Measures of central tendency

Measures of CentralMeasures of Central
TendencyTendency

Concept of Central TendencyConcept of Central Tendency
 A measure of central tendency is a typicalA measure of central tendency is a typical
value around which other figures congregatevalue around which other figures congregate
- Simpson & Kalfa- Simpson & Kalfa
OROR
An average is a single value which is used toAn average is a single value which is used to
represent all of the values in the series.represent all of the values in the series.

Measures of Central Tendency
Mean (mathematical
average)
Median (positional
average)
Mode (positional
average)
Arithmetic Mean Geometric mean Harmonic mean
Simple Arithmetic Mean
Weighted Arithmetic Mean
Mean of Composite Group

BasicsBasics
• MeanMean AverageAverage
• MedianMedian Mid positional valueMid positional value
• ModeMode Most frequentlyMost frequently
occurring valueoccurring value
44

Arithmetic MeanArithmetic Mean
Ungrouped (Raw) DataUngrouped (Raw) Data
sObservationofNumber
sObservationofSum
=x
n
xi∑
=

EXAMPLEEXAMPLE
Table 4.1 : Equity Holdings of 20 Indian Billionaires
( Rs. in Millions)
2717 2796 3098 3144 3527
3534 3862 4186 4310 4506
4745 4784 4923 5034 5071
5424 5561 6505 6707 6874

ExampleExample
For the above data, the A.M. is
2717 + 2796 +…… 4645+….. + 5424 + ….+ 6874
= --------------------------------------------------------------------------
20
= Rs. 4565.4 Millions
x

Arithmetic MeanArithmetic Mean
Grouped DataGrouped Data
 N=N= = Total frequency= Total frequency
 Here, xHere, xii is the mid value of the class interval.is the mid value of the class interval.
∑
∑=
i
i
f
f ix
x
∑=
n
i
if
1

exampleexample
 Calculate arithmeticCalculate arithmetic
mean from themean from the
following frequencyfollowing frequency
distribution of marks atdistribution of marks at
a test in statistics.a test in statistics.
MarksMarks No.ofNo.of
studentsstudents
2525 22
3030 33
3535 44
4040 88
4545 99
5050 44
5555 33
6060 22

 The details of the monthly salary of 100The details of the monthly salary of 100
employees of a firm are given below:employees of a firm are given below:
Monthly salary (in Rs.) No. of employees
1000 18
1500 26
2000 31
2500 16
3000 5
5000 4

 In grouped data, the middle value of each group is theIn grouped data, the middle value of each group is the
representative of the group bz when the data arerepresentative of the group bz when the data are
grouped, the exact frequency with which each valuegrouped, the exact frequency with which each value
of the variable occurs in the distribution is unknown.of the variable occurs in the distribution is unknown.
 We only know the limits within which a certainWe only know the limits within which a certain
number of frequencies occur.number of frequencies occur.
 So, we make an assumption that the frequenciesSo, we make an assumption that the frequencies
within each class are distributed uniformly over thewithin each class are distributed uniformly over the
range of class interval.range of class interval.

ExampleExample
 A company manufactures polythene bags. TheA company manufactures polythene bags. The
bags are evaluated on the basis of theirbags are evaluated on the basis of their
strength by buyers. The strength depends onstrength by buyers. The strength depends on
their bursting pressures. The following datatheir bursting pressures. The following data
relates to the bursting pressure recorded in arelates to the bursting pressure recorded in a
sample of 90 bags. Find the average burstingsample of 90 bags. Find the average bursting
pressure.pressure.

exampleexample
Bursting
pressure
( 1 )
No. of
bags
( fi
) ( 2 )
Mid Value of
Class Interval
( xi
) ( 3 )
Fi
xi
( 4 )
Col.(4) = Col.(2) x Col.
(3)
5-10 10 7.5 75
10-15 15 12.5 187.5
15-20 20 17.5 350
20-25 25 22.5 562.5
25-30 20 27.5 550
Sum Σ fi
=90 Σ fi
xi =1725

values of Σ fi
and Σ fi
xi
, in formula
= 1725/90
= 19.17
∑
∑=
i
i
f
f ix
x

EXAMPLE (short cut method)EXAMPLE (short cut method)
 Calculate the mean ofCalculate the mean of
the followingthe following
distribution ofdistribution of
monthly wages ofmonthly wages of
workers in a factory :workers in a factory :
MonthlyMonthly
wages(inwages(in
Rs.)Rs.)
No. ofNo. of
workersworkers
100-120100-120 1010
120-140120-140 2020
140-160140-160 3030
160-180160-180 1515
180-200180-200 55

