We had learnt how to present a statistical data precisely and meaningfully. This is to enable the statistician to study the data well and draw inferences from the,‘data. But by presentation only, it cannot be possible. Hence after presenting the data, it should be condensed. Condensation of data is essential in statistical analysis because, a large number of figures not only confusing but difficult to analyse also. In order to reduce the complexity of data and to make them comparable, it is essential that various characteristics which are being compared are reduced to one figure each. If, for example, a comparison is made between the marks obtained by 45 students belonging to Batch-A of plus-1 and 48 students belonging to Batch-B of plus-1 of your school in the first term examination, it would be impossible to arrive at any conclusion, if the two series relating to these marks are directly compared. On the other hand, if each of these series is represented by one figure, comparison would be easier. It is, obvious that a figure which is used to represent a whole series should neither have the lowest value in the series nor the highest value, but a value somewhere between these two limits, possibly in the centre, where most of the items of the series gather. Such a figure is called the average or a measure of central tendency.

The average represents the whole series and as such, its value always lies between the highest and the lowest values and generally it is located in the centre or middle of the distribution. Thus, central tendency summarises the data in a single value in such a way that this single value can represent the entire data. A measure of central tendency is a single figure that is computed from a given series to give a central value about the entire series. Hence a measure of central tendency may be defined as a typical value around which the values of a distribution congregate.

• It should be easy to understand
• It should be simple to compute
• It should be based on all the items
• It should not be unduly affected by extreme items
• It should be rigidly defined
• It should be capable of further algebraic treatment
• It should have sampling stability

1. ARITHMETIC MEAN
1. Simple Mean
2. Combined Mean
3. Weighted Mean
2. MEDIAN
3. MODE

On the basis of the type of data series that has provided to us (ie, Individual, Discrete, Continuous), it will be convenient if we use appropriate formula for finding averages in each of these series.

There are three methods by which Simple mean can be calculated in each of these three series.

They are :

1. Direct Method
2. Assumed Mean Method
3. Step Deviation Method

Calculation of arithmatic mean can be studied under two heads.

1. Arithmetic Mean for Ungrouped Data
2. Arithmetic Mean for Grouped Data

Arithmetic Mean for Ungrouped data can be calculated using the following methods:

1. Direct Method
2. Assumed Mean Method
3. Step Deviation Method

STEPS

1. Find the sum of observations (∑X)
2. Take the number of observations (N)
3. Use the formula \( X̅ = {{{\frac{ΣX}{N}} }} \)

$$ A.M = {{{\frac{60+75+72+68+80+65}{6}} }} $$ $$ = {{{\frac{420}{6}} }} = 70 $$

STEPS

1. Take an assumed mean A
2. Take the deviation of each X from the assumed mean. ie., d = X – A
3. Find the sum of the deviations to get Σd
4. Use the formula \( X̅ = A + {{{\frac{Σd}{N}} }} \) ; where N is the total number of observations

Let us find the arithmetic mean of the following 20 observations using assumed mean method.

2500, 6500, 3000, 5500, 4500, 6000, 3500, 3000, 5500, 5000, 2000, 4500, 3500, 3000, 4500, 6500, 4000, 3000, 2500, 4500

Table 5.1
X	d=X-A=X-4000
2500	-1500
6500	2500
3000	-1000
5500	1500
4500	500
6000	2000
3500	-500
3000	-1000
5500	1500
5000	1000
2000	-2000
4500	500
3500	-500
3000	-1000
4500	500
6500	2500
4000	0
3000	-1000
2500	-1500
4500	500
N=20	Σd = (-10000)+13000=3000

$$ {X̅=}A+\frac{Σd}{N} = 4000+\frac{3000}{20} $$

$$=4000 + 150 = 4150 $$

Complexity of calculations in finding arithmatic mean can nfurther be reduced by using step deviation method.

STEPS

1. Take an assumed mean A
2. Take the deviation of each X from the assumed mean. ie., d = X – A
3. Devide the deviation d by common factor c, i.e., \( d ‘ = {{{\frac{d}{c}} }} = {{{\frac{X – A}{c}} }} \)
4. Then find Σd’
5. Use the formula \( X̅ = A + {{{\frac{Σd’}{N}} }} × c \) ; where N is the total number of observations

Let us find the arithmatic mean of the following 10 obserations using step deviation method.

45, 30, 65, 70, 40, 25, 45, 25, 55, 40, 20, 50

Assumed mean is taken as 50.

Let c = 5 (we decide value of c only after seeing the column for d in the table)

Table 5.2
X	d=X-A=X-50	\( \mathbf {d ‘ = {{{\frac{d}{c}} }} = {{{\frac{d}{5}} }}} \)
45	-5	-1
30	-20	-4
65	15	3
70	20	4
40	-10	-2
25	-25	-5
45	-5	-1
25	-25	-5
55	5	1
40	-10	-2
20	-30	-6
50	0	0
N = 12		Σd’ = (-26) + 8 =-18

$$ X̅ = A + {{{\frac{Σd’}{N}} }} × c = {50+\frac{-18}{12}} × 5 $$

$$= 42.5 $$

STEPS

1. Multiply the frequency against each observation with the value of observation to get fx
2. Add the column of fx to get Σfx
3. Use the formula \( X̅ = {{{\frac{Σfx}{Σf}} }} \)

From the following data relating to the monthly income of 50 persons, determine the average monthly income using assumed mean method

Table 5.3
Income	Number of Persons
1200	2
1500	10
1800	15
2000	7
2300	5
2600	4
3400	3
4200	3
5000	1

Now we can create a table to find fx.

Table 5.4
Income	Number of Persons	fX
1200	2	2400
1500	10	15000
1800	15	27000
2000	7	14000
2300	5	11500
2600	4	10400
3400	3	10200
4200	3	12600
5000	1	5000
	Σf = 50	Σfx = 108100

$$ X̅ = {{{\frac{Σfx}{Σf}} }} = {\frac{-108100}{50}} $$

$$= 2162 $$

Assumed mean method is applying just to simplify the calculations.

STEPS

1. Take an assumed mean A
2. Take the deviation ‘d’ of each X from the assumed mean. ie., d = X – A
3. Multiply d with f to get fd
4. Add the column of fd to get Σfd
5. Add all the frequencies to get Σf
6. Use the formula \( X̅ = A + {{{\frac{Σfd}{Σf}} }} \)

From the following data relating to the monthly income of 50 persons, determine the average monthly income using assumed mean method.

Table 5.5
Income	Number of Persons
1200	2
1500	10
1800	15
2000	7
2300	5
2600	4
3400	3
4200	3
5000	1

Now we can create a table to find d and fd.

