module 4
TRANSCRIPT
Unit - 4MEASURES OF CENTRAL TENDENCY
AVERAGESMass data are colleted, classified, tabulated systematically. The data so presented is analyzed further to
bring its size to a single representative figure. Descriptive statistics which describes the presented data in a single number. It is concerned with the analysis of frequency distribution or other from of presentation mathematically by which a few constants or representative numbers are arrive.
CONCEPT OF CENTRAL TENDENCY:One of the main characteristics of numerical data is central tendency. It is found that the observations
tend to cluster around a point. This point is called the central value of the data. The tendency of the observations to concentrate around a central point is known as central tendency.
The statistical measures which tell us the position of central value or central point to describe the central tendency of the entire mass of data is known as measure of central tendency or Average of first order
Meaning of Statistical Averages:-An average is a figure which represents the large number of observations in a concise or single numerical
data. If is a typical size which describes the central tendency.According to CLARK AND SCHAKADE “Average is an attempt to find one single figure to describe
whole group of figures”According to COXTON & COWDEN “An average is a single value within the range of the data that is
used to represent all of the values in the series”A single simple expression in which the net result of huge mass of unwieldy numerical data or a
frequency distribution is concentrated and which is used to represent the whole data is called a statistical average.
Objects of Statistical average:An average is of great significance in all the fields of human knowledge, because it depicts the
characteristics of the whole group of data understudy.Following are the objectives of objectives of computing the statistical averages
1. To give or present the complex data in a simple manner and concise form.2. To facilitate the data for comparative study of two different series.3. To study the mass data from the sample4. To establish relationship between the two series 5. To provide basis for decision making6. To calculate the representative single value from the given data.
Requisites of a good and ideal Average.Any statistical average to be good and ideal average must possess some of the characteristics as it is a
single value representing a single value representing a group of values. Following are the requisite properties of a good and ideal average.
1. It should be easily understood .2. It should be simple in calculation.3. It should be based on all the observations.4. It should not be unduly affected by the extreme values.5. It should be rigidly defined.6. It should be capable of further algebraic treatment.7. It should have sampling stability.Thus a statistical average should have all the above requisites to be an ideal and good average.
Limitations of Averages:Although an average is useful in studying the complex data and is very widely used in almost all the
spheres of human activity, it is not without limitation that restrict scope and applicability. Following are the limitations of statistical averages.
49
1. The extreme values, if any, will affect the averageble figure disproportionately.2. The composition of the data cannot be viewed with the help of the average.3. The average does not represent always the characteristics of individual items.4. The average gives only a representative figure of the mass, but fails to depict the entire picture of the
data.5. An average may give us a value that does not exist in the data.6. some times an average might give very absurd results.In spite of the limitations, the statistical averages still are useful measures which play an important role in
analyzing the mass data.
TYPES OF STATISTICAL AVERAGES:Broadly speaking, there are five types of statistical averages which are commonly used in practice. They are.
1. Arithmetic Mean.2. Geometric Mean.3. Harmonic Mean.4. Median.5. Mode.
Arithmetic Mean Arithmetic Mean is the most widely used measurement which represents the entire data. Generally it is
termed as an average to a layman. It is the quantity obtained by dividing the sum of the values of the items in a variable by their number.
Arithmetic Mean may be two types1. Simple Arithmetic Mean.2. Weighted Arithmetic [Not included in your syllabus]
Simple Arithmetic MeanIt is the quotient of the sum of the values divide by their number.
Symbolically = x1 +x2 +x3 -----------+ xn = x n n
Where x = The sum of observations ‘n’ = number of observations = is symbol of arithmetic mean
INDIVIDUAL SERIES:Calculation of arithmetic mean in individual series.
DIRECT METHODThe process of computing the arithmetic mean in this case involves the following steps:-
a. Add all the values of the variable x and obtain xb. Divide this total of observations x by the number of observations ‘n’
Illustration = 01The monthly income of 5 persons is as given below Rs.1132, 1140, 1144, 1136, 1148 find out arithmetic
mean.Solution
Calculation of Arithmetic Mean
Serial NumberMonthly of income
In Rs. = x n = 5700 5 = 1140
12345
11321136114011441148
n=5 x 5700
50
Illustration = 02 Calculate mean from the following data:-
Roll No. 1 2 3 4 5 6 7 8 9 10Marks 40 56 55 78 58 60 73 35 43 48
SolutionCalculation of Mean
Roll Number Marks
= x n= 540 10 = 54
12345678910
35404348505558607378
N = 10 540
SHORT CUT METHODThe arithmetic mean can also be calculated by shortcut method. This method reduced the amount of
calculation. It involves the following steps.1. Assume any one value as an assumed mean which is also arbitrary mean [‘A’ = Assumed mean]2. Take the deviations of each item from the assumed mean (A) and denote this deviations by dx. ie
dx =x – A.3. Obtain the sum of these deviations i.e. dx4. Use the formula = A+ dx
N
Illustration = 03The marks obtained by ten students in an examination are as given below. Calculate mean by shortcut
method.Roll No. 1 2 3 4 5 6 7 8 9 10Marks 43 48 65 57 31 60 37 48 78 59
Solution A= 48, dx = x - A
Roll No Marksx – 48
dx12345678910
31374348485759606578
-17-11-5000911121730
n = 10dx =46 -33
+79
51
= A + dx n= 48 +46
10 = 48+4.6 = 52.6
Average marks of 10 students= 52.6
Illustration = 04Calculate mean wage of workers working in a factory.
Sl.No. 1 2 3 4 5 6 7 8 9 10Wages in Rs. 400 500 550 780 580 600 730 650 430 480Solution
Sl. NoWagesIn Rs.
x – 550dx
= A + dxn
= 550 + 200/10= 550 + 20= 570Average wages = Rs.570
-390 + 590
12345678910
400430480500550580600650730780
- 150- 120- 70- 50
03050100180230
n = 10200dx
DISCRETE SERIESCalculation of mean in discrete Series
Direct MethodThe process of computing arithmetic by direct method in case of discrete series involves the following
steps1. Represent the variable by x and the frequency by ‘f’2. Multiply each frequency with the corresponding value of the variable i.e. find fx3. obtain the total of these products i.e. obtain fx4. Find the total frequency i.e. N = f5. Divide the total obtained by the total frequency. i.e. = fx N
Illustration = 5Calculate mean from the following data
Variable: 50 52 55 56 60 63 64 65 67 70Frequency 5 6 10 15 20 12 8 11 9 10
Variablex
Frequencyf
fx
=fx N = 6438
106 = 60.72
50525556606364656770
5610152012811910
2503125508401200756512715603700
=fx
52
A = 550 dx = x - A
SHORT – CUT METHOD:-The process of computing arithmetic mean by short – cut method in case of discrete series involves the
following steps.1. Take an assumed mean (A) out of variable x2. Take the deviations of the values of x from the assumed mean (A) and denote the deviations by dx – x
– A3. Multiply dx with the respective frequency and obtain the total of these products. That is obtain
=fdx 4. Use the formula =A + fdx N
ILLUSTRATION = 06From the following frequency distribution, find out mean height of the students.
Height in inches x: 64 65 66 67 68 69 70 71 72 73No. of Students f 1 6 10 22 21 17 14 5 3 1
Solution A= 68, dx = x – AHeight
XF
x – 68dx
fdx
64656667686970717273
161022211714531
- 4-3-2-1012345
-4-18-20-220172815125
= A +fdx N
= 68 + 13/100= 68 + 0.13 = 68.13
Average height of student = 68.13inches.
N = 100 fdx 13 -64 +17
STEP – DEVIATION METHOD
Steps involved:-1. Take an assumed mean (A) out of variable x2. Take deviations of the values x from the assumed mean3. Divide each dx by c, where c is the common factorial each x i.e. dx = x –A
c4. Multiply this dx with the respective frequency and take and total fdx5. Use the formula. = A = fd xc N
ILLUSTRATION = 07Calculate the arithmetic mean for the following data.
Wages in Rs: 150 200 250 300 350 400 450 500 550 600Workers: 10 20 30 20 40 10 20 10 8 2
53
SolutionA= 350 C=50 dx = x – A
CWages in Rs. x
Workersf
x – 35050dx
Fdx
= A + fdx xC N
= 350 + -58 x50 170
= 350 – 2900/170= 350 – 17.05
= 332.95Meanage
150200250300350400450500550600
102030204010201082
-4-3-2-1012345
-40-60-60-2001040303210
170 fdx -58 -180 + 122
CONTINUOUS SERIESCalculation of arithmetic mean in case of continuous series
Direct MethodSteps involved:-1. Obtain the mid – points or mid x of each class2. Multiply these mid – points by the respective class frequency and obtain the total fm3. Use the formula = fm N
ILLUSTRATION = 08Calculate arithmetic mean for the following frequency distribution.