 The followingThe following
frequency distributionfrequency distribution
represents the timerepresents the time
taken in seconds totaken in seconds to
serve customers at a fastserve customers at a fast
food take away.food take away.
Calculate the mean timeCalculate the mean time
taken by to servetaken by to serve
customerscustomers
Time taken (in
seconds)
frequencies
40-60 6
60-80 12
80-100 15
100-120 12
120-140 10
140-160 5

Weighted Arithmetic Mean
 It takes into account the importance of eachIt takes into account the importance of each
value to the overall data with the help of thevalue to the overall data with the help of the
weights.weights.
 Frequency i.e. the no. of occurrence indicatesFrequency i.e. the no. of occurrence indicates
the relative importance of a particular data in athe relative importance of a particular data in a
group of observations.group of observations.
 Used in case the relative importance of eachUsed in case the relative importance of each
observation differs or when rates, percentagesobservation differs or when rates, percentages
or ratios are being averaged.or ratios are being averaged.

 The weighted AM of the n observations:The weighted AM of the n observations:
 AM is considered to be the best measure of centralAM is considered to be the best measure of central
tendency as its computation is based on each andtendency as its computation is based on each and
every observation.every observation.
∑
∑=
wi
wi ix
x

ExampleExample
 5 students of a B.Sc. (Hons)5 students of a B.Sc. (Hons)
course are marked by using thecourse are marked by using the
following weighing scheme :following weighing scheme :
 Mid-term = 20%Mid-term = 20%
 Project = 10%Project = 10%
 Attendance = 10%Attendance = 10%
 Final Exam = 60%Final Exam = 60%
Calculate the average marks in theCalculate the average marks in the
examination.examination.
Marks of the students in variousMarks of the students in various
components are:components are:
StudStud
entent
midtmidt
ermerm
ProjeProje
ctct
AttenAtten
dncednce
FinFin
alal
11 6565 7070 8080 8080
22 4848 5858 5454 6060
33 5858 6363 6565 5050
44 5858 7070 5454 6060
55 6060 6565 7070 7070

 A professor is interested inA professor is interested in
ranking the following fiveranking the following five
students in the order of merit onstudents in the order of merit on
the basis of data given below:the basis of data given below:
 Attendance average will count forAttendance average will count for
20% of a student’s grade; the20% of a student’s grade; the
homework 25%; assignment 35%;homework 25%; assignment 35%;
midterm examination 10% andmidterm examination 10% and
final examination 10%. Whatfinal examination 10%. What
would be the students ranking.would be the students ranking.
Stud
ent
Atte
nda
nce
Hom
ewo
rk
Assi
gnm
ent
Midt
erm
final
A 85 89 94 87 90
B 78 84 88 91 92
C 94 88 93 86 89
D 82 79 88 84 93
E 95 90 92 82 88

Mean of composite groupMean of composite group
 If two groups contain respectively, nIf two groups contain respectively, n11 and nand n22
observations with mean Xobservations with mean X11 and Xand X22, then the, then the
combined mean (X) of the combined group of ncombined mean (X) of the combined group of n11+n+n22
observations is given by :observations is given by :
21
2211
12
nn
XnXn
X
+
+
=

ExampleExample
 There are two branches of a companyThere are two branches of a company
employing 100 and 80 employeesemploying 100 and 80 employees
respectively. If arithmetic means of therespectively. If arithmetic means of the
monthly salaries paid by two branches aremonthly salaries paid by two branches are
Rs. 4570 and Rs. 6750 respectively, find theRs. 4570 and Rs. 6750 respectively, find the
A.M. of the salaries of the employees of theA.M. of the salaries of the employees of the
company as a whole.company as a whole.

 A factory has 3 shifts :- Morning, evening andA factory has 3 shifts :- Morning, evening and
night shift. The morning shift has 200 workers,night shift. The morning shift has 200 workers,
the evening shift has 150 workers and nightthe evening shift has 150 workers and night
shift has 100 workers. The mean wage of theshift has 100 workers. The mean wage of the
morning shift workers is Rs. 200, the eveningmorning shift workers is Rs. 200, the evening
shift workers is Rs. 180 and the overall meanshift workers is Rs. 180 and the overall mean
of the workers is Rs. 160. Find the mean wageof the workers is Rs. 160. Find the mean wage
of the night shift workers.of the night shift workers.