Table 5.6
Income	Number of Persons	d = X-A = X-2300	fd
1200	2	-1100	-2200
1500	10	-800	-8000
1800	15	-500	-7500
2000	7	-300	-2100
2300	5	0	0
2600	4	300	1200
3400	3	1100	3300
4200	3	1900	5700
5000	1	2700	2700
	Σf = 50		Σfd=(-19800)+12900 = -6900

$$ X̅ = A + {{{\frac{Σfd}{Σf}} }} = 2300 + {\frac{-6900}{50}} $$

$$= 2162 $$

Complexity of calculations in finding arithmatic mean can nfurther be reduced by using step deviation method. Under step deviation method, as in the individual series case, we divide the deviations d by a common factor say c. We denote that as d’.

STEPS

1. Take an assumed mean A
2. Take the deviation of each X from the assumed mean. ie., d = X – A
3. Devide the deviation d by common factor c, i.e., \( d ‘ = {{{\frac{d}{c}} }} = {{{\frac{X – A}{c}} }} \)
4. Multiply each d’ with frequency to get fd’
5. Add the column of fd’ to get Σfd’
6. Add all the frequencies to get Σf
7. Use the formula \( X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c \) ; where N is the total number of observations

From the following data relating to the monthly income of 50 persons, determine the average monthly income using step deviation method.

Table 5.7
Income	Number of Persons
1200	2
1500	10
1800	15
2000	7
2300	5
2600	4
3400	3
4200	3
5000	1

Now we can create a table to find d, d’ and fd’. We can take assumed mean as 2300 and c as 100. C is taken as 100 after getting d column.

$$ X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c = 2300 + {\frac{-69}{50}} × 100$$

$$= 2162 $$

STEPS

1. Find the mid value of each class, denoted by m
2. Multiply each class frequency with m to get fm
3. Add the column fm to get Σfm
4. Use the formula \( X̅ = {{{\frac{Σfm}{Σf}} }} \)

Note :-

1. For finding arithmetic mean, it is not necessary to convert inclusive type to exclusive type. However, in case of finding median or; mode the coiwersion .is necessary-
2. If intervals are having unequal width, then also the same method can be followed.
3. If the distribution is exdlusive type, then, the class interval ot a class is got by the formula (upper limit – lower limit)
4. If the distribution is inclusive type, then, there is a gap between the upper limit of the class and the lower limit of the next class. In that case, the class interval of the class is got by the formula, (upper limit – lower limit + gap).

Let us find arithmetic mean of the following distribution by direct method.

Table 5.9
Marks	Number of Students
0-10	8
10-20	12
20-30	15
30-40	17
40-50	25
50-60	20
60-70	16
70-80	13
80-90	10
90-100	4

Let us create a table showing m and fm.

Table 5.10
Marks	Number of Students	m	fm
0-10	8	5	40
10-20	12	15	180
20-30	15	25	375
30-40	17	35	595
40-50	25	45	1125
50-60	20	55	1100
60-70	16	65	1040
70-80	13	75	975
80-90	10	85	850
90-100	4	95	380
	Σf = 140		Σfm = 6660

$$ X̅ = {{{\frac{Σfm}{Σf}} }} = {\frac{6660}{140}} $$

$$= 47.57 $$

Assumed mean method is applying just to simplify the calculations.

STEPS

1. Find the mid value (m) of each class
2. Take an assumed mean, A
3. Take deviation of each m from A, i.e., d = m – A
4. Multiply d with f to get fd
5. Add all fd to get Σfd
6. Add all f to get Σf
7. Apply the formula \( X̅ = A + {{{\frac{Σfd}{Σf}} }} \)

Find arithmetic mean of the following distribution by assumed mean method.

Table 5.11
Marks	Number of Students
0-10	8
10-20	12
20-30	15
30-40	17
40-50	25
50-60	20
60-70	16
70-80	13
80-90	10
90-100	4

Let us create a table showing m, d and fd. Assumed mean is taken as 45.

Table 5.12
Marks	Number of Students	m	d = m-45	fd
0-10	8	5	-40	-320
10-20	12	15	-30	-360
20-30	15	25	-20	-300
30-40	17	35	-10	-170
40-50	25	45	0	0
50-60	20	55	10	200
60-70	16	65	20	320
70-80	13	75	30	390
80-90	10	85	40	400
90-100	4	95	50	200
	Σf = 140			Σfd = -1150 + 1510 = 360

$$ X̅ = A + {{{\frac{Σfd}{Σf}} }} = 45 + {\frac{360}{140}} $$

$$= 47.57 $$

In this method, in order to simplify calculations we take a common factor for the data.

STEPS

1. Find the mid value (m) of each class
2. Take an assumed mean A
3. Take the deviation of each m from A, i.e., d = m-A
4. Divide d by a common factor c to get d’. \( d ‘ = {{{\frac{d}{c}} }} = {{{\frac{m – A}{c}} }} \)
5. Multiply d’ with f to get Σfd’
6. Add all the frequencies to get Σf
7. Apply the formula \( X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c \)

EXCLUSIVE CLASS

Find arithmetic mean of the following distribution using step deviation method.

Table 5.13
Marks	Number of Students
0-10	8
10-20	12
20-30	15
30-40	17
40-50	25
50-60	20
60-70	16
70-80	13
80-90	10
90-100	4

To find arithmetic mean through step deviation method, we need to find m, d, d’ and fd’.

Table 5.14
X	f	m	d = m-45	\( \mathbf {d ‘ = {{{\frac{d}{c}} }} = {{{\frac{d}{10}} }}} \)	fd’
0-10	8	5	-40	-4	-32
10-20	12	15	-30	-3	-36
20-30	15	25	-20	-2	-30
30-40	17	35	-10	-1	-17
40-50	25	45	0	0	0
50-60	20	55	10	1	20
60-70	16	65	20	2	32
70-80	13	75	30	3	39
80-90	10	85	40	4	40
90-100	4	95	50	5	20
	Σf = 140				Σfd’ = (-115) + 151 = 36

$$ X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c = 45 + {\frac{36}{140}} × 10 $$

$$= 47.57 $$

INCLUSIVE CLASS

Find arithmetic mean of the following distribution using step deviation method. The given distribution is in inclusive format.

Table 5.15
Marks	Number of Students
0-9	1
10-19	5
20-29	12
30-39	15
40-49	24
50-59	20
60-69	9
70-79	8
80-89	4
90-99	2

Here the classes are inclusive type. It is not neccessary to convert them into exclusive, because mid-points remain the same whether or not the adjustment is made. To find arithmetic mean through step deviation method, we need to find m, d, d’ and fd’

Table 5.16
X	f	m	d = m-45	\( \mathbf {d ‘ = {{{\frac{d}{c}} }} = {{{\frac{d}{10}} }}} \)	fd’
0-9	8	4.5	-50	-5	-5
10-19	12	14.5	-40	-4	-20
20-29	15	24.5	-30	-3	-36
30-39	17	34.5	-20	-2	-30
40-49	25	44.5	-10	-1	-24
50-59	20	54.5	0	0	0
60-69	16	64.5	10	1	9
70-79	13	74.5	20	2	16
80-89	10	84.5	30	3	12
90-99	4	94.5	40	4	8
	Σf = 140				Σfd’ = (-115) + 45 = -70

$$ X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c = 54.5 + {\frac{-70}{100}} × 10 $$

$$= 47.57 $$

UNEQUAL CLASSES

Let us Find arithmetic mean of the following distribution using step deviation method. Here the distribution is provided with unequal class intervals. If classes are having unequal width, it is not necessary to convert them to equal width.