Marks obtained
0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 –80 80 - 90
No. of Student
5 4 6 20 12 10 10 8 5
SolutionMarks
0.xf mid x
mFm
= fm N = fm N
= 3700/80= 46.25
Mean Marks = 46.25
0 – 1010 -2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 90
5462012101085
51525354555657585
2560150700540550650600425
N = 80 3700 fm
SHORT – CUT METHOD CUM STEP – DEVIATIONThe process of computing arithmetic mean by short – cut method in case of continuous series involves the
following steps.1. Obtain the mid – point of each class2. Take the assumed mean out of midx3. Deduct assumed mean from the mid – point of each class & find out the deviations dx = x –A4. Multiply the respective frequencies of each class by deviations and obtain the total fdm5. Use the formula = A + fdm x C
N
54
ILLUSTRATION = 09Find arithmetic mean from the following frequency distribution.
Profit per Shop 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 –70 70 -80No. of shops ‘f’ 12 20 18 40 25 10 12 13
SolutionA = 45, C = 10, dxn = x – A
C
X fx– 45 /10
mid xDxn fdm = A + fdm x C
N= 45 + -111 x 10
150 = 45 – 1110/150 = 45 – 7.4
= 37.6
0 – 1010 - 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
1220184025101213
515253545556575
-4-3-2-10123
-48-60-36-400102439
N = 150 fdm -111 -184 + 73‘C’ = Common factor or class interval.
ILLUSTRATION = 10The following frequency distribution represents the age of the persons interviewed, calculate average age.
Age in years x 80 – 89 70 – 79 60 – 69 50 – 59 40 – 49 30 – 39 20 – 29 10 - 19No. of persons f 2 2 6 20 56 40 42 32
SolutionNote : Given values are arranges in ascending order
A= 44.5, C= 10, dm = x – A C
x = 44.5/10X f midx Dm fdm
= A + fdm x C N
= 44.5 + 1740/200 = 44.5 – 8.7 = 35.8
10 - 1920 – 2930 – 3940 – 4950 – 5960 – 6970 – 7980 – 89
3242405620622
14.524.534.544.554.564.574.584.5
-3-2-101234
-96-84-400201268
N = 200 fdm -174 -220 +46
ILLUSTRATION = 11 Calculate arithmetic mean from the following data
Variable (x) Frequency (f)Less than 05Less than 10Less than 15Less than 20Less than 25Less than 30Less than 35
72045759599100
Solution First of all we have to convert the given cumulative frequency table into an ordinary frequency Table as
under.
55
A = 175, C = 5, dm = x – A/Cx – 17.5/5
X f mid x Dm fdm = A + fdm x C N
= 17.5 + -41 x5 100
= 17.5 – 205/100 = 17.5 – 2.05 = 15.45
0 –55 –10
10 - 1515 – 2020 – 2525 – 3030 – 35
71325302041
2.57.512.517.522.527.532.5
-3-2-10123
-21-26-2502083
N = 100 fdx -41 -72 +31
ILLUSTRATION : 12From the following data calculate the average marks of 65 students.
Marks: Less than 10 20 30 40 50 60 70 80No. of Students 4 10 15 25 30 35 45 65Solution
Note: we will first convert the dat in simple series from the given cumulative frequencies.A= 35, C = 10, dm= x – A
CMarks
XStudents
fMidx
(x – 35)/10dx
fdm = A + fdm x C N= 35 + 96 x 10 65= 35 +(960/65)= 35 + 14.77= 49.77 Average Marks
0 – 1010 - 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
410-4=615 –10 =525 –15=1030 –25=535 –30=545 –35=1065 –45=20
515253545556575
-3-2-101234
-12-12-505103080
65 fdm 96 -29 + 125
ILLUSTRATION : 13From the following information pertaining to 150 workers. Calculate average wage paid to workers.
Wages in Rs. more than 75 85 95 105 115 125 135 145No of Workers 150 140 115 95 70 60 40 25
Solution The lower limit of the first class in 75, the class interval would be 75 – 85, 85 – 95 and soon
A = 110, C =10, dm = x - A C
WagesIn Rs.
X
No. of Workersf
Mid x(x – 110)/10
dxnfdm
= A + fdm xCN
= 110 + 95 x 10 150
= 110 + (950/150)= 110 + 6.33= 116.33 Average wage
75 – 8585 –95
95 – 105105 –115115 –125125 – 135135 – 145145 – 155
150 –140 = 10140 – 115 =20115 –95 = 2095 – 70 = 2570 – 60 = 1060 – 40 = 2040 – 25 = 1525 – 0 = 25
8090100110120130140150
- 3 - 2 - 101234
-30-50-200104045100
N = 150 fdm 95 -100 + 195
ILLUSTRATION: 14Calculate average marks of 70 students
56
Marks more than 20 30 35 40 50 60 65 70 80No. of Students 70 63 55 40 30 18 10 7 0
Solution:-Convert the more than frequency table into ordinary frequency distribution & then proceed as usual.
A = 45, C = 5, dm = x – A C
MarksX
Frequencyf
Mid x(x – 45)/5
dmfdm
= A + fdm xC N
= 45 + 37 x 5 70= 45 + (185/70)= 45 +2.643= 47.643 Marks
20 – 3030 – 3535 – 4040 – 5050 – 6060 – 6565 – 7070 – 80
70 – 63 = 0763 – 55 = 0855 – 40 = 1540 – 30 = 1030 – 18 = 1218 – 10 = 810 – 7 = 37 – 0 = 7
2532.537.54555
62.567.575
- 4-2.5-1.5
02
3.54.56.0
-28-20
-22.2502428
13.542.0
N =70 fdm 37 - 70.5 + 107.5
Open – End ClassIn a grouped frequency distribution, if the lower limit of the first class and the upper limit of last class are
not known, it is difficult to find the Arithmetic Mean under such circumstances, we have to assume that the width of the open classes are equal to the common width of closed classes.
It will be clear from the following example.
ILLUSTRATION = 15Calculate Arithmetic Mean from the following data
Wages in Rs.
Below 100 100 – 109 110 – 119 120 – 129 130 – 139 140 – 149 150 & above
No of Workers
10 15 25 40 28 12 20
Solution:A= 124.5, C= 10, dm = x – A
CX f mid x dx fdm = A + fdm x C
N= 124.5 + 27 x 10 15= 124.5 + (270/150)= 124.5+ 1.8= 126.3 Average wage
90 – 99100 – 109110 – 119120 – 129130 – 139140 – 149150 – 159
10152540281220
94.5104.5114.5124.5134.5144.5154.5
-3-2-10123
-30-30-250282460
N = 150 fdm 27 - 85 + 112
MERITS OF ARITHMETIC MEAN DEMERITS OF ARITHMETIC MEAN1. It is easy to understand and easy to calculate.2. It is based on all the observations.3. It is capable of further algebraic treatment.4. It is rigidly defined and determinate .5. It is least affected by fluctuations of sampling.
AM is as stable as possible.
1. The mean is unduly affected by the extreme items.
2. It is un realistic. 3. It may lead to a false conclusion.4. It cannot be calculated in case of open – end
classes.5. It may not be represented in the actual data.
GEOMETRIC MEAN
57
The Geometric mean is obtained by finding out the products of the different items of a series and getting the root of the product corresponding to the number of items.
The Geometric mean of a set of N observations is the nth root of their product
It may be defined as the nth root of the product of N values.
This can be put in the following formulaG.M = The product of N valuesG.M = n 1 x 2 x 3 …………xn
When the number of observations exceeds two the computations are simplified through the use of logarithms then.
GM = Antilog of logx n
USES OF GEOMETRIC MEAN1. This G.M. is highly useful in averaging ratios percentages and rate of increase between two periods2. G.M. is important in the construction of index Numbers.3. In economic and social sciences, where we want to give more weight to smaller items and smaller
weight to large items, G.M is appropriate.
Calculation of Geometric Mean – individual series Steps1. Find out the logarithms of each value2. Add all the values of logx i.e. logx3. The sum of log (logx) is divided by the number of items - logx
n4. Find out the antilog of the quotient this is the GM of the data.