Properties of A.M.Properties of A.M.
 If a constant amount is added or subtracted from each value inIf a constant amount is added or subtracted from each value in
the series, mean is also added or subtracted by the samethe series, mean is also added or subtracted by the same
constant amount. E.g. Consider the values 3,5,9,15,16constant amount. E.g. Consider the values 3,5,9,15,16
A.M. = 9.6A.M. = 9.6
If 2 is added to each value, thenIf 2 is added to each value, then A.M. = 11.6 = 9.6 + 2.A.M. = 11.6 = 9.6 + 2.
Thus, mean is also added by 2.Thus, mean is also added by 2.
 Sum of the deviations of a set of observations say x1, x2, , xnSum of the deviations of a set of observations say x1, x2, , xn
from their mean is equal to zero.from their mean is equal to zero.
 A.M. is dependent on both change in origin and scale.A.M. is dependent on both change in origin and scale.
 The sum of the squares of the deviations of a set ofThe sum of the squares of the deviations of a set of
observation from any number say A is least when A is X.observation from any number say A is least when A is X.

Merits and demerits ofMerits and demerits of
Arithmetic Mean
Advantages Disadvantages
(i) Easy to understand and
calculate
(ii) Makes use of full data
(iii) Based upon all the
observations.
(i ) Unduly influenced by extreme
values
(ii) Cannot be
calculated from the data with
open-end class.e.g. below 10
or above 90
(iii) It cannot be obtained if a single
observation is missing.
(iv) It cannot be used if we are
dealing with qualitative
characteristics which cannot be
measured quantitatively;
intelligence, honesty, beauty

Harmonic MeanHarmonic Mean
The harmonic mean (H.M.) is defined as the reciprocal
of the arithmetic mean of the reciprocals of the
observations.
For example, if x1
and x2
are two observations, then the
arithmetic means of their reciprocals viz 1/x1
and 1/ x2
is
{(1 / x1
) + (1 / x2
)} / 2
= (x2
+ x1
) / 2 x1
x2
The reciprocal of this arithmetic mean is 2 x1
x2
/ (x2
+
x1
). This is called the harmonic mean.
Thus the harmonic mean of two observations x1
and x2
is 2 x1
x2
-----------------

 In general, for the set of n observations XIn general, for the set of n observations X11,X,X22……..X……..Xnn,,
HM is given by :HM is given by :
 And for the same set of observations with frequenciesAnd for the same set of observations with frequencies
ff11,f,f22……..f……..fnn, HM is given by:, HM is given by:
∑
=
ix
n
HM
1
∑
=
i
i
x
f
n
HM

 HM gives the largest weight to the smallestHM gives the largest weight to the smallest
item and the smallest weight of the largestitem and the smallest weight of the largest
itemitem
 If each observation is divided by a constant, KIf each observation is divided by a constant, K
then HM is also divided by the same constant.then HM is also divided by the same constant.
 If each observation is multiplied by a constant,If each observation is multiplied by a constant,
K then HM is also multiplied by the sameK then HM is also multiplied by the same
constant.constant.
 It is used in averaging speed, price of articles.It is used in averaging speed, price of articles.

 If time varies w.r.t. a fixed distance then HM determines theIf time varies w.r.t. a fixed distance then HM determines the
average speed.average speed.
 If distance varies w.r.t. a fixed time then AM determines theIf distance varies w.r.t. a fixed time then AM determines the
average speed.average speed.
 EXAMPLE : If a man moves along the sides of a square withEXAMPLE : If a man moves along the sides of a square with
speed v1, v2, v3, v4 km/hr, the average speed for the wholespeed v1, v2, v3, v4 km/hr, the average speed for the whole
journey =journey = 44
(1/v1)+(1/v2)+(1/v3)+(1/v4)(1/v1)+(1/v2)+(1/v3)+(1/v4)

EXAMPLEEXAMPLE
 In a certain factory a unit of work isIn a certain factory a unit of work is
completed by A in 4 min, by B in 5 min, bycompleted by A in 4 min, by B in 5 min, by
C in 6 min, by D in 10 min, and by E in 12C in 6 min, by D in 10 min, and by E in 12
minutes.minutes.
 What is the average no. of units of workWhat is the average no. of units of work
completed per minute?completed per minute?

