

Range is defined as the difference between the largest item and the smallest item in the set of observations.

$$\begin{aligned} \text{Range} &= L - S \\ \text{Coefficient of Range} &= \frac{L-S}{L+S} \end{aligned}$$

where $L$ = largest item and $S$ = smallest item.

Quartile Deviation

The measure of dispersion depending upon the lower and upper quartiles is known as the quartile deviation.

The difference between the upper and lower quartiles is known as the interquartile range. Half of the interquartile range is known as the semi-interquartile range or quartile deviation.

$$\begin{aligned} Q.D. &= \frac{Q_3 - Q_1}{2} \\ \text{Coefficient of Q.D.} &= \frac{Q_3 - Q_1}{Q_3 + Q_1} \end{aligned}$$

Quartile deviation is the most suitable absolute measure of dispersion for open-end classes.

Mean Deviation

Mean deviation is defined as the arithmetic mean of the deviations of the items from mean, median, or mode, when all deviations are considered positive.

Individual series

$$\begin{aligned} \text{M.D. from mean} &= \frac{\sum |x - \bar{x}|}{n} \\ \text{M.D. from median} &= \frac{\sum |x - M_d|}{n} \\ \text{M.D. from mode} &= \frac{\sum |x - M_0|}{n} \end{aligned}$$

Discrete and continuous series

$$\begin{aligned} \text{M.D. from mean} &= \frac{\sum f|X-\bar{X}|}{N} \\ \text{M.D. from median} &= \frac{\sum f|x-M_d|}{N} \\ \text{M.D. from mode} &= \frac{\sum f|x-M_0|}{N} \end{aligned}$$

where $n$ = number of observations and $N$ = sum of frequencies.

$$\begin{aligned} \text{Coefficient of M.D. from mean} &= \frac{\text{M.D. from mean}}{\text{Mean}} \\ \text{Coefficient of M.D. from median} &= \frac{\text{M.D. from median}}{\text{Median}} \\ \text{Coefficient of M.D. from mode} &= \frac{\text{M.D. from mode}}{\text{Mode}} \end{aligned}$$

Note: Mean deviation from median is calculated generally if the question is just asked to calculate the mean deviation.

Standard Deviation

Standard deviation is defined as the positive square root of the mean of the square of the deviations taken from the arithmetic mean. It is denoted by $\sigma$.

Individual series

Direct method:

$$\sigma = \sqrt{\frac{\sum(x-\bar{x})^{2}}{n}} = \sqrt{\frac{\sum x^{2}}{n} - \left(\frac{\sum x}{n}\right)^{2}}$$

Short-cut (assumed mean) method:

$$\begin{aligned} \sigma &= \sqrt{\frac{\sum d^{2}}{n} - \left(\frac{\sum d}{n}\right)^{2}} \\ d &= x - a \end{aligned}$$

where $a$ = assumed mean.

Step-deviation method:

$$\sigma = \sqrt{\frac{\sum d^{2}}{n} - \left(\frac{\sum d}{n}\right)^{2}} \times h$$

Discrete series

$$\begin{aligned} \sigma &= \sqrt{\frac{\sum f(x-\bar{x})^{2}}{N}} = \sqrt{\frac{\sum f x^{2}}{N} - \left(\frac{\sum f x}{N}\right)^{2}} \\ \sigma &= \sqrt{\frac{\sum f d^{2}}{N} - \left(\frac{\sum f d}{N}\right)^{2}} \end{aligned}$$

Continuous series

$$\begin{aligned} \sigma &= \sqrt{\frac{\sum f(x-\bar{x})^{2}}{N}} = \sqrt{\frac{\sum f x^{2}}{N} - \left(\frac{\sum f x}{N}\right)^{2}} \\ \sigma &= \sqrt{\frac{\sum f d^{\prime 2}}{N} - \left(\frac{\sum f d^{\prime}}{N}\right)^{2}} \times h \\ d^{\prime} &= \frac{x-a}{h} \end{aligned}$$

where $a$ = assumed mean and $h$ = class size or common factor.

Variance and relative measures based on $\sigma$

The square of the standard deviation is known as the variance. It is denoted by $\sigma^{2}$ or $\mu_2$.

Coefficient of S.D. $= \dfrac{\sigma}{\bar{X}}$
Variance $= \sigma^{2}$
Coefficient of variation (C.V.) $= \dfrac{\sigma}{\bar{X}} \times 100\%$

C.V. is used to compare two or more variables. Less C.V. represents more consistency; more C.V. represents more variability. If a question is asked about variability, uniformity, consistency, or homogeneity, calculate C.V.