ILLUSTRATION : 01Calculate the Geometric mean of the following
Values : 50 72 54 82 93
Solution:Values
XLogx G.M = Antilog of logx
n = -------“------ 9.1710/5= ------“ ------ 1.8342= 68.26
5072548293
1.69901.85731.73241.91381.9685
N = 5 9.1710
ILLUSTRATION – 2
Calculate geometric mean from the following data.Values x: 6.5 169.0 11.0 112.5 14.2 75.5 35.5 215.0
SolutionValues
XLog
xG.M = Antilog of logx n
58
= ---------“------- 13.0461 8
= --------“--------1.6308= 42.74
6.5169.511.0112.514.275.535.5215.0
0.81292.22791.04142.05111.15231.87791.55022.3325
N = 8 13.0461
ILLUSTRATION = 03 : Calculate Geometric Mean of the followingValues: 15 250 15.7 1.57 105.7 10.5 1.06 25.7 0.257 10
SolutionValues:x logx G.M = Antilog of logx
n = ----------“-------- 9.8561 10= -----------“----------0.98561G.M = 9.674
n = 10
1525015.71.57105.710.51.0625.70.25710.0
1.17612.39791.19590.19592.02401.02120.02531.40991.40991.0000
logx 9.856110 –1 = 9
ILLUSTRATION :04Calculate Geometric Mean of the following data
Values x 3834 382 63 8 0.4 0.03 0.009 0.0005
Solution:Values x log x G.M = Antilog of logx
n= --------“------- 1.6062
8= --------“ ------- 0.2008GM = 1.588
3834382638
0.40.030.0090.0005
3.58932.58211.79930.90311.60212.47713.95424.6990
N = 8 1.6062 5 + 6 = 11 –10 = 1
ILLUSTRATION :05Find the geometric mean from the following data
Values 0.8974 0.0570 0.0081 0.5677 0.0002 0.0984
SolutionValues Log G.M = Antilog of logx
59
X x N -12 + 3= -----------“------ -9.6655
6= ------------“--- 2.6109GM = 0.04083
0.89740.05700.00810.56170.00020.0984
1.95302.75593.90851.75414.30102.9930
N = 5 9.6655 4 – 13 = - 9ILLUSTRATION :06
Calculate the G.M of the followingValues 375 0.05 0.5672 0.0854 0.005
SolutionValues x Logx GM = Antilog of logx
n (-5 + 2) = ------“---- 3.6573 5= -------“------ -1.5314= GM = 0.3399
3750.05
0.56720.08540.005
2.57402.69901.75382.93153.6990
N = 5 3.6573 5 – 8 = -3ILLUSTRATION = 07
Find the average rate of increase in population which in the first decade has increased by 20%, in the next by 30% and in the third by 45%Solution
Geometric mean is more appropriate to find out an average increased in population over three decades.
DecadeRate of increase
Population at the end each decade
[Base =100]Log x
GM = A. Log of logx n
= A.L of 6.3545 3 = A.L of 2.1181= 131.3
FirstSecondThird
20%30%45%
120130145
2.07922.11392.1614
n = 3 6.3545The average percentage increase in population over three decades is 31.3% (131.3 – 100)
Merits of Geometric Mean:1. It is rigidly defined.2. It is based on all the observations.3. It is suitable for further mathematical treatment.4. It is not much affected by fluctuations of sampling.5. It gives comparatively more weight to small values.
Demerits1. It is comparatively difficult to calculate and not simple to understand.2. If any observation is zero, then geometric mean becomes zero.3. It is not defined for negative values.4. It cannot be obtained by inspection.5. It may not be represented in the actual data.
HARMONIC MEANHarmonic mean like geometric mean is a measure of central tendency in solving special types of problem
involving variable expressed in time rates. Harmonic Mean is the reciprocal of the arithmetic mean of the reciprocal of the given observations.
The reciprocal of a number is that value which is obtained dividing one by the value.It is defined as the reciprocal of the average of the individual items.
Symbolically HM = . n . 1/x +1/x2 +1/x3 …………+ 1/xn
USES:
60
It is useful for computing the average rate of increase of profit of a concern or average speed at which a journey has been performed. The rate usually indicates the relation between two different types of measuring units that can be expressed reciprocally.
INDIVIDUAL SERIES:Calculation of Harmonic Mean is individual series.
Steps1. Find out the reciprocal of each size i.e. 1/x2. Add all the reciprocals of all values 1/x3. Apply the formula
HM = n n = no of observations 1/x 1/x = Total reciprocal value of given variables.
ILLUSTRATION – 01Calculate the Harmonic Mean from the following data relating to incomes of 10 families
Family 1 2 3 4 5 6 7 8 9 10Income in Rs. 85 70 10 75 500 8 42 250 40 36
SolutionIncome in Rs.
xReciprocals
1/x
H.M = n . 1/x = 10/0.34631 = 28.87
857010755008422504036
0.011760.014260.100000.013330.002000.125000.023180.004000.025000.02778
1/x = 0.34631
ILLUSTRATION = 02Calculate Harmonic Mean from the following data.Values 2574 475 75 5 0.8 0.08 0.005 0.0009Solution
ValuesX
Reciprocals1/x
H.M = n . 1/x = . 8 . 1324.96581 = 0.0060378
2574475755
0.80.080.0050.0009
0.000380.002100.013330.200001.2500012.50000200.000001111.00000
n = 8 1/x = 1324.96581
ILLUSTRATION = 03Calculate the Harmonic Mean for the following data
Variable x 35 250 18.7 234.6 1.06 98.72 0.987
SolutionVariable
XReciprocal
1/x
61
H.M = n . 1/x = . 7 . 2.05701 = 3.402998
3525018.7234.61.0698.720.987
0.028570.004000.053480.004260.943400.010131.01317
n = 7 1/x 2.05701
Merits of Harmonic Mean Demerits of Harmonic Mean1. It is rigidly defined.2. It is based on all the observation of the series3. It is suitable in case of series having wide
dispersion.4. It is suitable for further mathematical
treatment.5. It gives less weight to large items and more
weight to small items.
1. It is difficult to calculate and is not understandable.
2. All the values must be available for computation.
3. It is usually a value which does not exist in series.
4. It cannot be used when any one of the item is 0 or negative, because when the item 0 is used as a divider the items will be zero.
MEDIANMedian may be defined as the value of that item which divides the series into two equal parts, one – half
containing values greater than it and the other half containing values less that it. Therefore the series has to be arranged in ascending or descending order, before finding the median. The Median is a positional average and the term position refers to the place of a value in a series.
The definitions of Median given by different authors are as follows.
“Median of a series is the value of the item actual or estimated when a series is arranged in order of magnitude which divides the distribution into two parts” – SECRIST
“ The Median is that value of the variable which divides the group into two equal parts, one part containing all values greater and the other all values less than median. – By L.R. Conner.
The Median, as its name indicates, is the value of the middle item in a series, when items are arranges according to magnitude.
Calculation of Median:INDIVIDUAL SERIES
Steps1. Arrange the observations in ascending or descending order of magnitude2. Locate the middle value3. use the formula Median = the size of n + 1 th item
2ILLUSTRATION = 01
Calculate Median from the following data Marks Obtained 53 48 69 78 82 93 45 68 64 75
SolutionArrange the values in ascending order
Marks Obtained
Median = the size of n + 1 th item 2
62
X
50% below
68.5 Median
50% above
= -----“------ 10 + 1 = 5.5th item 2
= 5th item is 68+ difference of 68 & 69= 68 + 0.5 of 69 – 68= 68 + 0.50 of 1= 68.5
45485364686975788293
N = 10
ILLUSTRATION =02Calculate the Value of Median from the following data relating to wages in Rs.
Wages in Rs. 125 150 120 160 175 148 182 134 145
SolutionArrange the values in ascending order
WagesIn Rs. x
Median
Median = the size of n + 1 th item 2
= -------“------- 9 + 1 th item 2
= -------“------- 10/2 = 5th itemM = 148
120125134145148150160175182
N = 9
ILLISTRATION = 03In a batch of 15 students, 5 students failed in a test the marks of 10 students, who got passing marks were
as follows.9 6 7 8 8 9 6 5 5 7
What was the median Marks of all the 15 students
Solution:In case of only 10 student marks were given but failed student marks were given. We have to assume that
failed candidates must have secured less than 5 marks i.e. minimum marks for pass in the test.Marks Obtained
63
By Students x Median = the size of n + 1 th item 2
= ------“----- (15+1)/2 = 16/2 = 8th item= 8th item is 6
Median Marks of all the 15 students is 6
023335566778899
Median = 6
n = 15
DISCRETE SERIESCalculation of Median in case of discrete Series involves the following steps.
1. Obtain the total frequency f = N2. Obtain cumulative frequency 3. Locate median by using the formula
Median = The size of N + 1 th item’s value 2
ILLUSTRATION = 04Calculate median for the following data.