Table 5.17
Marks	Number of Students
0-5	8
5-15	12
15-20	15
20-40	17
40-50	25
50-60	20
60-65	16
65-80	13
80-90	10
90-100	4

To find arithmetic mean through step deviation method, we need to find m, d, d’ and fd’.

$$ X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c = 55 + {\frac{-556}{100}} × 2.5 $$

$$= 45.07 $$

OPEN END CLASSES

Find arithmetic mean of the following distribution using step deviation method.

Table 5.19
Marks	Number of Students
Less than 15	8
15-30	15
30-45	32
45-60	26
60-75	13
Above 75	6

Here the given table has two open end classes, namely. the first and the last. In the case of open end classes, we cannot find out the arithmetic mean unless we make an assumption about the limits of the open end classes. For that, look at the class intervals of the classes following the first class and preceding the last class. The class interval of the second class is 15. Hence by assuming the class interval of the first class also as 15, we get the lower limit of the first class as 0. Similarly, since the class interval of the class preceding to the last class is 15, by assuming the class interval of the last class as 15, we get the upper limit of the last class as 90.

Now we get the adjusted table as given below.

Table 5.20
Marks	Number of Students
0-15	8
15-30	15
30-45	32
45-60	26
60-75	13
75-90	6
Total	100

Now we can simply find arithmetic mean by using any method. Here , we used step deviation method to find arithmetic mean.

$$ X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c = 52.5 + {\frac{-183}{100}} × 5 $$

$$= 43.35 $$

MORE THAN CUMULATIVE FREQUENCIES

Find arithmetic mean of the following distribution using step deviation method.

Table 5.22
Wage	Number of Labourers
Above 0	675
Above 10	625
Above 20	550
Above 30	450
Above 40	275
Above 50	150
Above 60	75
Above 70	25

Here the frequencies given are not the simple frequencies. They are more than cumulative frequencies. Hence we need to convert them to simple frequency. The simple frequency of the first class = 675 — 625 = 50. For the second class, simple frequency = 625 — 550 = 75. For the third class, simple frequency = 550 — 450 = 100; and so on. Also we should write the classes accordingly.

Now we get the adjusted table as given below.

Table 5.23
X	f
0-10	50
10-20	75
20-30	100
30-40	175
40-50	125
50-60	75
60-70	50
70-80	25
Total	675

Now we can find arithmetic mean using step deviation method.

$$ X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c = 35 + {\frac{125}{675}} × 10 $$

$$= 36.85 $$

LESS THAN CUMULATIVE FREQUENCIES

Find arithmetic mean of the following distribution using step deviation method.

Table 5.25
Mark	Number of Students
Less than 10	4
Less than 20	16
Less than 30	40
Less than 40	76
Less than 50	96
Less than 60	112
Less than 70	120
Less than 80	125

By inspection, we can see that the frequencies given are not the simple frequencies. They are less than cumulative frequencies. In order to calculate arithmetic mean, we need simple frequency of each class. So first we need to convert cumulative frequencies to simple frequency. The simple frequency of the first class is 4 itself. For the second class, simple frequency = 16 — 4 = 12. For the third class, simple frequency = 40 — 16 = 24; and so on. Also we should write the classes accordingly.

Now we get the adjusted frequencies as in the table given below.

Table 5.26
X	f
0-10	4
10-20	12
20-30	24
30-40	36
40-50	20
50-60	16
60-70	8
70-80	5
Total	125

Now we can find arithmetic mean using step deviation method.

Table 5.27
X	f	m	d = m-35	\( \mathbf {d ‘ = {{{\frac{d}{c}} }} = {{{\frac{d}{10}} }}} \)	fd’
0-10	4	5	-30	-3	-12
10-20	12	15	-20	-2	-24
20-30	24	25	-10	-1	-24
30-40	36	35	0	0	0
40-50	20	45	10	1	20
50-60	16	55	20	2	32
60-70	8	65	30	3	24
70-80	5	75	40	4	20
	Σf = 125				Σfd’=(-60)+96 = 36

$$ X̅ = A + {{{\frac{Σfd’}{Σf}} }} × c = 35 + {\frac{36}{125}} × 10 $$

$$= 37.88 $$

COMBINED MEAN

Consider a sample 20, 23, 15, 21, 28 and 19. We can call this as sample – I. Its arithmetic mean is :

Consider another sample 32, 28, 19 and 13. Let us call it as sample – II. Its arithmetic mean is :

Let us find out the arithmetic mean of combined sample.

We can find the combined mean of samples without knowing the individual observations. We only need to know the sizes and means of each sample to find out the combined mean. Suppose that we do not know the individual observations of sample-I and sample- II above, and we only know the sizes of samples and their means. Let size of sample-I as n₁ and size of sample-II as n₂, ; Also, we denote the mean of sample- I as X&#772₁ and mean of sample- II as X&#772₂,. Then the combined mean X&#772 of samples.1 and 2 can be determined using the formula:

Let us find combined mean for given data as n₁ = 6, n₂ = 4, X̅₁ = 21 and X̅₂ = 23.

CORRECTION IN MEAN

While calculating mean, sometimes we may consider numbers wrongly by mistake. This will lead to wrong results. But, when we realized the mistake, the results may be corrected without doing the problem afresh. The process of finding out the correct mean value is very simple. First we multiply the incorrect mean by the total number of observations. The incorrect values are then subtracted from that and the correct values are added, Then divide the number you have got by the total number of observations. The below given example will illustrate the method.

The average mark secured by 50 students was calculated as 48. later on, it was found that a mark 60 was misred as 16. Find the correct average mark secured by the students.

WEIGHTED MEAN

The arithmetic mean we had studied is simple arithmetic mean. In the calculation of simple arithmetic mean each item of the series is considered equally important. But there may be cases where all items may not have equal importance. Some of them may be comparatively more important than others. For example, if we are finding out the change in the cost of living of a certain group of people and if we merely find the simple arithmetic average of the prices of the commodities consumed by them, the average would be unrepresentative. All the items of consumption are not equally important. The price of salt may increase by 100 per cent but this will not affect the cost of living to the extent to which it would be affected, if the price of rice goes up only by 10 per cent. In such cases simple average is not suitable and we give different weights to each item while computing average. Arithmetic mean computed by assigning different weights to each item is called weighted arithmetic mean.

Let x₁, x₂, …. , x_n be n items with weights w₁, w₂, … , w_n respectively. Then the weighted arithmetic mean is:

$$ { X̅_w =\frac{w_1 x_1 + w_2 x_2 +…+ w_n x_n}{w_1 + w_2 +…+ w_n} = \frac{Σwx}{Σw}} $$

Let us find weighted arithmatic mean of the given data.