Combined Standard Deviation

$$\begin{aligned} \sigma_{12} &= \sqrt{\frac{n_1(\sigma_1^{2}+d_1^{2})+n_2(\sigma_2^{2}+d_2^{2})}{n_1+n_2}} \\ d_1 &= \bar{x}_1 - \bar{x}_{12},\quad d_2 = \bar{x}_2 - \bar{x}_{12} \\ \bar{x}_{12} &= \frac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2} \end{aligned}$$

For three groups:

$$\begin{aligned} \sigma_{123} &= \sqrt{\frac{n_1(\sigma_1^{2}+d_1^{2})+n_2(\sigma_2^{2}+d_2^{2})+n_3(\sigma_3^{2}+d_3^{2})}{n_1+n_2+n_3}} \\ d_1 &= \bar{x}_1 - \bar{x}_{123},\quad d_2 = \bar{x}_2 - \bar{x}_{123},\quad d_3 = \bar{x}_3 - \bar{x}_{123} \\ \bar{x}_{123} &= \frac{n_1\bar{x}_1+n_2\bar{x}_2+n_3\bar{x}_3}{n_1+n_2+n_3} \end{aligned}$$

Combined coefficient of variation:

$$C.V._{12}=\frac{\text{Combined S.D.}}{\text{Combined Mean}}$$

Correction of wrong items

$$\begin{aligned} \text{Wrong }\sum X^{2} &= n\{(\text{Wrong }\sigma)^{2}+(\text{Wrong }\bar{X})^{2}\} \\ \text{Correct }\sum X^{2} &= \text{Wrong }\sum X^{2}-(\text{Wrong items})^{2}+(\text{Correct items})^{2} \\ (\text{Correct }\sigma)^{2} &= \frac{\text{Correct }\sum X^{2}}{n}-\left(\frac{\text{Correct }\sum X}{n}\right)^{2} \end{aligned}$$

Lorenz’s curve is listed among the methods of measuring dispersion as a graphical approach to showing inequality of distribution (for example of income or wealth) relative to a line of equal distribution.

Worked Examples

Comparing variability with C.V. (factories)

	Factory A	Factory B
Average weekly wage $\bar{x}$	$\bar{x}_1 = 4600$	$\bar{x}_2 = 4900$
Standard deviation $\sigma$	$\sigma_1 = 50$	$\sigma_2 = 40$
Number of workers $N$	$N_1 = 100$	$N_2 = 80$

Combined mean and combined S.D.:

$$\begin{aligned} \bar{x}_{12} &= \frac{100\times 4600 + 80\times 4900}{180} = 4733.33 \\ d_1 &= 4600-4733.33 = -133.33 \\ d_2 &= 4900-4733.33 = 166.67 \\ \sigma_{12} &= \sqrt{\frac{100(50^{2}+(-133.33)^{2})+80(40^{2}+166.67^{2})}{180}} = 155.96 \end{aligned}$$

$$\begin{aligned} CV(A) &= \frac{50}{4600}\times 100 = 1.09\% \\ CV(B) &= \frac{40}{4900}\times 100 = 0.82\% \end{aligned}$$

Factory A has greater variability in the distribution of weekly wages due to higher C.V. ($1.09 > 0.82$).

C.V. for firms A and B

	Firm A	Firm B
Number of workers	$N_1=500$	$N_2=600$
Average wage	$\bar{x}_1=1500$	$\bar{x}_2=1550$
Variance	$\sigma_1^{2}=100$	$\sigma_2^{2}=121$
Standard deviation	$\sigma_1=10$	$\sigma_2=11$

$$\begin{aligned} CV_A &= \frac{10}{1500}\times 100 = 0.67\% \\ CV_B &= \frac{11}{1550}\times 100 = 0.71\% \end{aligned}$$

There is greater variability in individual wages in firm B due to higher C.V. than that of firm A.

Combined mean $\bar{x}_{12} = 1527.27$; with $d_1=-27.27$ and $d_2=22.73$, combined variance $\sigma_{12}^{2} = 731.29$.

Boys and girls — combined mean and S.D.

Total students $N=150$, combined mean $\bar{x}_{12}=60$ kg. Boys: $\bar{x}_1=70$, $\sigma_1=10$. Girls: $\bar{x}_2=55$, $\sigma_2=15$.

$$\begin{aligned} 60 &= \frac{N_1\times 70+(150-N_1)\times 55}{150} \\ N_1 &= 50,\quad N_2 = 100 \\ d_1 &= 10,\quad d_2 = -5 \\ \sigma_{12} &= \sqrt{\frac{50(10^{2}+10^{2})+100(15^{2}+(-5)^{2})}{150}} = 15.28 \end{aligned}$$

Open-end income — quartile deviation

For open-end income data with $N=400$, $Q_1$ lies in class 120–160 and $Q_3$ in 160–200:

$$\begin{aligned} Q_1 &= 120+\frac{100-35}{130}\times 40 = 140\ ({\rm in\ 000}) = 140{,}000 \\ Q_3 &= 160+\frac{300-165}{145}\times 40 = 197.2414\ ({\rm in\ 000}) = 197{,}241.6 \\ Q.D. &= \frac{197{,}241.6-140{,}000}{2} = 28{,}620.70 \end{aligned}$$

Lowest income of the richest 20% ($P_{80}$): Rs.\ 205,714.3. Highest income of the poorest 20% ($P_{20}$): Rs.\ 133,846.2. Limits of the middle 50% of families: Rs.\ 140,000 and Rs.\ 197,241.6 ($P_{25}=Q_1$, $P_{75}=Q_3$).

Important Questions

Board exam questions asked on this unit, by exam year (marks in brackets). Short answers are given where worked above.