Wages in Rs. 255 257 258 260 265 268 270 275 280 281No. of Workers 10 15 12 24 30 26 15 12 10 6
SolutionCumulative Frequency
WagesIn Rs x
No. of Workers
FCf
80.5
Median = The size of n + 1 th items Value 2 = ------“------ 160 + 1 = 161
2 2 = 80.5Median = 265
255257258260265268270275280281
1015122430261512106
1025376191117132144154160
N = 160
ILLUSTRATION = 05Calculate the value of Median from the following data,
Marks Obtained 45 50 55 60 65 70 75 80 85 90No. of students f 15 28 15 12 25 35 25 15 20 10
SolutionMarks
xf cf
M = The size of n + 1 th items Value 2 = The size of 200 + 145 15 15
64
505560657075808590
281512253525152010
43587095130155170190200
100.5
2 = 200/2 = 100.5M = 70
Median = 70
N= 200
CONTINUOUS SERIESSteps
1. Make the classes in exclusive form, if they are in intensive form.2. Find total frequency N = f3. Find out median number m = N/24. Obtain cumulative frequency5. Find out with the help of m median class6. Apply the formula to locate the median
Median = l1+ l2 – l1 (m – c) f
Where m = N/2Where l1 & l2 represents lower and upper class limits of median classf = Frequency of the median class c = Cumulative frequency of the class preceding the median class
ILLUSTRATION – 06Calculate Median for the following data.
Income in Rs 250 – 300 300 – 350 350 – 400 400 – 450 450 – 500 500 – 550No. of persons 18 12 24 36 22 8
Solution:-Income in
Rs. xf cf
Median = l1+ l2 – l1 (m –c) f
= Where m = N/2 = 120/2 = 60= 400 + 450 –400(60 –54)
36= 400 + 50 x 6 36= 400 + 300 36= 400 + 8.33M = 408.33
250 –300300 – 350350 – 400400 – 450450 – 500500 – 550
18122436228
18305490112120
N = 120
ILLUSTRATION = 07
Calculate the value of Median from the following data Monthly Income
501-600 601- 700 701- 800 801 –900 901-1000 1001-1100 1101-1200 1201-1300
Number of families
5 15 80 120 200 100 60 20
Solution: Note: Since the variable are in the inclusive form, it has to be convert into exclusive formL1 = ½ of 1 ‘1’ is the difference between the upper limit of preceding class & the lower limit of nextL2 + ½ of 1 class
65
X f cf
300
Median = l1+ l2 – l1 (m –c) f
Where m = N/2 = 600/2 = 300Median class = 900.5 – 1000.5= 900.5 + 1000.5 + 900.5 (300 – 200)
200= 900.5 + 100 x 100 200= 950.5
500.5 – 600.5600.5 – 700.5700.5 – 800.5800.5 – 900.5900.5 – 1000.51000.5 – 1100.51100.5 – 1200.51200.5 – 1300.5
515801202001006020
520100220420520580600
N = 600
ILLUSTRATION = 08From the following data the median marks
Marks 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99Frequency 7 15 18 25 30 20 16 7 2
SolutionNote: Convert the inclusive series into exclusive form.
MarksX
f cf
70
Median = l1+ l2 – l1 (m –c) f
Where m=N/2 = 140/2 =70Median = 49.5 – 59.5= 49.5 + 59.5 + 49.5 (70 – 65) 30= 49.5 + 10 x 5
30= 49.5 + 1.67Median = 51.17 Marks
9.5 – 19.519.5 – 29.529.5 – 39.539.5 – 49.549.5 – 59.559.5 – 69.569.5 – 79.579.5 – 89.589.5 – 99.5
715182530201672
722406595115131138140
N= 140
ILLUSTRATION = 9Compute Median from the following data.
Mid Value 115 125 135 145 155 165 175 185 195Frequency 6 25 48 72 116 60 38 22 3
Solution:Note: The given problem is a continuous frequency distribution Mid – values of the class – intervals are
given.The Difference between two mid – values is 10 Divide this 10 by 2, to get half class interval i.e. 10/2 = 5This 5 is to be deducted from each mid value to get lower limit and 5 is added to get the upper limit of
class.
Valuesx
F cf Median = l1+ l2 – l1 (m –c) f
Where m = N/2 = 390/2 = 195Median Class = 150 – 160= 150 + 160 – 150 (195 –151)
110 – 120120 – 130130 – 140
62548
63179
66
140 – 150150 – 160160 – 170170 – 180180 – 190190 – 200
721166038223
151267327365387390
185 116= 150 + 10 x 44
116= 150 +(440/116)= 150 + 3.79= 153.79
N = 390
ILLUSTRATION = 10Marks Obtained less than 10 20 30 40 50 60 70 80No. of Students 4 16 40 76 96 112 120 125
SolutionLet us convert the data from less than frequency distribution into normal distribution.
Marksx
F cf
62.5
Median = l1+ l2 – l1 (m –c) f
Where m = N/2 = 125/2 = 62.5= 30 + 40 – 30 x (62.5 – 40)
36= 30 + (10/36) x 22.5= 30 + (225 / 36)= 30 + 6.25M = 36.25
0 – 1010 - 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
4122436201685
416407696112120125
N = 125
ILLUSTRATION = 11The marks obtained by 65 students in statistics are shown in the table given, calculate Median marks.
Marks No. of StudentsMore than 20%More than 30%More than 40%More than 50%More than 60%More than 70%
65634040187
SolutionMarksX
StudentsF
cf
32.5
Median = l1+ l2 – l1 (m –c) f
Where m = N/2 = 65/2 = 32.5Median class = 50 – 60= 50 + 60 – 50 (32.5 – 25)
22= 50 + (10/22)x 7.5= 50 + 3.4= 53.4 marks
20 – 3030 – 4040 – 5050 – 6060 – 70 70 – 80
223022117
22525475865
N = 65
Merits of Median 1. It is easy to understand easy to compute.2. It is quite rigidly defined.3. It is eliminates the effect of extreme items.4. It is amenable to further algebraic process.
67
5. Median can be calculated even from qualitative phenomena6. Its value generally lies in the distribution.
Demerits 1. Typical representative of the observations cannot be computed if the distribution of items is irregular 2. It ignores the extreme items.3. It is only a locative average, but not computed average.4. It is more effected by fluctuations of sampling than in Mean.5. Median is not amenable to further algebraic manipulation.
QUARTILESThe quartiles are also positional averages like the median. As the median value divides the entire
distribution into two equal parts, the quartile divide the entire distribution into four equal parts.A measure while divides an array into four equal parts is known as quartile. Basically there are three such
points – Q1, Q2 & Q3 – termed as three quartiles. The first quartile (Q1) or lower quartile has 25% of the items of the distribution below it and 75% of the items are greater than it.
Incidentally Q2 the second quartile, coincide with the median, has 50% of the observations above it and 50% of the observations below it.
The upper quartile or third quartile (Q3) has 75% of the items of the distribution below it and 25% of the items are above it.
These quartiles are very helpful in understanding the formation of a series. They tell us how various items are spread round the median. Their special utility Rs. in a study of the dispersion of items.
The working principle for computing the quartile is basically the same as that of computing the median.
Calculation of QuartilesQ1 = The size of (n + 1) th item
4Q3 = The size of 3(n +1) th item
4INDIVIDUAL SERIES:-ILLUSTRATION – 01
From the following data given below, calculate the first quartile and third quartilesValues: 6 10 8 14 12 16 22 20 18 Solution
Value rearranged in ascending orderValues
XQ1 = The size of n + 1 th item
4 = The size of (9 + 1) = 10/4 = 2.5th item 4 = 2nd item + 0.5 of difference between 8 & 10 = 8 + 0.5 of 2(10 – 8)
Q1 = 9Q3 = The size of 3(n + 1) th item
4= -------“------- 3(9+1) = (3x10)/4 = 30/4 = 7.5th item
4= 7 item + 0.5 of difference between 18 & 20 = 18 + 0.5 of 2 (20 - 18)Q3 = 19
6810121416182022
n = 9
ILLUSTRATION : 02Find the quartiles from the following data
Values 19 27 24 39 57 44 56 50 59 67 62Solution
Values arranged in ascending order
68
Valuesx
Q1 = The size of n + 1 th item
4 = The size of (11 + 1)/4 = 12/4 = 3rd item Q1 = 27
Q2 or Median = the size of n + 1 th item 2 = the size of (11 + 1)/2 = 12/2 = 6th itemQ2 or M = 50
Q3 = The size of 3(n+1) th item n = The size of 3 (11 + 1) 3 x 12 = 9th item 4 4Q3 = 59
1924273944505657596267
Q1
Q2 or M
Q3
n = 11
ILLUSTRATION = 03A manager while in specting his departments found the number absentees in 10 Departments as follows
calculate Q1 & Q3
21 12 15 17 18 18 20 19 0 8Solution
Given values re – arranged in ascending orderValues
xQ1= The size of n + 1 th item
4 = The size of (10 + 1)/4 = 11/4 = 2.75th item = 6 + 0.75 (12 –6) = 6 + 0.75 of 6 + 4.5 = 10.5
Q3 = The size of 3(n +1)th item 4
= ----“--- 3(10 + 1) = 3 x 11 = 8.25th item 4 4 = 8th item + ¼ (9th item – 8th item) = 19 + ¼ of 1 (20 – 19)Q3 = 19.25
061215171818192021
N = 10
DISCRETE SERIESCalculation of Quartiles from discrete series.