Table 5.28
Item	Amount	Weight
1	35	3
2	27	7
3	65	1
4	47	4
5	30	9

Now we need to create a table with w and wx as given below.

Table 5.29
x (Amount)	w (Weight)	wx
35	3	105
27	7	189
65	1	65
47	4	188
30	9	270
	Σw 24 = 24	Σwx = 817

$$ { X̅_w =\frac{Σwx}{Σw} = \frac{817}{24}} = 34.04 $$

Let us practice an another example.

An examination in the subjects, English, Mathematics and Social Science was held to decide the award of a scholarship. The marks obtained by the top three candidates are given below. We can find out who is going to win the scholarship.We can find out the possibility of change in the selection of student, if we use simple arithmetic mean for selection.

Let us create a table showing wx for each student.

Now we can calculate weighted and simple mean

If weighted mean is considered, student – 3 will get scholarship. But if we consider simple mean, then, student -1 will be selected for scholarship.

AN INTERESTING PROPERTY OF AM

It is interesting to note that the sum of deviations of items in a series about arithmetic mean is always equal to zero.

Symbolically, Σ(X – X̅) = 0

However, arithmetic mean is affected by extreme items. That is, any large value on either end, can push it up or down.

Consider 6 items 20, 23, 15, 21, 28 and 19. Its arithmetic mean is 21. Their deviation from the arithmetic mean are -1, 2, -6, 0, 7, and -2 respectively.

Then, the sum of the deviations = (-1) + 2 + (-6) + 0 + 7 + (-2) = 0

• It is the most popular average
• It is easy-to understand
• It is based on all items of the series
• It is not very much affected by sampling fluctuations
• It is capable of further algebraic treatment
• It is not a positional value
• It is rigidly defined so as to avoid ambiguity

• It cannot be determined by inspection
• It cannot be determined graphically
• It is affected by extreme values
• It is not suitable for averaging ratios or percentages
• It cannot be determined if one of the observations is missing
• It may not be a number in the series
• For open-end classes, assumptions may be made about the class interval

Median is also a measure of central tendency. As the name itself suggests, it is the value of the middle item of a series arranged in ascending or descending order of magnitude. Thus if there are seven items in a series arranged in ascending or descending order of magnitude, then median will be the magnitude of the 3″ item. This item would divide the series in two equal parts; one part containing values less than the median value and the other part containing values above the median value. Thus, median is the middle most element of a series when it is arranged in the order of the magnitude of the items. If however there are even number of items in a series, then we cannot find a central item which divides the series in to two equal parts. For example, if there are eight items in a series, then, there is no single item in the middle, but two items, namely, the 4″ and the 5″. Then we take the arithmetic mean of the two middle items as the median. We can see that as against arithmetic mean, which is based on all items of the distribution, the median is only a positional value depends on the position occupied by the item.

Thus we can say that,

1. Median refers to the middle value in a distribution
2. It has a middle position in a series
3. It is also called positional average
4. It will not be affected by extreme items
5. It splits the observations into two halves

STEPS

1. Arrange the data in ascending or descending order
2. Locate the \( {{{(\frac{N + 1}{2})}^{th} }} item \, if \, N \, is\, odd \)
3. Find the mean of the \( {{{(\frac{N}{2})}^{th} }} \) item and the next item, if N is even

Median can be easily computed by sorting the data from smallest to largest and counting the middle value.

DATA OF ODD NUMBERS

Let us find the median of the following series.

8, 21, 12, 5, 32, 9, 25, 23, 5

First we arrange the items in ascending order.

5, 5, 8, 9, 12, 21, 23, 25, 32

Here there are 9 items in this series. That is N=9, which is odd.

Hence, \( Median \, =\,value \, of\, {{{(\frac{N + 1}{2})}^{th} }} item \)

\( =\,value \, of\, {{{(\frac{9 + 1}{2})}^{th} }} item \)

\( =\,value \, of\, {{{(\frac{10}{2})}^{th} }} item \)

= value of 5^th item = 12

Let us compute the median from Weekly wages of 7 workers.

Wages	100	148	80	9	155	200	145

Now we can create a table showing wages arranged in ascending order

Sl.No	Wages arranged in ascending order
1	80
2	90
3	100
4	145
5	148
6	155
7	200

Median = \( {{{(\frac{N + 1}{2})}^{th} }} item \)

\( ={{{(\frac{7 + 1}{2})}^{th} }} item \) = 4^th item.

$$ {\biggl[\frac{1\,2\,3\,\,\,4\,\,\,5\,6\,7}{|\,|\,|\,\,\,|\,\,\,|\,|\,|\,}\biggl]} $$

Size of 4^th item = 145, hence we can say that median wage is 145

DATA OF EVEN NUMBERS

Let us find median of the following series

2400, 1500, 1750, 3200, 1350, 2400, 3600, 1500, 2250, 2800.

First we arrange the items in ascending order.

1350, 1500, 1500, 1750, 2250, 2400, 2400, 2800, 3200, 3600.

Here there are 10 items. That is N = 10, which is even.

Since N is even, median = mean of the \( {{{(\frac{N}{2})}^{th} }} \) item and the next item

= mean of the \( {{{(\frac{10}{2})}^{th} }} \) item and the next item

= mean of the 5^th item and the 6,^th item

= mean of 2250 and 2400

= \( {{{(\frac{2250 + 2400}{2})} }} \) = 2325.

Let us find median of the following series using another method

2400, 1500, 1750, 3200, 1350, 2400, 3600, 1500, 2250, 2800.

First we arrange the items in ascending order.

1350, 1500, 1500, 1750, 2250, 2400, 2400, 2800, 3200, 3600.

Median = mean of the \( {{{(\frac{N + 1}{2})}^{th} }} \) item

= \( {{{(\frac{10 + 1}{2})}^{th} }} \) item

= \( {{{(\frac{11}{2})}^{th} }} \) item =5.5^th item

= 5^th item + 0.5 (6^th item – 5^th item)

= 2250 + 0.5 (2400 – 2250)

= 2250 + 0.5 × 150 = 2325.

In discrete series items are grouped and frequencies are given. There will not be any classes. For finding the median of a discrete series, first we arrange the items in ascending or descending order as we did in individual series. Then we find the less than cumulative frequencies. Now it is easy to locate the value of the \( {{{(\frac{N + 1}{2})}^{th} }} \) item.

STEPS

1. Arrange the data in ascending or descending order of magnitude
2. Take the cumulative frequencies
3. Median = the value of the \( {{{(\frac{N + 1}{2})}^{th} }} \) item
4. Look at the cumulative frequency column
5. Locate the value corresponding to \( {{{(\frac{N + 1}{2})}^{th} }} \) if it is a whole number, otherwise corresponding to higher integer.

Let us find median for the following data

Income	Number of persons
100	24
140	25
75	15
200	20
260	6
190	30

We need to arrange the data in ascending order and also need to find cumulative frequency (cf). It is shown in the below given table.