ExampleExample
 The profit earned by 19The profit earned by 19
companies is givencompanies is given
below:below:
calculate the HM ofcalculate the HM of
profit earned.profit earned.
ProfitProfit
(lakhs)(lakhs)
No. ofNo. of
companiescompanies
20-2520-25 44
25-3025-30 77
30-3530-35 44
35-4035-40 44

Geometric MeanGeometric Mean
Neither mean, median or mode is the appropriate average in
calculating the average % rate of change over time. For this G.M. is
used.
The Geometric Mean ( G. M.) of a series of observations with x1
, x2
,
x3
, ……..,xn
is defined as the nth
root of the product of these
values . Mathematically
G.M. = { ( x1
)( x2
)( x3
)…………….(xn
) } (1/ n )
It may be noted that the G.M. cannot be defined if any value of x is
zero as the whole product of various values becomes zero.

 When the no. of observation is three or more then to simplifyWhen the no. of observation is three or more then to simplify
the calculations logarithms are used.the calculations logarithms are used.
log G.M. = log X1 + log X2 + ……+ log Xnlog G.M. = log X1 + log X2 + ……+ log Xn
NN
G.M. = antilog (log X1 + log X2 + ……+ log Xn)G.M. = antilog (log X1 + log X2 + ……+ log Xn)
NN
For grouped data,For grouped data,
G.M. = antilog (f1log X1 + f2log X2 + ……+ fnlog Xn)G.M. = antilog (f1log X1 + f2log X2 + ……+ fnlog Xn)
NN

Geometric meanGeometric mean
 GM is often used to calculate the rate ofGM is often used to calculate the rate of
change of population growth.change of population growth.
 GM is also useful in averaging ratios, rates andGM is also useful in averaging ratios, rates and
percentages.percentages.

EXAMPLEEXAMPLE
 A machinery is assumed to depreciate 44% in value in firstA machinery is assumed to depreciate 44% in value in first
year, 15% in second year and 10% in next three years, eachyear, 15% in second year and 10% in next three years, each
percentage being calculated on diminishing value. What ispercentage being calculated on diminishing value. What is
the average % of depreciation for the entire period?the average % of depreciation for the entire period?
 Compared to the previous year the overhead expenses wentCompared to the previous year the overhead expenses went
up by 32% in 2002; they increased by 40% in the next yearup by 32% in 2002; they increased by 40% in the next year
and by 50% in the following year. Calculate the average rateand by 50% in the following year. Calculate the average rate
of increase in the overhead expenses over the three years.of increase in the overhead expenses over the three years.

ExampleExample
 The annual rate of growth for a factory for 5 years isThe annual rate of growth for a factory for 5 years is
7%,8%,4%,6%,10%respectively.What is the average7%,8%,4%,6%,10%respectively.What is the average
rate of growth per annum for this period.rate of growth per annum for this period.
 The price of the commodity increased by 8% from 1993The price of the commodity increased by 8% from 1993
to 1994,12%from 1994 to 1995 and 76% from 1995 toto 1994,12%from 1994 to 1995 and 76% from 1995 to
1996.the average price increase from 1993 to 1996 is1996.the average price increase from 1993 to 1996 is
quoted as 28.64% and not 32%.Explain and verify thequoted as 28.64% and not 32%.Explain and verify the
result.result.
3636

Combined G.M. of Two Sets ofCombined G.M. of Two Sets of
DataData
If G1
& G2
are the Geometric means of two sets
of observations of sizes n1 and n2, then the
combined Geometric mean, say G, of the
combined series is given by :
n1
log G1
+ n2
log G2
log G = -------------------------------
n1
+ n2

ExampleExample
 The GM of two series of sizes 10 and 12 areThe GM of two series of sizes 10 and 12 are
12.5 and 10 respectively. Find the combined12.5 and 10 respectively. Find the combined
GM of the 22 observations.GM of the 22 observations.

Combined G.M. of Two Sets ofCombined G.M. of Two Sets of
DataData
10 log 12.5 + 12log 10
log G = -------------------------------
10 + 12
22.9691
= ------------ = 1.04405
22
Therefore,
G = antilog 1.04405 = x
Thus the combined average rate of growth for the period of 22
years is x%.

Relationship Among A.M. G.M. andRelationship Among A.M. G.M. and
H.M.H.M.
The relationships among the magnitudes of the three
types of Means calculated from the same data are as
follows:
(i) H.M. ≤ G.M. ≤ A.M.
i.e. the arithmetic mean is greater than or equal
to the geometric which is greater than or equal to the
harmonic mean.
( ii ) G.M. =
i.e. geometric mean is the square root of the product of
arithmetic mean and harmonic mean.
( iii) H.M. = ( G.M.) 2
/ A .M.
... MHMA *

medianmedian
 It is aIt is a positional average.positional average.
 Middle most value of the distribution, which divides theMiddle most value of the distribution, which divides the
distribution into two equal parts.distribution into two equal parts.
 whenever there are some extreme values in the data,whenever there are some extreme values in the data,
calculation of A.M. is not desirable.calculation of A.M. is not desirable.
 Further, whenever, exact values of some observationsFurther, whenever, exact values of some observations
are not available, A.M. cannot be calculated.are not available, A.M. cannot be calculated.
 In both the situations, another measure of location calledIn both the situations, another measure of location called
Median is used.Median is used.