Steps1. Find the less than type cumulative frequencies2. Calculate quartiles using the formula
Q1 = The size of N+1 th item’s value 4Q2 = The size of N + 1 th items value
2Q3 = The size of 3(N + 1) th items value
4ILLUSTRATION = 04
Locate median and quartile from the following dataSize of Shoe 4.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 8.0No. of pairs 20 36 44 50 80 30 20 16 14
SolutionVariable
Xf Cf
77.75
Median = The size of N + 1 th item’s value 2
= The size of 310 + 1 = 311 = 155.5 2 2
4.04.5
2036
2056
69
5.05.56.06.57.07.58.0
44508030201614
100150230260280296310
155.5233.25
M = 6.0Q1 = The size of N+1 th item’s values 4 = The size of (310 + 1)/4 = 77.75Q1
= 5.0Q3 = The size of 3(N+ 1) th item’s values
4 = The size of 3(310 +1) = 233.5
4Q3 = 6.5
N 310
ILLUSTRATION = 05Calculate the value of the median and quartiles from the following data
Marks 25 30 35 40 45 50 55 60 65 70 75 80 85No. of Students 12 18 22 18 43 17 20 10 30 20 10 15 5
SolutionMarks
XStudents
fcf
60.25
120.5
180.75
M = The size of N + 1 th items values 2
= The Size of (240 + 1)/ 2 = 241/ 2 = 120.5M = 50Q1 = The size of N+1 th item’s values 4 = The Size of (240+1)/4 = 241/4 = 60.25Q1 = 40
Q3 = The size of 3(N + 1) th items values4
= The size of 3(240 + 1) = 3 x 241 = 180 .75 4 4
Q3 = 65
25303540455055606570758085
1218221843172010302010155
12305270113130150160190210220235240
240CONTINUOUS SERIES
Steps1. Find out the cumulative frequency2. Find out the first quartile item q1 = N/43. Find out the third quartile item q3 = 3N/4
Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 f
Q2 or M = l1 + l2 – l1 (m – c) where m = N/2 f
Q3 = l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 f
ILLUSTRATION = 06Calculate the value of the Median and Quartiles from the following data.
Values x 4–6 6 – 8 8–10 10–12 12-14 14–16 16-18 18-20 20-22 22-24 24-26 26-28Frequency f 10 5 20 15 12 25 14 26 32 18 20 10
Solution
X F cf Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 = 207/4 = 51.75 f4 – 6 10 10
70
6 – 88 – 1010 - 1212 – 1414 – 1616 – 1818 – 2020 – 2222 – 2424 – 2626 – 28
520151225142632182010
1535506287101127159177197207
51.75
103.5155.25
Lower quartile class 12 – 14= 12 + (14-12) (51.75 – 50) 12= 12 +(2/12) x 1.75 = 12 + (3.5/12) = 12 + 0.29Q1 = 12.29Q2 or M = l1 + l2 – l1 (m – c) where m = N/2
f = 207/2 = 103.5Median class 18 – 20 M = 18 + 20 –18 (103.5 – 101)
26M = 18 + (2/26) x 2.5 = 18 + (5/26) = 18 + 0.19M = 18.19207
Q3 = l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 = (3x207)/4 = 155.25 f
Upper Quartile lies between 20 – 22= 20 + 22 – 20 (155.25 – 127)
32= 20 + (2/32) x 28.25= 20 + (56.50/32)= 20 + 1.765= 21.765
ILLUSTRATION = 07Find the median and quartiles from the following data.
Size 11 – 15 16 –20 21 – 25 26 –30 31 –35 36 – 40 41 – 45 46 – 50Frequency 7 10 13 26 35 22 11 5
Solution:Convert inclusive series into exclusive series
X f cf
32.2564.596.75
10.5 – 15.5 15.5 – 20.5 20.5 – 25.5 25.5 – 30.530.5 – 35.5 35.5 – 40.540.5 – 45.545.5 – 50.5
71013263522115
717305691113124129
N 129
ILLUSTRATION = 08Form the following data, calculate median & quartiles
Wages Below 6 6 – 10 11 – 15 16 – 20 21 – 25 26 – 30 31 – 35 36 – 40 TotalWorkers 03 10 20 30 20 9 5 3 100
Solution: Convert inclusive series into exclusive series
X f cf
71
Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 = 100/4 = 25f
= 10.5 + 15.5 – 10.5 (25 – 13) = 10.5 + (5/20) x 12 20Q1 = 13.5
Q1 = Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N = 129 = 32.25 f 2 4
= 25.5 + 30.5 – 25.5 (32.25 – 30) 26Q1 = 25.93
Median = l1 + l2 – l1 (m – c) where m = N/2 f
= 30.5 +35.5 – 30.5(64.5 – 56) 35
M = 31.71Q3 =l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 = (3x129) = 96.75
f 4= 35.5 + (40.5 – 35.5) (96.75 – 91)
22Q3 = 36.81
255075
0.5 – 5.55.5 – 10.510.5 – 15.515.5 – 20.520.5 – 25.525.5 – 30.530.5 – 35.535.5 – 40.5
0310203020953
03133363839297100
N 100
ILLUSTRATION = 09Calculate median and quartiles from the following data
Expenses Less than 5 10 15 20 25 30 35 40 45No. of Students 5 8 28 37 67 70 90 96 100
Solution Convert the cumulative frequency distribution into an ordinary frequency distribution.
ExpenditureX
f cf
25
50
75
0 – 55 – 1010 - 1515 – 2020 – 2525 – 3030 – 3535 - 4040 – 45
532093032064
58283767709096100
N 100
ILLUSTRATION = 10Calculate the Values of Median and quartiles from the following data
Age More than 0 10 20 30 40 50 60 70 80 90 100No. of persons 80 77 72 65 55 43 28 16 10 8 0
SolutionConvert the given cumulative frequency distribution into an ordinary table.
X f Cf
72
Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 = 80/4 = 20f
= 30 + 40 – 30 (20 – 15) = 30 + (10/10)x 5 10Q1 = 35
Q1 = l1 + l2 – l1 (q1 – c) Where q1 = N/4 = 100/4 = 25f
= 10 + 15 – 10 (25 – 8) = 10 + (5/20) x 17 20Q1 = 14.25
Median = l1 + l2 – l1 (m – c) where m = N/2 = 100/2 = 50 f
= 20 +25 – 20(50 – 37) 30
= 20 +(5/30) x 13M = 22.16
Q3 =l1+ l2 – l1 (q3 – c) Where q3 = 3N/4 = (3x100)/4 = 75 f
= 30 + 35 – 30 (75 – 70) 20
= 30 + (5/20) x 5Q3 = 31.25
20
4060
0 – 1010 - 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 9090 – 100100 – 110
357101215126280
38152537526470728080
N 80
MODEMode is the most common item of a series. Mode is the values which occurs the greatest numbers of
frequency in a series. It is derived from the French word ‘Lamode’ meaning the fashion. Mode is the most fashionable or typical value of a distribution, because it is repeated the highest number of times in a series.
According to Croxton and Cowden “ The mode of a distribution is the value at the point around which the item tend to be most heavily concentrated”.
Mode is defined as the value of the variable which occurs most frequently in a distribution.The chief feature of mode is that it is the size of that item which has the maximum frequency and is also
affected by the frequencies of the neighboring items.Calculation of Mode – Individual series
Mode can often be found out by mere inspection in case of individual observations.
The data have to be arranged in the form of an array so that the value which has the highest frequency can be known.For example : 10 persons have the following income Rs.850, 750, 600, 825, 850, 725, 600, 850, 640, 850 hear Rs.850 repeats four times. Therefore model salary is Rs.850.