Ascending order	Number of persons (f)	cf
75	15	15
100	24	39
140	25	64
190	30	94
200	20	114
260	6	120

Median = size of \( {{{(\frac{N + 1}{2})}^{th} }} \) item

= \( {{{(\frac{120 + 1}{2})}^{th} }} \) item

= 60.5^th item

60.5 is included in cumulative frequency 64 and therefore median = 140.

In continuous series data is given in frequency classes. For finding the median, first we identify the median class. Median class is the class in which the value corresponding to the frequency \( {{{(\frac{N}{2})} }} \) may lie. After identifying the median class, we use the following formula for finding median:

Where,

L = Lower limit of the median class

N = Total frequency

cf = Cumulative frequency of the class preceding the median class

f = Frequency of the median class

h = Class width of the median class

STEPS

1. Take the cumulative frequencies
2. Find \( {{{(\frac{N}{2})} }} \)
3. Find the median class, which is the class corresponding to \( {{{(\frac{N}{2})} }} \)
4. Find the values of L, cf, and h
5. Use the formula Median = \( { L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

If classes are not in exclusive form, we need to convert them into exclusive form.

Let us find median of the following frequency distribution.

Marks	Number of students
0-10	1
10-20	3
20-30	7
30-40	12
40-50	21
50-60	16
60-70	9
70-80	4
80-90	2

First we constructs the cumulative frequency table. Here N is 75 and \( {{{(\frac{N}{2})} }} \) = 37.5. Mark the median class. That is the class corresponding to the cumulative frequency 38. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class.

Marks	Number of students	cf
0-10	1	1
10-20	3	4
20-30	7	11
30-40	12	23
40-50 (median class)	21	44
50-60	16	60
60-70	9	69
70-80	4	73
80-90	2	75
	N = 75

L = 40, \( {{{(\frac{N}{2})} }} \)= 37.5, cf = 23, f = 21, h = 10

Median = \( { L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

Median = \( { 40 + \frac{{37.5} – {23}}{21} × 10} \) = 46.9

INCLUSIVE CLASS

Let us find median from inclusive type classes

X	Frequency
1-5	22
6-10	34
11-15	53
16-20	60
21-25	48
26-30	26
31-35	18
36-40	14

In order to find median from distribution with inclusive classes, we have to convert the classes into the exclusive form along with the construction of the cumulative frequency table. In this distribution N = 275 and \( {{{(\frac{N}{2})} }} \) = 137.5. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 138. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class.Let us create a table with exclusive classes.

Inclusive Class	Converted to Exclusive	f	cf
1-5	0.5-5.5	22	22
6-10	5.5-10.5	34	56
11-15	10.5-15.5	53	109
16-20	15.5-20.5	60	169
21-25	20.5-25.5	48	217
26-30	25.5-30.5	26	243
31-35	30.5-35.5	18	261
36-40	35.5-40.5	14	275

L = 15.5, \( {{{(\frac{N}{2})} }} \) = 137.5, cf = 109, f = 60, h = 5

Median = \( { L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

Median = \( { 15.5 + \frac{{137.5} – {109}}{60} × 5} \) = 17.83

OPEN END CLASS

Let us find median from open end type classes

X	Frequency
Less than 100	40
100-200	89
200-300	148
300-400	64
400 and above	39

In the above distribution, the first and last classes are open end classes. But as in the calculation of AM, it is not needed to make assumption about their class intervals. We can just leave them as they are. In this distribution N = 380 and \( {{{(\frac{N}{2})} }} \) = 190. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 138. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Let us create a table with cumulative frequencies.

X	Frequency	cf
Less than 100	40	40
100-200	89	129
200-300	148	277
300-400	64	341
400 and above	39	380

L = 200, \( {{{(\frac{N}{2})} }} \) = 190, cf = 129, f = 148, h = 100

Median = \( { L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

Median = \( { 200 + \frac{{190} – {129}}{148} × 100} \) = 241.21

LESS THAN CUMULATIVE FREQUENCIES

Let us find median from the following distribution.

Value	Frequency
Less than 10	4
” ” 20	16
” ” – 30	40
” ” 40	76
” ” 50	96
” ” 60	112
” ” 70	120
Less than 80	125

In the above given distribution cumulative frequencies are given. We have to find the simple frequency of each class, along with the construction of the cumulative frequency table. The classes are 0-10, 10-20, etc. Simple frequency of first class is 4 itself. Simple fréquency of second class is 16 — 4 = 12 and simple frequency of third class is 40 — 16 = 24: and so on. In this distribution N = 125 and \( {{{(\frac{N}{2})} }} \) = 62.5. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 63. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Now we can create a table with exclusive class showing normal frequency.

Value	Converted to Exclusive	f	cf
Less than 10	0-10	4	4
” ” 20	10 – 20	12	16
” ” 30	20 – 30	24	40
” ” – 40	30 40	36	76
” ” 50	40 – 50	20	96
” ” 60	50 – 60	16	112
” ” 70	60 – 70	8	120
Less than 80	70 – 80	5	125

L = 30, \( {{{(\frac{N}{2})} }} \) = 62.5, cf = 40, f = 36, h = 10

Median = \( { L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

Median = \( { 30 + \frac{{62.5} – {40}}{36} × 10} \) = 36.25

MORE THAN CUMULATIVE FREQUENCIES

Let us find median from the following distribution.

Value	Frequency
More than 10	50
” ” 20	43
” ” 30	28
” ” 40	13
More than 50	4

In the above given distribution, more than cumulative frequencies are given. We have to find the simple frequency of each class, and less than cumulative frequencies. These classes are 10-20, 20-30, etc. Simple frequency of first class is 50-43 = 7. Simple frequency of second class is 43-28 = 15 and simple frequency of third class is 28-13 = 15, and so on. In this distribution N = 50 and \( {{{(\frac{N}{2})} }} \) = 25. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 25. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Let us create a table with simple frequencies and cumulative frequencies.

Value	Converted to Exclusive	f	cf
More than 10	10-20	7	7
” ” 20	20-30	15	22
” ” 30	30-40	15	37
” ” 40	40-50	9	46
More than 50	50-60	4	50

L = 30, \( {{{(\frac{N}{2})} }} \) = 25, cf = 22, f = 15, h = 10

Median = \( { L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

Median = \( { 30 + \frac{{25} – {22}}{15} × 10} \) = 32

WITH MIDPOINTS OF CLASSES

Let us find median from the following distribution.

Mid values	Frequencies
2.5	4
7.5	12
12.5	18
17.5	10
22.5	7
27.5	4

In the above given distribution, only mid values are given. We have to find corresponding classes and cumulative frequencies for determining the median. These classes are 0-5, 5-10, etc. In this distribution N = 55 and \( {{{(\frac{N}{2})} }} \) = 27.5. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 28. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Let us create a table with classes and cumulative frequencies.