 The median divides the data into two parts such that theThe median divides the data into two parts such that the
number of observations less than the median are equal to thenumber of observations less than the median are equal to the
number of observations more than it.number of observations more than it.
 This property makes median very useful measure when theThis property makes median very useful measure when the
data is skewed like income distribution amongdata is skewed like income distribution among
persons/households, marks obtained in competitivepersons/households, marks obtained in competitive
examinations like that for admission to Engineering / Medicalexaminations like that for admission to Engineering / Medical
Colleges, etc.Colleges, etc.
 Before median calculation, the arrangement of the data inBefore median calculation, the arrangement of the data in
either ascending or descending order is a must.either ascending or descending order is a must.

Graphical Method of Finding theGraphical Method of Finding the
MedianMedian
 If we draw both the ogives viz. “Less Than “ and “If we draw both the ogives viz. “Less Than “ and “
More Than”, for a data, then the point of intersectionMore Than”, for a data, then the point of intersection
of the two ogives is the Median.of the two ogives is the Median.
0
5
10
15
20
25
Median
Less Than Ogive
More Than Ogive

Computation of MedianComputation of Median
 For a simple series :For a simple series :
step 1 : arrange obs. in ascending or descending orderstep 1 : arrange obs. in ascending or descending order
step 2 : let n = no. of obs.step 2 : let n = no. of obs.
(i) n is odd(i) n is odd
median =median = observationobservation
(ii) n is even(ii) n is even
median = mean ofmedian = mean of andand obs.obs.thn
)
2
( thn
)1
2
( +
thn
)
2
1
(
+

 For ungrouped frequency distributionFor ungrouped frequency distribution
Step 1 : calculate the cumulative frequenciesStep 1 : calculate the cumulative frequencies
Step 2 : N=Step 2 : N=
(i) N is odd(i) N is odd
median = size ofmedian = size of observationobservation
(ii) N is even(ii) N is even
median = mean of the sizes ofmedian = mean of the sizes of andand
obs.obs.
∑ if
thN
)
2
1
(
+
thN
)
2
(
thN
)1
2
( +

Median - Ungrouped DataMedian - Ungrouped Data
First the data is arranged in ascending/descending order.
In the earlier example relating to equity holdings data of 20 billionaires given in
Table 4.1, the data is arranged as per ascending order as follows
2717 2796 3098 3144 3527 3534
3862 4187 4310 4506 4745 4784 4923
5034 5071 5424 5561 6505 6707 6874
Here, the number of observations is 20, and therefore there is no middle
observation. However, the two middle most observations are 10th
and 11th
. The
values are 4506 and 4745. Therefore, the median is their average.
4506 + 4745 9251
Median = ----------------- = -----------
2 2
= 4625.5
Thus, the median equity holdings of the 20 billionaires is Rs.4625.5 Millions.

ExampleExample
 Obtain the median forObtain the median for
the following frequencythe following frequency
distribution:distribution:
Obs.Obs. ff
11 88
22 1010
33 1111
44 1616
55 2020
66 2525
77 1515
88 99
99 66

Median – Grouped frequencyMedian – Grouped frequency
distributiondistribution
The median for the grouped frequency distribution is also defined as
the class corresponding to the c.f. just greater than N/2, and is
calculated from the following formula:
( (N/2) –fc
)
Median = Lm
+ ----------------- × hm
fm
where,
•Lm
is the lower limit of 'the median class internal
•fm
is the frequency of the median class interval
•fc
is the cumulative frequency of the class preceding the median class
•hm
is the width of the median class-interval
•N is the number of total frequency

ExampleExample
Wages (Rs.) No. of
employees
Cumulative
frequency
2000-3000 3 3
3000-4000 5 8
4000-5000 20 28
5000-6000 10 38
6000-7000 5 43