In certain cases that there may not be a mode or there may be more than one more for example.a. 40, 44, 57, 78, 48 -------------No modeb. 45, 55, 50, 45, 55 -------------Bimodal
When we calculate the mode from the given data, if there is only one mode in the series, it is called unimodal, if there are two modes, it is called Bimodal.
DISCRETE SERIES:-In a discrete series mode can be located in two ways
1. By Inspection2. By Grouping
Inspection Method:-When there is a regularity and homogeneity in the series, then there is a single mode which can be
located at a glance by looking into the frequency column for having maximum frequency.
ILLUSTRATION = 01Find mode from the following series.
73
Height in cm 150 160 170 180 190 200 210No. of persons 2 4 8 15 6 5 3Solution:
By inspection of the frequency, it is noted that the maximum frequency is 15 which corresponds to the value 180. Hence, Mode is 180cm.
GROUPING METHOD:When there are irregularities in the frequency distribution or two or more frequencies are equal then it is
not obvious that which one is the maximum frequency. In such cases, we use the method of grouping to decide which one may be considered as maximum frequency. That is we try to find out single mode by using grouping method. This method involves the following steps.
a. Prepare grouping tableb. Prepare analysis tablec. Find mode
a. Preparing a grouping table:Steps: construct a table of 6 columns.
Column 1:- Given frequenciesColumn 2:- The given frequencies are added in two’sColumn 3:- The frequencies are added in two’s leaving out the first frequencyColumn 4:- The given frequencies are added in three’s Column 5:- The given frequencies are added in three’s leaving out the first frequencyColumn 6:- The given frequencies are added in three’s leaving out the first two frequencies.
After making these columns, the maximum frequency is underlined or round off
b. Construction of analysis Table:-The values containing the maximum frequencies are noted down for each column and are written in a
table called analysis Table.
c. Location of Mode – The value of the variable which occurs maximum number of times in the analysis table, is the value of mode.
ILLUSTRATION = 02Calculate mode from the following data
Variable 8 9 10 11 12 13 14 15Frequency 5 6 8 7 9 8 9 6Solution
Here maximum frequency 9 belongs to two values of the items 12 & 14. However due to irregular distribution of frequencies, we use the method of grouping to decide which one may be considered maximum frequency.
ValuesSum of Frequencies Analysis Table
F 1 2 3 4 5 6 Tally bars Repeat8 5
11
15
17
15
14
16
17
19
24
21
26
24
23
9 6 1 110 8 1 1 211 7 11 11 1 512 9 11 11 413 8 1 1 1 314 915 6
Separate analysis TableColumns no Size of items containing maximum frequency
74
10 11 12 13 14 Since 12 occurs maximum number of times. Hence the value 12 is the mode.1 12 14
2 12 133 12 13 144 11 12 135 -- 13 146 10 11 12
1 2 5 4 3
ILLUSTRATION = 03Calculate the mode from the following
Size : 10 11 12 13 14 15 16 17 18Frequency: 10 12 15 19 20 8 4 3 2Solution:
Grouping TableSize F Sum of 2 Fre Sum of 3 Fre Analysis Table
1 2 3 4 5 6 Tally bars Repeat10 10
22
34
28
7
27
39
12
5
37
47
9
46
32
54
15
11 12 1 112 15 1 11 313 19 1 1 1 1 1 514 20 1 11 1 415 8 1 116 417 318 2
ANALYSIS TABLESize of item containing maximum frequency
Column 11 12 13 14 15 The mode is 13 as size of item repeats five times. But through inspection we say the mode is 14, because the size 14 occurs 20 times. But this wrong decision is revealed by analysis Table.
1 142 12 133 13 144 13 14 155 11 12 136 12 13 14
1 3 5 4 1CONTINUOS SERIES – CALCULATION OF MODE:Steps: -
1. Determine the model class (in exclusive). The class having the maximum frequency is called model class. This is done either by inspection or by grouping method.
2. Determine the value of mode by applying the formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
Z = ModeL1 = Lower limit of the model classL2 = Upper limit of the model classf1 = Frequency of the model classf0 = Frequency of the pre – model classf2 = Frequency of the class succeeding the model class.
Note: - This formula is applicable only in case of equal and exclusive class – intervals in ascending order. If the mode lies in the first class interval, than f0 is taken as zero. If mode lies in the last class interval then f2 is taken as zero.ILLUSTRATION = 04
The distribution of wages in a factory is as follows calculate the mode.Wages in Rs. 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70
75
No. of workers 6 9 10 16 12 8 7Solution: - By inspection, the maximum frequency is 16, Hence the model class is 30 – 40Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 30 + 16 - 10 (40 –30) 2 x 16 – 10 – 12
= 30 + (6/32 – 22) x 10 = 30 + (60 / 10) = 36ILLUSTRATION = 05
Determine the mode in the following frequency distribution.Income in Rs 100-200 200-300 300-400 400-500 500-600 600-700 700-800No. of persons 30 70 80 100 20 30 20Solution
By inspection, the maximum frequency is 100, Hence the modal class is 400 – 500
Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 400 + 100 - 80 (500 - 400) 2 x 100 – 80 – 20
Mode = 420
ILLUSTRATION = 06Calculate mode from the following series
Variable 0 – 5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35Frequency 1 2 10 4 10 9 2Solution:
Since the frequencies of class 10 – 15 and 20 – 25 are 10 each, we have to decide the model class by grouping
Valuef Sum of two Sum of three Analysis table1 2 3 4 5 6 Tally bar Repeat
0 – 5 13
14
19
12
14
11
13
23
16
21
24 1 1 11 1 1 1111 1 1 1 1 1
23631
5 – 10 210 – 15 1015 – 20 420 – 25 1025 – 30 930 – 35 2
Separate AnalysisCN 10 – 15 15- 20 20 – 25 25 – 30 30 – 351 10 –15 20 –252 20 – 25 25 – 303 15 –20 20 – 254 15 –20 20 – 25 25 – 305 20 – 25 25 – 30 30 – 356 10 –15 15– 20 20 – 25
2 3 6 3 1Therefore model class is 20 – 25, with maximum number of times;
Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 20 + 10- 4 (25 - 20) 2 x 10 – 4 – 9
= 20 + (6/ 20 - 13) x 15 = 24.28Calculate Mode in inclusive series
ILLUSTRATION = 07Calculate mode for the following data
Classes 0 – 9 10 – 19 20 – 29 30 – 39 40 – 49 50 – 59 60 – 69 70 - 79
76
Frequency 1 2 6 6 12 8 5 3 Solution – Convert inclusive series into exclusive series
Class F- 0. 5 – 9.59.5 – 19.519.5 – 29.529.5 – 39.539.5 – 49.549.5 – 59.559.5 – 69.569.5 – 79.5
126612853
To Calculate mode in case of un – equal class intervals.ILLUSTRATION = 08
Calculate the value of mode from the following dataClasses 0 – 5 5 – 7 7 – 9 9 – 10 10 – 13 13-15 15-16 16-19 19-20 20-21 21 -25Frequency 2 3 4 2 10 5 3 4 3 2 3Solution:-
Re – arrange the above series into a series with equal class intervals.Classes 0 – 5 5 – 10 10 – 15 15 – 20 20 - 25Frequency 5 9 15 10 5
By inspection, the maximum frequency is 15, hence the modal class is 10 – 1.Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 10 + 15- 9 (15 – 10) 2 x 15 – 19 – 10
= 10 + (6/ 30 -19) x 8 Z = 12. 73
ILLUSTRATION – 09Find the mode of the following data. Also calculate the value of Median –
Height in inches 56 –62 63 – 65 66 – 68 69 – 71 72 – 84Frequency 15 54 60 81 24Solution: -
Since the class intervals are not equal and cannot be converted into equal class – intervals. So we use empirical formula Z = 3m – 2Mean to find mode.