X	f	cf
0 – 5	4	4
5 – 10	12	16
10 – 15	18	34
15 – 20	10	44
20 – 25	7	51
25 – 30	4	55

L = 10, \( {{{(\frac{N}{2})} }} \) = 27.5, cf = 16, f = 18, h = 5

Median = \( { L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

Median = \( { 10 + \frac{{27.5} – {16}}{18} × 5} \) = 13.2

WITH UNEQUAL CLASS INTERVALS

Let us find median from the following distribution.

Values	Frequencies
0 – 10	7
10 – 20	13
20 – 50	24
50 – 70	48
70 – 80	26
80 – 100	12

In the above given distribution, class intervals are unequal. But the frequencies need not to be adjusted to make the class intervals equal. Bu it should be remember that the value of h in the formula is the class interval of the median class. In this distribution N = 130 and \( {{{(\frac{N}{2})} }} \) = 65. As we did in the previous work, mark the median class. That is the class corresponding to the cumulative frequency 65. Then mark the frequency of the median class. And then mark the cumulative frequency preceding to the median class. Let us create a table with cumulative frequencies.

Values	Frequencies	cf
0 – 10	7	7
10 – 20	13	20
20 – 50	24	44
50 – 70	48	92
70 – 80	26	118
80 – 100	12	130

L = 50, \( {{{(\frac{N}{2})} }} \) = 65, cf = 44, f = 48, h = 20

Median = \( { L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

Median = \( { 50 + \frac{{65} – {44}}{48} × 20} \) = 58.75

We had studied the method of finding median graphically. In order to find the median graphically, we draw the less than ogive and greater than ogive in the same graph and from the point of intersection of the ogives we draw a line perpendicular to the x-axis. Then, the point where the perpendicular touches the x-axis will give the value of the median. We can also very well locate median by drawing a single ogive. For example, if we draw the less than ogive, then take \( {{{\frac{N}{2}} }} \) on the y-axis and draw a perpendicular from y-axis to meet the ogive. From the point where it meets the ogive, draw another perpendicular on the x-axis. That point on the x-axis will give the value of the median.

Let us find median of the following distribution using less than ogive.

Values	Frequencies
0 – 2	2
2 – 4	4
4 – 6	5
6 – 8	8
8 – 10	7
10 – 12	4

We can create a table showing cumulative frequencies.

Values	Frequencies	cf
0 – 2	2	2
2 – 4	4	6
4 – 6	5	11
6 – 8	8	19
8 – 10	7	26
10 – 12	4	30
	N = 30

Using above given cumulative frequency table we loacate median. You can see how median is located using less than ogive from the below given graph.

1. It is easy to compute and understand
2. It gives best results with open-end classes
3. It is not influenced by the magnitude of extreme deviations
4. It is the most appropriate average in dealing with qualitative data
5. The value of median can be determined graphically

1. It is only a positional average
2. It may not be the true representative of the series in many cases.
3. It is not based on all items
4. The value of median is affected by sampling fluctuations
5. It is not capable of algebraic treatment

Now we understood that median of a distribution is that value which divides the distribution in to two equal parts. We can also find three values which divide the distribution in to four equal parts. These three values are called the quartiles. Likewise we can find nine values which divide the distribution in to ten equal parts. These nine values are called deciles. In a similar manner we can define percentiles. Percentiles are ninety nine values which divide the distribution in to hundred equal parts.

Quartiles are those values which divide the series into four equal parts. There are only three quartiles. We denote them by Q₁, Q₂, and Q₃. Q₁ is called the lower quartile, which is the value such that (1/4)^th of the total observations fall below it and (3/4)^th above it. Q₂ is the same as median. Q₃ is called the upper quartile, which is the value such that (3/4)^th of the total observations fall below it and (1/4)^th above it. The method of determining quartiles is same as that of median.

STEPS

1. Arrange the data in ascending order
2. \( {\mathbf Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th} \) item
3. \( {\mathbf Q_1 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th} \) item

DATA OF ODD NUMBERS

Find the first quartile Q₁, the median (Q₂) and the third quartile Q₃ of the following series.

12, 6, 21, 15, 9, 20, 16, 13, 8, 7, 9, 11, 5, 8, 10

Let us arrange the data in ascending order

5, 6, 7, 8, 8, 9, 9, 10, 11, 12, 13, 15, 16, 20, 21

Here N = 15

\( {\mathbf Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th} \) item

\( { = value\, of\, the\, \Biggl[\frac {{15} + {1}}{4}}\Biggl]^{th} \) item

= value of the 4^th item = 8.

\( {\mathbf Q_2 = \Biggl[\frac {{N} + {1}}{2}}\Biggl]^{th} \) item

\( { = value\, of\, the\, \Biggl[\frac {{15} + {1}}{2}}\Biggl]^{th} \) item

= value of the 8^th item = 10.

\( {\mathbf Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th} \) item

\( { = value\, of\, the\, \Biggl[\frac {({15} + {1})}{4}}\Biggl]^{th} \) item

= value of the 12^th item = 15.

DATA OF EVEN NUMBERS

Find the first quartile Q₁, the median (Q₂) and the third quartile Q₃ of the following series.

166, 167, 171, 160, 133, 175, 156, 158, 175, 142, 120, 133

Let us arrange the data in ascending order

120, 133, 133, 142, 156, 158, 175, 166, 160, 167, 171, 175, 175

\( {\mathbf Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th} \) item

\( { = value\, of\, the\, \Biggl[\frac {{12} + {1}}{4}}\Biggl]^{th} \) item

= value of the 3.25^th item

= value of the 3^rd item + 0.25 (4^th item – 3^rd item)

= 133 + 0.25 (142 – 133)

= 133 + 0.25(9)

= 135.25.

\( {\mathbf Q_2 = \Biggl[\frac {{N} + {1}}{2}}\Biggl]^{th} \) item

\( { = value\, of\, the\, \Biggl[\frac {{12} + {1}}{2}}\Biggl]^{th} \) item

= value of the 6.5^th item

= value of the 6^th item + 0.5 (7^th item – 6^rd item)

= 158 + 0.5 (166 – 158)

= 158 + 0.5(8)

= 162.

\( {\mathbf Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th} \) item

\( { = value\, of\, the\, \Biggl[\frac {({12} + {1})}{4}}\Biggl]^{th} \) item

= value of the 9.75^th item

= value of the 9^th item + 0.75 (10^th item – 9^th item)

= 167 + 0.75 (171 – 167)

= 167 + 0.75(4)

= 170.

STEPS

1. Arrange the data in ascending order
2. Find the cumulative frequencies
3. \( \mathbf {Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th} \) item
4. \( \mathbf {Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th} \) item

Let us find the first quartile Q₁, the median (Q₂) and the third quartile Q₃ of the following series.