2078 Exam

The coefficient of variation of a distribution is found to be 20% and the variance of the distribution is 36. Find the mean value of the distribution. [2 marks]

$$\begin{aligned} \sigma^{2}&=36\Rightarrow\sigma=6 \\ C.V.&=\frac{\sigma}{\bar{X}}\times 100=20 \\ \bar{X}&=\frac{6}{20}\times 100=30 \end{aligned}$$
The following two groups of samples describe the age of the students in the regular morning MBS programme and evening MBS programme of National College. If homogeneity of the class is a positive factor for learning, which of the two programmes will be easier to teach? [10 marks]

Evening MBS 24 30 28 23 25 22 26 27 28 25

Morning MBS 28 27 34 33 29 27 28 29 33 26

Calculate C.V. for each series; the programme with the smaller C.V. is more homogeneous (easier to teach).

Evening MBS	24	30	28	23	25	22	26	27	28	25
Morning MBS	28	27	34	33	29	27	28	29	33	26

2079 Exam

A factory produces two types of electric lamps, A and B. In an experiment relating to their life, the following results were obtained: [15 marks]

Life (in years) Number of lamps

Model A Model B

0 – 2 5 4

2 – 4 11 30

4 – 6 26 12

6 – 8 10 8

8 – 10 8 6

(a) Find which model of electric lamp has greater uniformity of life. Give the reason. (b) Calculate the combined standard deviation.

Life (in years)	Number of lamps
0 – 2	5	4
2 – 4	11	30
4 – 6	26	12
6 – 8	10	8
8 – 10	8	6

2080 Exam

If the third quartile and median are 30 and 22 respectively, find the first quartile, assuming the distribution to be symmetrical. [2 marks]

$$\begin{aligned} \text{For a symmetrical distribution: } Q_3 - M_d &= M_d - Q_1 \\ 30-22 &= 22-Q_1 \\ Q_1 &= 14 \end{aligned}$$

2081 Exam

Calculate the coefficient of variation of a distribution if the mean is found to be 200 and the variance of the distribution is 36. [2 marks]

$$C.V.=\frac{6}{200}\times 100=3\%$$
Find the coefficient of quartile deviation of a distribution when the upper and lower quartiles are 90 and 55 respectively. [2 marks]

$$\text{Coefficient of Q.D.}=\frac{90-55}{90+55}=\frac{35}{145}=0.2414$$
Find the appropriate measure of dispersion from the following income table: [10 marks]

Monthly income (Rs.) Number of persons

Below 1000 15

1000 – 1999 500

2000 – 2999 550

3000 – 3999 300

4000 – 4999 200

5000 and above 150

The appropriate measure is quartile deviation, because the series has open-end classes.
The following two samples describe the age (years) of students in regular programmes of a university: [15 marks]

MBS 25 30 28 25 23 22 26 27 28 24

MBA 26 27 34 33 29 27 28 29 33 28

(a) If homogeneity of age is a positive factor for the teaching-learning process, which of the two programmes will be easier to teach? (b) Calculate the combined standard deviation.

Monthly income (Rs.)	Number of persons
Below 1000	15
1000 – 1999	500
2000 – 2999	550
3000 – 3999	300
4000 – 4999	200
5000 and above	150

MBS	25	30	28	25	23	22	26	27	28	24
MBA	26	27	34	33	29	27	28	29	33	28

2082 Exam

Calculate the coefficient of variation of a distribution when mean = 40 and standard deviation = 11. [2 marks]

$$C.V.=\frac{11}{40}\times 100=27.5\%$$
The following samples describe the marks of students in Statistics in the BBS programme of two sections. [15 marks]

Section A 54 60 58 53 55 52 56 57 58 55

Section B 58 57 54 53 59 57 58 59 63 56

(a) State which section of students secures more uniform marks. (b) Find the combined standard deviation.

Section A	54	60	58	53	55	52	56	57	58	55
Section B	58	57	54	53	59	57	58	59	63	56

2083 Exam

The profit distributions of two companies A and B are given below: [15 marks]

Profit (million Rs.) 0 – 2 2 – 4 4 – 6 6 – 8 8 – 10 10 – 12

No. of items A 2 7 11 20 7 3

No. of items B 5 4 9 13 15 4

(a) An investor wants to invest according to the profit of the company. Which company is more consistent on the basis of profit? (b) Also find the combined standard deviation.

	Factory A	Factory B
Average weekly wage \(\bar{x}\)	\(\bar{x}_1 = 4600\)	\(\bar{x}_2 = 4900\)
Standard deviation \(\sigma\)	\(\sigma_1 = 50\)	\(\sigma_2 = 40\)
Number of workers \(N\)	\(N_1 = 100\)	\(N_2 = 80\)

	Firm A	Firm B
Number of workers	\(N_1=500\)	\(N_2=600\)
Average wage	\(\bar{x}_1=1500\)	\(\bar{x}_2=1550\)
Variance	\(\sigma_1^{2}=100\)	\(\sigma_2^{2}=121\)
Standard deviation	\(\sigma_1=10\)	\(\sigma_2=11\)

Business Statistics

Business Statistics