X f cf mid value dx = x – 67 fdx = A + fdx x c N = 67 + (225/234)x1 = 67 + 0.96 = 67.96
55.5 – 62.562.5 – 65.565.5 – 68.568.5 – 71-571.5 – 84.5
1554608124
1569129210234
5964677078
- 8-30311
-120-162
0243264
N= 234 fdx= 225Median = L1 + L2 – L1 (m – c) Where m = N/2 = 234/2 = 117.0
fMedian class 65.5 – 68.5= 65.5 + 68.5 – 65.5 (117.0 – 69)
60= 65.5 + (3/60) x 48 = 67.9
Mode = 3 Median – 2 Mean Z = 3M - 2
= 67.9 x 3 – 2x 67.96 Z = 67.78
77
By Inspection, the maximum frequency is 12, hence the modal class in 39.5 – 49.5Formula Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 39.5 + 12 - 6 (49.5 – 39.5) 2 x 12 – 6 – 8
= 39.5 + (6/ 2x 12 – 6 - 8) x 10= 39.5 + (60/ 10)
Z = 45.5
ILLUSTRATION = 10Find mode of the following frequency distribution
Class 14 –15 16 – 17 18 – 20 21 – 24 25 – 29 30 - 34 35 – 39Frequency 6 14 15 11 11 10 9 Solution
Since the class intervals are not equal, we may use the empirical formula to find mode. Z = 3M - 2
Convert into exclusive seriesA = 22.5, C = 1, dx(x – A)/c
ClassX
f cf Mid value dx =(x – A)/c Fdx
= A + fdx x c N = 22.5+ (98/76)x1 = 23.78
13.5-15.515.5-17.517.5-20.520.5-24.524.5-29.529.5-34.534.5-39.5
614151111109
6203546576776
14.516.519.522.527.032.037.0
-8-6-30
4.59.514.5
-48-84-450
49.595.0130.5
N= 76 -177 + 275.9 fdx =98
Median = L1 + L2 – L1 (m – c) Where m = N/2 = 76/2 = 38 f
Median class 20.5 – 24.5= 20.5 + 24.5 – 20.5 (38 – 35)
11= 20.5 + (4/11) x 3 = 21.59
Mode = 3 Median – 2 Mean Z = 3M - 2
= 3 x 21.59 – 2x 23 .78 Z = 17.21
ILLUSTRATION = 11Calculate mode of the following frequency distribution
Mid Value 55 65 75 85 95 105 115Frequency 8 10 16 14 10 5 2Solution
Prepare class intervals with the help of the mid value. By inspection maximum frequency is 16 Hence Modal class is 70 - 80
X f .’. Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 70 + 16 – 10 (80 – 70) 2 x16 – 10 –14 = 70 + (6/ 32 –24)x 10 Z = 77.5
50 – 6060 – 7070 – 8080 – 9090 – 100100 – 110110 – 120
81016141052
To calculate mode in case of cumulative frequency distributionILLUSTRATION = 12
Calculate mode from the following data.
78
Marks more than 0 10 20 30 40 50 60 70 80 90 100Students 80 77 72 65 55 43 28 16 10 8 0Solution: Convert the given cumulative frequency distribution to ordinary table.
X No. of Students Since maximum frequency is 15, Hence modal class is 50 – 60
Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 50 + 15 – 12 (60 – 50)2 x 15 – 12 –12
= 50 + (3 / 30 –24)x10Z = 55
0 – 1010 - 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 9090 – 100
80 – 77 = 377 – 72 = 572 – 65 = 765 – 55 = 1055 – 43 = 1243 – 28 = 1528 – 16 = 1216 – 10 = 610 – 8 = 28 – 0 = 0
ILLUSTRATION = 13Calculate Median and mode from the following data
Marks less than 20 30 40 50 60 70 80 90 100No. of Students 15 40 72 128 206 246 267 290 300Solution
Convert the given cumulative frequency distribution into an ordinary tablex No. of Students cf
10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 8080 – 9090 – 100
1540 – 15 = 2572 – 40 = 32128 – 72 = 56206 –128 = 78246 – 206 = 40267 – 246 = 21290 – 267 =23300 – 290 = 10
154072128206246267290300
N = 300
Merits of Mode Mode as a measure of central tendency has some merits.
1. It is simple to understand and it is easy to calculate. 2. Generally it is not affected by end values.3. It can be determined graphically & it can be found out by inspection.4. It is usually an actual value as it occurs most frequently in the series.5. It is the most representative average.6. Its value can be determined in an open end class interval without ascertaining the class limit.
Demerits of mode1. It is not suitable for further mathematical treatment. 2. It may not give weight to extreme items.3. In a bimodal distribution there are two modal classes and it is difficult to determine the value of the mode.
and it is difficult to determine the value of the mode.4. It is difficult to compute, when are both positive and negative items in a series.5. It is not based on all the observations of a given series.6. It will not give the aggregate value as in average.
RELATIONSHIP BETWEEN 3’M’S OR EMPIRICAL RELATIONSHIPIn a symmetrical distribution. Mean Median and Mode will coincide i.e. = M = Z. But in an
asymmetrical (skewed) distribution, these values will be different when the distribution is moderately skewed and has greater concentration in the lower values >Median >Z, it means the distribution is positively skewed. If the distribution concentrated in higher values, the tail is towards the lower values, then it is gentatively skewed.
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Median = L1 + L2 – L1 (m – c) Where m = N/2 = 300/2 =150 f
Median class is 50 – 60 = 50 + 60 – 50 (150 – 128) = 50 + (10 /78)x 22 = 52.82 78By inspection maximum frequency is 78. Hence modal class is 50 - 60Z = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 50 + 78 – 56 (60 – 50)2 x 78 – 56 – 40
= 50 + (220/60) = 53.67
In a moderately asymmetrical distribution the difference between and Z is three times of the differences between and median.
Symbolically (Karl Pearson’s)Mode = 3 Median – 2 Mean Z = 3M - 2
When the value of mode is ill – defined for a given series in such case, Mode value can be estimated by using the above formula.
ILLUSTRATION – 15If in a moderately asymmetrical frequency distribution, the values of Median and Mean are 72 and 78
respectively estimate the value of the mode.Solution.
The value of the mode is estimated by applying the following formulaMode = 3 Median – 2 Mean
= 3 x 72 – 2 x78= 216 – 156
Z = 60
ILLUSTRATION = 16The mean and mode in a moderately symmetrical distribution are 24.6 and 26.1 respectively. Find the
median.
Solution :- Formula Z = 3M - 226.1 = 3M – 2 x 24.626.1 = 3M – 49.2-3M = -49.2 – 26.1M = -75.3/-3
M = 25.1
ILLUSTRATION = 17If mode is 15.99 and median is 15.73, find the most probable value of the Mean, by using 3M’s formula
Z = 3M - 215.99 = 3 x 15.73 - 215.99 = 47.19 – 22 = 47.19 - 15.99, = 31.20/2, = 15.60
CALCULATION OF ARITHMETIC MEAN, MEDIAN, MODE AND QUARTILESPROBLEM = 01
From the following data calculate the values of Mean, Median, Mode and Quartiles.Classes 0 –3 3 –6 6 –10 10–11 11-15 15-19 19-20 20-24 24-30 30-33 33-37 37-40Frequency 04 12 24 32 44 52 52 28 20 08 04 10
Solution :A = 17, C = 1, dx = (x – c)/cx F Cf midx dx fdm = A + fdm x c
N = 17 + (49/290) x 1 = 17.168
0 –33 –6
6 – 1010 - 11
04122432
04164072
1.54.58.010.5
-15.5-12.5-9.5-6.5
-62-150-228-208
80
11 –1515 –1919 – 2020 –2424 –3030 –3333 – 3737 – 40
4452522820080410
116168220248268276280290
13.017.019.522.027.031.535.038.5
-4.00.02.55.010.014.518.021.5
-1720
13014020011672215
Median = L1 + L2 – L1(m –c ) f
Where m = N/2 = 290/2 = 145Median class is 15 - 19
N= 290 fdm= 49 -824 +873Note: To Calculate the Value of mode, the given series must be re – arranged with equal class intervals
ValuesX
FBy Inspection, the highest frequency is 180 so modal class is 10 –20
0 – 1010 - 2020 –3030 – 40
401804822
N= 290FormulaZ = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
= 10 + 180 – 40 (20 – 10) 2 x 180 – 40 –48= 10 +(140/360-85)x10 = 10+ (1400/275) = 15.09