Values	Frequencies
5	8
8	13
12	20
15	22
17	11
20	6

Here N = 80,

\( \mathbf {Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th} \) = 20.25

\( \mathbf {Q_2 = \Biggl[\frac {{N} + {1}}{2}}\Biggl]^{th} \) = 40.5

\( \mathbf {Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th} \) = 60.75

Now we can create a table showing cumulative frequencies

Values	Frequencies	cf
5	8	8
8	13	21 (Q1 lies here)
12	20	41 (Q2 lies here)
15	22	63 (Q3 lies here)
17	11	74
20	6	80
	N = 80

\( \mathbf {Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th} \) item

= value of the 21^st item = 8

\( \mathbf {Q_2 = \Biggl[\frac {{N} + {1}}{2}}\Biggl]^{th} \) item

= value of the 41^st item = 12

\( \mathbf {Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th} \) item

= value of the 61^st item = 15

STEPS

1. Find the cumulative frequency
2. Find Q₁ and Q_{3 classes}
3. \( { \mathbf Q_1 =\, size\, of\,\frac {{N}}{4}}^{th} \) item
4. \( { \mathbf Q_3 =\, size\, of\,\frac {3{N}}{4}}^{th} \) item

After locating these classes, the values of Q₁ and Q₃ can be interpolated.

$$ \mathbf Q_1 \,= \,{ L + \frac{\frac{N}{4} – {cf}}{f} × h} $$

Where,

L = the lower limit of the lower quartile class

f = freqwuency of the lower quartile class

h = magnitude of the lower quartile class

cf = the cumulative frequency of the class preceding the lower quartile class $$ \mathbf Q_3 \,= \,{ L + \frac{\frac{3N}{4} – {cf}}{f} × h} $$

Where,

L = the lower limit of the upper quartile class

f = freqwuency of the upper quartile class

h = magnitude of the upper quartile class

cf = the cumulative frequency of the class preceding the upper quartile class

Let us find the first quartile Q₁, the median (Q₂) and the third quartile Q₃ of the following series.

Age	Frequencies
0-5	2
5-10	5
10-15	9
15-20	13
20-25	18
25-30	24
30-35	19
35-40	17
40-45	10
45-50	3

We can create a table with cumulative frequency to find Q₁, Q₂, and Q₃.

Age	Frequencies	cf
0-5	2	2
5-10	5	7
10-15	9	16
15-20	13	29
20-25	18	47
25-30	24	71
30-35	19	90
35-40	17	107
40-45	10	117
45-50	3	120
	N = 120

N = 120, \( { \frac {{N}}{4}} \) = 30

Lower Quartile Class = 20 – 25

L = 20, \( {\frac {{N}}{4}} \) = 30, cf = 29, f = 18, h = 5

\( \mathbf Q_1 \,= \,{ L + \frac{\frac{N}{4} – {cf}}{f} × h} \)

= \( { 20 + \frac{{30} – {29}}{18} × 5} \)

= 20.28

N = 120, \( { \frac {{N}}{2}} \) = 60

Median Class = 25 – 30

L = 25, \( {\frac {{N}}{2}} \) = 60, cf = 47, f = 24, h = 5

\( \mathbf Q_2 \,= \,{ L + \frac{\frac{N}{2} – {cf}}{f} × h} \)

= \( { 25 + \frac{{60} – {47}}{24} × 5} \)

= 27.7

N = 120, \( { \frac {3{N}}{4}} \) = 90

Upper Quartile Class = 30 – 35

L = 30, \( {\frac {3{N}}{4}} \) = 90, cf = 71, f = 19, h = 5

\( \mathbf Q_3 \,= \,{ L + \frac{\frac{3{N}}{4} – {cf}}{f} × h} \)

= \( { 30 + \frac{{90} – {71}}{19} × 5} \)

= 35

Like, Arithmetic Mean and Median, Mode is also a measure of central tendency. It is most common item of a series. It represents the most typical value of a series. It is the value which occurs the largest number of times in a series. Mode is the value around which there is the greatest concentration of values. In other words, it is the item having the largest frequency. In some cases, there may be more than one point of concentration of values and the series may be bi-modal or multi-modal. When one value occurs more frequently than any other value, the distribution is called unimodal.

The word mode is derived from the French word ‘la mode’ which means fashion or the most popular phenomenon. Mode, is thus the most popular item of a series around around which there is the highest frequency density. It is denoted by M_o.

There can be more than one mode in a distribution. A distribution with a single modal value is called Unimodal.

A distribution with two model value is called Bimodal and with more than two is Multimodal.

It may so happen that there may be no mode at all in a distribution when no value appears more frequently than any other value. For example, in a series 1, 1, 2, 2, 3, 3, 4, 4, there is no mode.

Let us discuss the methods of finding mode in diffrent series of data.

Comparing to mean and median, computation of mode is easy. In individual series, mode is that value which repeats highest number of times. It is often found by mere inspection.

Let us find the mode of the following series.

2, 5, 7, 3, 6, 4, 5, 8, 2, 7, 5, 10, 9, 5

Here the item 2 repeats two times; 5 repeats four times and 7 repeats two times. All the other items occurs once. So the most frequent item is 5; and therefore the mode is 5.

Let us find the mode of the another series.

2, 5, 7, 3, 6, 4, 3, 5, 8, 2, 3, 7, 5, 10, 5, 9, 3.

Here the item 2 repeats two times; 5 repeats four times; 7 repeats two times and 3 repeats four times. All the other items occur once. When there are two or more values having the same maximum frequency, mode is said to be ill-defined. We cannot find the modal value of this series in the said way. In such case; we use a formula for finding mode, which is known as the empirical formula.

Here, Mean = \( {{{\frac{N}{2}} }} \)

= \( {{{\frac{2+5+7+3+6+4+3+5+8+2+3+7+5+10+5+9+3}{17}} }} \)

= \( {{{\frac{87}{17}} }} \)

= 5.1

In order to find median we need to arrange the data in ascending order.

2, 2, 3, 3, 3, 3, 4, 5,5, 5, 5, 6, 7, 7, 8, 9, 10

Median = value of the \( \Biggl[{{{\frac{N+1}{2}} }}\Biggl]^{th} \) item

= \( \Biggl[{{{\frac{17+1}{2}} }}\Biggl]^{th} \) item

= value of the 9^th item

= 5

∴ Mode = 3 median – 2 mean

= (3 × 5) – (2 × 5.1)

= 15 – 10.2

= 4.8

In discrete series, mode is determined just by inspection. The item having highest frequency is taken as mode.

Let us find mode of the following distribution.

X	Frequencies
5	7
10	12
15	15
20	18
25	13
30	10
35	5
40	2

The value having highest frequency is 20; and therefore, 20 is the modal value.

I n continuous series, mode lies in the class having highest frequency. Hence the modal class may be determined either by inspection or by grouping table. Then mode is determined using the formula:

L = lower limit of the modal class

D₁ = difference between the frequencies of the modal class and the class preceding it (ignoring the sign)

D₂ = difference between frequencies of the modal class and the class succeeding it (ignoring the sign); and

h = class interval of the modal class

Let us calculate the mode of the following series.

X	Frequencies
0 – 5	7
5 – 10	10
10 – 15	12
15 – 20	18
20 – 25	13
25 – 30	8
30 – 35	5
35 – 40	2

The class having highest frequency is 15 – 20

∴ Modal class = 15 – 20

Now, mark the modal class. Then mark frequencies just above and below the modal class.