Here, N=43, N/2 = 21.5
Cumulative frequency just >21.5 is 28
So, median class is 4000-5000
Further,
Lm
= 4000
fm
= 20
fc
= 8
wm
= 1000
Therefore,
(21.5 –8) x 1000
Median = 4000 + -------------------------
20
= 4000 + 675
= 4675
Thus,median wages is Rs. 4675

EXAMPLEEXAMPLE
 Calculate the medianCalculate the median
from the followingfrom the following
data pertaining to thedata pertaining to the
profits (in crore Rs.)profits (in crore Rs.)
of 125 companies:of 125 companies:
ProfitsProfits No. of companiesNo. of companies
Less than 10Less than 10 44

Median
(i) Simple to understand
(ii) Extreme values do not have
any impact
(iii) Can be calculated even if
values of all observations are not
known or data has open-end
class intervals
(iv) Used for measuring qualities
and factors which are not
quantifiable like to find the
average intelligence among a
group of people.
(v) Can be approximately
determined with the help of a
graph (ogives)
(i) In case of even number of obs.
Median cannot be
determined exactly.
(ii) Not amenable for
mathematical calculations
(iii) No based on all the
observations.

ModeMode
 Mode is the value which occurs most frequently in aMode is the value which occurs most frequently in a
set of obs. And around which the other items of the setset of obs. And around which the other items of the set
cluster densely.cluster densely.
 In other words, it is the value which occurs withIn other words, it is the value which occurs with
maximum frequency.maximum frequency.
distributiondistribution
Uni-modalUni-modal BimodalBimodal MultimodalMultimodal
(single mode)(single mode) (two modal value)(two modal value) (more than(more than
two mode)two mode)

 Mode is more favourable than Mean/MedianMode is more favourable than Mean/Median
whenever we need to determine the mostwhenever we need to determine the most
typical size or value.typical size or value.
 Eg: Most common size of shoes, mostEg: Most common size of shoes, most
commonly purchased vehiclecommonly purchased vehicle

Determination of ModeDetermination of Mode
 By InspectionBy Inspection
 By method of GroupingBy method of Grouping
 By Interpolation FormulaBy Interpolation Formula
By Inspection : When the FD is regular, we can determineBy Inspection : When the FD is regular, we can determine
mode just by inspection.mode just by inspection.
Ex:Ex:
– Since the FD in this example is fairly regular, therefore,Since the FD in this example is fairly regular, therefore,
– mode of this distribution is = 5.mode of this distribution is = 5.
X 1 2 3 4 5 6 7 8 9 10
F 4 6 8 12 17 11 7 6 5 3

 Given the following series, determine theGiven the following series, determine the
modal value:modal value:
 4141 4242 4545 4444 4545 4848 5050 4545 4747
5151 5656

 By method of GroupingBy method of Grouping
This method is used when the FD is not regular.This method is used when the FD is not regular.
We can understand this through the followingWe can understand this through the following
example,example,
 In this variate we notice following things. 1.In this variate we notice following things. 1.
Irregular FD (Sudden rise from 20 to 100)Irregular FD (Sudden rise from 20 to 100)
 Mode in this case, therefore, cannot beMode in this case, therefore, cannot be
obtained by Inspection.obtained by Inspection.
X 10 11 12 13 14 15 16 17 18 19
F 8 15 20 100 98 95 90 75 50 30

 To obtain mode in this example, we haveTo obtain mode in this example, we have
to write the following Grouping table,to write the following Grouping table,
X F(1) F(2) F(3) F(4) F(5) F(6)
10 8
23 4311 15
35
135
12 20
120
218
13 100
198
293
14 98
193
28315 95
185
260
16 90
165
215
17 75
125
155
18 50
8019 30

 The highest frequency total in each of the 6The highest frequency total in each of the 6
columns of the previous table is identified andcolumns of the previous table is identified and
analyzed in the following table,analyzed in the following table,
Since the values 14 and 15 occur max. no. ofSince the values 14 and 15 occur max. no. of
times, therefore the mode is ill defined.times, therefore the mode is ill defined.
Columns 10 11 12 13 14 15 16 17 18 19
1 1
2 1 1
3 1 1
4 1 1 1
5 1 1 1
6 1 1 1
Total 0 0 0 3 4 4 2 1 0 0

 The following table gives the measurements ofThe following table gives the measurements of
collar sizes of 230 students in a university.collar sizes of 230 students in a university.
Determine the modal size of the collarDetermine the modal size of the collar
Collar
sizes(c
m)
32 33 34 35 36 37 38 39 40 41
No. of
student
s
7 14 30 28 35 34 16 14 36 16

ModeMode
6161

 Mode by Interpolation
fm
- f0
Mode = Lm
+ ----------------- × i
2 fm
- f0
- f2
where ,
 Lm
is the lower point of the modal class interval
 fm
is the frequency of the modal class interval
 f0
is the frequency of the interval just before the modal interval
 f2
is the frequency of the interval just after the modal interval
 i is the width of the modal class interval

Equity Holding DataEquity Holding Data
6262
Class Interval Frequency Cumulative
frequency
2000-3000 2 2
3000-4000 5 7
4000-5000 6 13
5000-6000 4 17
6000-70000 3 20

the modal interval i.e., the class interval with the the modal interval i.e., the class interval with the
maximum frequency (6) is 4000 to 5000.  Further,maximum frequency (6) is 4000 to 5000.  Further,
                   LLmm   = = 40004000
                   h h =  =   10001000
                   ffmm   =  =   66
                   ff00   =  =   55
ff22 == 44
ThereforeTherefore
6363

                                        ( 6 – 5( 6 – 5  ))
Mode    =   Mode    =    4000 +  -------------------- 4000 +  -------------------- ×× 1000  1000
                                                                          2 2 ×× 6 – 5 – 4 6 – 5 – 4
       =   =    4000 +    (1/3)4000 +    (1/3)×× 1000  1000
         =   =    4000 + 333.34000 + 333.3
         =   =    4333.34333.3
Thus the modal equity holdings of the billionaires is Rs. Thus the modal equity holdings of the billionaires is Rs.
4333.3 Millions.4333.3 Millions.
6464

 Example : The FD of marks obtained byExample : The FD of marks obtained by
60 students of a class is given below :60 students of a class is given below :
Calculate mode :Calculate mode :
Marks 30-34 35-39 40-44 45-49 50-54 55-59 60-64
Frequency 3 5 12 18 14 6 2

EXAMPLEEXAMPLE
 Calculate the mean, Calculate the mean,
median and mode median and mode
from the following from the following
data pertaining to the data pertaining to the
profits (in crore Rs.) profits (in crore Rs.)
of 100 companies:of 100 companies:
ProfitsProfits No. of companiesNo. of companies
Less than 10 Less than 10 44

Empirical Relationship amongEmpirical Relationship among
Mean, Median and ModeMean, Median and Mode
In a moderately skewed distributions, it is found that the following In a moderately skewed distributions, it is found that the following
relationship, generally, holds good :relationship, generally, holds good :

Mean – Mode Mean – Mode =    3 (Mean – Median)=    3 (Mean – Median)

From the above relationship between, Mean, Median and Mode, if the From the above relationship between, Mean, Median and Mode, if the
values of two of these are given, the value of third measure can be values of two of these are given, the value of third measure can be
found out found out
For a symmetrical distribution,For a symmetrical distribution,
Mean = Median = ModeMean = Median = Mode

• Example : In a moderately symmetrical distribution –Example : In a moderately symmetrical distribution –
 The mode and median are 75 and 60 respectively.The mode and median are 75 and 60 respectively.
Find Mean.Find Mean.
• Solution : Using the empirical relation betweenSolution : Using the empirical relation between
Mean/Median/Mode, we can writeMean/Median/Mode, we can write
Mean – Mode = 3 (Mean – Median)Mean – Mode = 3 (Mean – Median)
0R0R
2 Mean = 3 Median – Mode2 Mean = 3 Median – Mode
Hence,Hence,
Mean = [3 Median – Mode] / 2 = [3(60) – 75] / 2 = 52.5Mean = [3 Median – Mode] / 2 = [3(60) – 75] / 2 = 52.5

SymmetricalSymmetrical
Mode Median Mean

7070
4333 4500 4565
(mode) (median) (mean)

Mode
7171
(i) Simple to understand
(ii) Extreme values do not have
any impact
(iii) Can be calculated when we
want to compare the consumer
prefernces for products.
(iv) It can be detected at just a
mere look at the graph.
(i) It is not based on all the
observations.
(ii) Incase of two,three or many
modal values data becomes
difficult to compare and
interpret.
(iii) It is not suitable for algebraic
treatment.