X	Frequencies
0 – 5	7
5 – 10	10
10 – 15	12
15 – 20	18
20 – 25	13
25 – 30	8
30 – 35	5
35 – 40	2

L = 15, D₁ = 18 – 12 = 6, D₂ = 18 – 13 = 5, h = 5

\( M_o \,= \,{ L + \frac{{D_1}}{{D_1}+{D_2}} × h} \)

\( = \,{ 15 + \frac{{6}}{{6}+{5}} × 5} \)

\( = \,{ 15 + \frac{{30}}{{11}}} \)

= 15 + 2.73

= 17.73

INCLUSIVE CLASS

Look at the below given distribution. It is a distribution having inclusive type classes. Let us find mode of the distribution.

X	Frequencies
0 – 9	22
10 – 19	34
20 – 29	53
30 – 39	85
40 – 49	48
50 – 59	26
60 – 69	18
70 – 79	14

We need to convert the classes into the exclusive form as in the given below table.

X	Frequencies
.5 – 9.5	22
9.5 – 19.5	34
19.5 – 29.5	53
29.5 – 39.5	85
39.5 – 49.5	48
49.5 – 59.5	26
59.5 – 69.5	18
69.5 – 79.5	14

The class having highest frequency is 29.5 – 39.5

∴ Modal class = 29.5 – 39.5

Now, mark the modal class. Then mark frequencies just above and below the modal class.

L = 29.5, D₁ = 85 – 53 = 32, D₂ = 85 – 48 = 37, h = 10

\( M_o \,= \,{ L + \frac{{D_1}}{{D_1}+{D_2}} × h} \)

\( = \,{ 29.5 + \frac{{32}}{{32}+{37}} × 10} \)

\( = \,{ 29.5 + \frac{{320}}{{69}}} \)

= 29.5 + 4.64

= 34.14

OPEN END CLASSES

Let us find mode of the following distribution.

X	Frequencies
Less than 100	40
100 – 200	89
200 – 300	148
300 – 400	64
400 and above	39

In the above given distribution, the first and last classes are open end classes. But as in the calculation of AM, it is not needed to make assumptions about their class intervals . We may just leave them as they are.

Let us create a table to show model classes, lower and upper frequencies of model class frequency.

X	Frequencies
Less than 100	40
100 – 200	89
200 – 300	148
300 – 400	64
400 and above	39

The class having highest frequency is 200 – 300

∴ Modal class = 200 – 300

Now, mark the modal class. Then mark frequencies just above and below the modal class.

L = 200, D₁ = 148 – 89 = 59, D₂ = 148 – 64 = 84, h = 100

\( M_o \,= \,{ L + \frac{{D_1}}{{D_1}+{D_2}} × h} \)

\( = \,{ 200 + \frac{{59}}{{59}+{84}} × 100} \)

\( = \,{ 200 + \frac{{5900}}{{143}}} \)

= 200 + 41.25

= 241.3

LESS THAN CUMULATIVE FREQUENCY

Let us find mode of the following frequency distribution.

Value	Frequencies
Less than 60	65
” ” 80	185
” ” 100	270
” ” 120	342
Less than 140	400

In the above given distribution, only cumulative frequencies are given. We have to find class and simple frequencies of each class. This is shown in the below given table.

Value	Frequencies
40 – 60	65
60 – 80	120
80 – 100	85
100 – 120	72
120 – 140	58

The class having highest frequency is 60 – 80

∴ Modal class = 60 – 80

Now, mark the modal class. Then mark frequencies just above and below the modal class.

L = 60, D₁ = 120 – 65 = 55, D₂ = 120 – 85 = 35, h = 20

\( M_o \,= \,{ L + \frac{{D_1}}{{D_1}+{D_2}} × h} \)

\( = \,{ 60 + \frac{{55}}{{55}+{35}} × 20} \)

\( = \,{ 60 + \frac{{1100}}{{90}}} \)

= 60 + 12.2

= 72.22

WITH UNEQUAL CLASS INTERVALS

Let us find mode of the following frequency distribution.

Value	Frequencies
0 – 30	9
30 – 40	8
40 – 50	15
50 – 60	9
60 – 70	3
70 – 90	4

In the above given distribution the class intervals are unequal. For finding mode, the frequencies should be adjusted to make the class intervals equal, Smallest class interval is 10. So make all the classes with class interval 10. Class 0-30 to be replaced by classes 0-10, 10-20 and 20-30 and the class 70-90 to be replaced by classes 70-80 and 80-90. The frequencies of the classes 0-10, 10-20 and 20-30 will be \( { \frac{{9}}{{3}}} = 3 \) each; and frequencies of the classes 70 – 80 and 80 – 90 will be \( { \frac{{4}}{{2}}} = 2 \).

We can create a table showing adjusted classes and frequencies in the below given table.

Value	Frequencies
0 – 10	3
10 – 20	3
20 – 30	3
30 – 40	8
40 – 50	15
50 – 60	9
60 – 70	3
70 – 90	4
70 – 80	2
80 – 90	2

The class having highest frequency is 40 – 50

∴ Modal class = 40 – 50

Now, mark the modal class. Then mark frequencies just above and below the modal class.

L = 40, D₁ = 15 – 8 = 7, D₂ = 15 – 9 = 6, h = 10

\( M_o \,= \,{ L + \frac{{D_1}}{{D_1}+{D_2}} × h} \)

\( = \,{ 40 + \frac{{7}}{{7}+{6}} × 10} \)

\( = \,{ 40 + \frac{{70}}{{13}}} \)

= 40 + 5.38

= 45.38

STEPS

1. Draw a histogram of the given data.
2. Draw two lines diagonally in the inside of the modal class bar, starting from each corner of the bar to the upper corner of the adjacent bar.
3. Then, draw a perpendicular line from the point of intersection to the X – axis, which gives us-the modal value.

The monthly profits in rupees of 100 business concerns are distributed as follows:

X	Frequencies
0 – 100	12
100 – 200	18
200 – 300	27
300 – 400	20
400 – 500	17
500 – 600	6

Let us find mode by drawing histogram for the above given distribution.

MERITS OF MODE

• Mode is the true representative value of a distribution.
• It is not affected by extreme items.
• It can be determined in open-end distributions.
• It can be used to describe qualitative phenomenon.
• It can also be determined graphically.

DEMERITS OF MODE

• In the case of bi-modal series, mode cannot be determined.
• It is not capable of algebraic treatment.
• It is not based on each and every item of the series.
• It is not a rigidly defined one.

Relative Position of Arithmetic Mean, Median and Mode

In a moderately asymmetrical distribution,

$$ Mode > Median >Mean $$

$$or$$ $$ Mode < Median < Mean $$

Median is always in between Mode and Mean.

karl Pearson has expressed a relationship for a moderately skewed distribution as follows:

$$ Mode = 3 Median – 2 Mean $$