QuartilesQuartiles
 Median divides the data into two parts such that 50 % of the Median divides the data into two parts such that 50 % of the
observations are less than it and 50 % are more than it. observations are less than it and 50 % are more than it.
Similarly, there are “Quartiles”. There are three Quartiles Similarly, there are “Quartiles”. There are three Quartiles
viz. Qviz. Q11 , Q , Q22 and Q and Q33. These are referred to as first, second and . These are referred to as first, second and
third quartiles. third quartiles.
 The first quartile , QThe first quartile , Q11, divides the data into two parts such , divides the data into two parts such
that 25 % ( Quarter ) of the observations are less than it and that 25 % ( Quarter ) of the observations are less than it and
75 % more than it. 75 % more than it.
 The second quartile, QThe second quartile, Q22, is the same as median, is the same as median. The third . The third
quartile divides the data into two parts such that 75 % quartile divides the data into two parts such that 75 %
observations are less than it and 25 % are more than it. observations are less than it and 25 % are more than it.
 All these can be determined, graphically, with the help of All these can be determined, graphically, with the help of
the Ogive curvethe Ogive curve

QuartilesQuartiles
data Q1
and Q3
are defined as values corresponding to
an observation given below :
Ungrouped Data Grouped Data
(arranged in ascending
or descending order)
Lower Quartile Q1
{( n + 1 ) / 4 }th
( n / 4 )th
Median Q2
{ ( n + 1 ) / 2 }th
( n / 2 )th
Upper Quartile Q3
{3 ( n + 1 ) / 4 } th
(3 n / 4 )th

QuartilesQuartiles
 For the grouped frequency distribution, jFor the grouped frequency distribution, jth th
quartile, j=1,2,3 is  quartile, j=1,2,3 is
given bygiven by
Where, L = lower limit of quartile classWhere, L = lower limit of quartile class
  CCf f = c.f. of the class preceding the quartile class= c.f. of the class preceding the quartile class
  f   =  frequency of the quartile classf   =  frequency of the quartile class
  h  = width of the quartile classh  = width of the quartile class
3,2,1;*
)4/(
=
−
+= jh
f
CjN
LQ
f
j

PercentilesPercentiles
 Divides the total obs. Into 100 equal parts.Divides the total obs. Into 100 equal parts.
Where, L = lower limit of percentile classWhere, L = lower limit of percentile class
  CCf f = c.f. of the class preceding the percentile class= c.f. of the class preceding the percentile class
  f   =  frequency of the  percentile classf   =  frequency of the  percentile class
  h  = width of the  percentile classh  = width of the  percentile class
99,....,2,1;*
)100/(
=
−
+= jh
f
CjN
LP
f
j

DecilesDeciles
 Just like quartiles divide the data in four parts, the deciles divide Just like quartiles divide the data in four parts, the deciles divide
the data into ten parts – first deciles ( 10% ) , second ( 20% ) , and the data into ten parts – first deciles ( 10% ) , second ( 20% ) , and
so on. And just as second quartile and median are the same, so so on. And just as second quartile and median are the same, so thethe
fiftieth decilefiftieth decile i.e.i.e. PP5050 and the median are the sameand the median are the same. .
 Median = QMedian = Q22 = P = P50 50 = D= D55
9,....,2,1;*
)10/(
=
−
+= jh
f
CjN
LD
f
j

Calculating exactly:Q1Calculating exactly:Q1

Using the formula:
16
X f CF
0 < 20 15 15
20 < 40 60 75
40 <100 25 100
N/4 = 25th
item
This is in the group 20 < 40
Lower limit (l) is 20
Width of group (i) is 20
Frequency of group (f) is 60
CF of previous group (F) is 15
Formula is:





 −
+=
f
FN
ilQ q
4
11
First Quartile





 −
+=
60
1525
20201Q
60
10
2020 ×+= 333.320 +=
= 23.3333
This means that 25% of the data is below 23.333

Q3Q3

17
Third Quartile
This is in the group 20 < 40
Lower limit (l) is 20
Frequency of group (f) is 60
X f CF
0 < 20 15 15
20 < 40 60 75
40 <100 25 100
3N/4 = 75th
item
Formula is:





 −
+=
f
FN
ilQ q
43
33





 −
+=
60
1575
20203Q
60
60
2020 ×+= 2020 +=
= 40
So 25% of the data is above this point

ExampleExample
 Calculate Q1,D7 and Calculate Q1,D7 and
P90 from the following P90 from the following
grouped data, related to grouped data, related to
profits of 100 profits of 100
companies in Rs. Lakhcompanies in Rs. Lakh

Profit (lakh)Profit (lakh) No. of No. of
companiescompanies
20-3020-30 44
30-4030-40 88
40-5040-50 1818
50-6050-60 3030
60-7060-70 1515
70-8070-80 1010
80-9080-90 88
90-10090-100 77

Measures of central tendency

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