Median = L1 + L2 – L1(m –c ) f
M = 15 + 19 – 15 (145 – 116) 52 = 15 + (4/52)x29 = 15 + 2.23 = 17.23Q1= L1+ L2 – L1 (q1–c) Where q1=N/4 = 290/4=72.5 fLower quartile class = 11 –15 = 11 + 15 – 11 (72.5 – 72) 44 = 11 + (4/44)x0.5 = 11 + (2.0/44) = 11.05Q3 = L1+ L2 – L1 (q1–c) f Where q3 = 3N/4 = (3x290)/4 =217.5Upper quartile class = 19 – 20= 19 + 20 – 19 (217.5 –168)
52= 19 + (1/52)x49.5 = 19+0.95 = 19.95
Problem = 02From the following data of the weekly wages of workers employed in a certain factory, construct a
frequency table with 0 – 10, 10-20 and soon. Find Mean, Median, Mode and quartileValues 35 56 10 8 37 50 35 12 61 53 40 30 62 64 7 31 5 49 8 15
12 35 24 9 13 22 28 53 69 45 20 60 19 44 40 57 48 27 41 375 5 53 35 63 33 17 42 25 44
Solution A = 35, C=10, dx = (x – A)/cValues
XTally bars f cf midx Dx Fdm
= A + fdm xc N = 35 + (-2/50) x 10 =35-(20/50)= 34.6
Where m = N/2 = 50/2 = 25 Median class is 30 – 40
0 – 1010 - 2020 –3030 –4040 –5050 –6060 –70
1111 11111 1111111 11111 11111111 11111111 111111
0608060909075
6142029384550
05152535455565
-3-2-10123
-18-16-6091415
N=50 fdm= -2Where m = N/2 = 50/2 = 25 Median class is 30 – 40 Median = L1 + L2 – L1(m –c )
fM = 30 + 40 – 30 (25 – 20) 9 = 30 + (10/9) x 5 = 35.55Since there are two highest frequencies in a given series, we have to find out mode through 3’M’s Formula
Z = 3M -2Z = 3x35.55 – 2x34.6
Q1= L1+ L2 – L1 (q1–c) Where q1=N/4 = 50/4=12.5 fLower quartile class = 10 – 20
= 10 + 20 – 10 (12.5 – 6) 8= 10 + (10/8)x 6.5 = 18.125
Q3 = L1+ L2 – L1 (q1–c) f Where q3 = 3N/4 = (3x50)/4 =37.5Upper quartile class = 40 – 50
81
Z = 106.65 – 69.2 Z = 37.45
= 40 + 50 – 40 (37.5 –29) 9
= 40 + (10/9) x 8.5 = 49.4
PROBLEM = 03From the following data, prepare a frequency distribution taking first class as 40 – 49, 50 – 59 and so an
Also calculate Mean, Median, Mode and Quartiles.42 74 40 60 82 115 41 61 75 83 63 76 53 100 7684 50 67 65 78 77 56 95 78 68 69 104 80 79 7954 73 59 81 100 79 66 49 77 90 84 76 42 64 6970 80 72 72 50 79 52 103 96 51 86 78 94 71 73
Solution: - A = 74.5, C = 10, dx = (x – A)/CValue
XTally bars f cf Midx dx fdxm
= A + fdm xc N = 74.5+ (-9/60) x 10 = 74.5 – (90/60) = 73
40 –4950 – 5960 – 6970 – 7980 – 8990 – 99
100 – 109110 – 119
11111111 1111111 11111111 1111 1111 11111111 111111111111
5810208441
513234351555960
44.554.564.574.584.594.5104.5114.5
-3-2-101234
-15-16-10088124
60 fdx -9Q1= L1+ L2 – L1 (q1–c) Where q1=N/4 = 60/4=15 fLower quartile class = 60 – 69L1 = 60 – 0.5= 59.5L2 = 69 +0.5= 69.5 = 59.5 + 69.5 – 59.5(15 – 13) 10= 59.5 + (10/10)x 2 = 61.5
Q3 = L1+ L2 – L1 (q1–c) f Where q3 = 3N/4 = (3x60)/4 =45Upper quartile class = 80 – 89 L1 = 80 – 0.5 = 79.5L2 = 89 +0.5 = 89.5= 79.5 + 89.5 – 79.5 (45 –43)
8= 79.5 + (10/8)x 2 = 82
Where m = N/2 = 60/2 = 30 Median class is 70 – 79Convert into exclusiveL1 = 70 – 0.5 = 69.5L2 = 79 + 0.5 = 79.5Median = L1 + L2 – L1(m –c )
fM = 69.5+ 79.5 – 69.5 (30 -23) 20 = 69.5 + (10/20)x7 = 73
Mode = L1 + f1 – f0 (L2 – L1) 2f1 – f0 – f2
Maximum frequency is 20, so modal class is 70 – 79 i.e. 69.5 – 79.5= 69.5+ 20 – 10 (79.5 – 69.5) 2 x 20 – 10 –8= 69.5+(10/40-18)x10 = 69.5+ (100/22) = 74
MODAL QUESTIONS – THEORETICAL(5, 10 & 15 Marks)1. What is the concept of measures of central tendency? What are its objectives and functions?2. What are the characteristics of an ideal average? How far arithmetic Mean, Median, Mode, Geometric
mean and Harmonic mean satisfies these characteristics3. Define different statistical averages and explain the merits and demerits.4. What do you mean by Median and quartiles? State their uses.5. Define geometric mean and Harmonic Mean. Also State their Merits & demerits.6. What purpose is served by an average? Explain relative merits and demerits of various types of statistical
averages.PRACTICAL PROBLEMS:Problem No. 1
The marks scored by 50 students in an examination in statistics are given below.30 45 48 55 39 25 31 12 18 21
82
54 59 51 33 43 44 10 38 19 2647 35 37 41 46 33 51 37 58 5817 19 23 26 29 38 57 36 35 4443 27 31 43 22 31 47 34 18 15
Prepare a frequency table taking a class interval of 10, taking first class as 10 – 20, 20 – 30 &so onAlso calculate the arithmetic mean, median, mode & quartiles.
[Answers = 35.6, Median = 36, Mode = 36.6, Q1= 25.625, Q3 = 45.9]Problem No. 2
Following are the marks (out of 100) obtained by 50 students in statistics.70 55 51 42 57 45 60 47 63 5383 65 89 82 55 64 50 62 65 7590 80 58 52 66 45 42 65 70 6193 59 49 41 75 52 46 42 75 6555 65 45 63 54 48 64 66 67 60
Make a frequency distribution taking a class interval of 10 marks, to start with 40 – 49, 50 – 59 & soonAlso calculate AM, M,Z, & Quartiles
[Answers = 67.5, M = 60.75, Z = 62.5, Q1 = 50.33, Q3 = 68.56]Problem No. 3:
The frequency distribution of weight in grams of mangoes of a given variety is given below. Calculate the arithmetic Mean, and ModeWeight in grams
410 –419 420 –429 430 –439 440 –449 450 –459 460 – 469 470 –479 Total
No. of Mangoes
14 20 42 54 45 18 7 200
[Answers : = 443.4, M = 443.94, Mode = 445.21]Problem No: 4
Find out the Median and the Mode for the following table: -No, of days absent less than 5 10 15 20 25 30 35 40 45No. of students 29 224 465 582 634 644 650 653 655[ Answers: Median = 12.1 days Mode = 11.35 days]Problem No. 5
Calculate the Mean, Median and Mode for the data given belowDaily wages in Rs
50-53 53-56 56-59 59-62 62-65 65-68 68-71 71-74 74-77 Total
No. of persons 03 08 14 30 36 28 16 10 05 150[Answers: = 63.82, Median = 63.67, Mode =63.29]
Problem No. 6a. The values of mode and median for a moderately skewed distribution are 64.2 and 68.6 respectively. Find the value of the mean – [Use Z = 3M - 2]
[Answer = 70.8]b. In a moderately asymmetrical distributions, the values of Mode and Mean are 32.1 and 35.4 respectively. Find the Median value – [Use Z = 3M - 2]
[Answer Median = 34.3]c. If the Mean and Median of a moderately asymmetrical series are 26.8 and 27.9 respectively, what would be its most probable mode? [Use Z = 3M –2X] [Answer Mode = 30.1]
Problem No. 7Calculate the Median and the quartiles from the following.
Mid value 15 25 35 45 55 65 75 85 95 105Frequency 18 29 46 62 96 56 42 30 16 10[Answers: M = 54.9, Q1= 41.33, Q3 = 69.42]Problem: 8
Calculate geometric Mean of the following:-Values 2574 475 5 0.8 0.005 0.0009
[Ans GM = 1.84]
83
Problem: 9Calculate the Mean, Median and Mode of the following distribution
Mid x 59 61 63 65 67 69 71 70Frequency 01 02 09 48 131 102 40 17
[Answers Mean= 67.9, Median = 67.75, Mode = 67.48]Problem: 10Values 2574 475 75 5 0.8 0.08 0.005 0.0009
[Ans. HM = 0.006039]Problem: 11
Calculate arithmetic Mean, Mode and Median for the following distributionMid Values 95 105 115 125 135 145 155 165 175Frequency 4 2 18 22 21 19 10 3 2
[Ans. = 132.62, Z = 128, M = 132.14]Problem No. 12
Calculate Mean, Median and mode form the following dataMarks less than 10 20 30 40 50 60 70 80No. of students 15 35 60 84 96 127 198 250
[ Answers = 50.4, Median = 59.35, Mode = 66.78]Problem No: 13
Calculate the Mean, Median and Mode from the following frequency distributionOut put in units No. of workers
Answers: Mean = 339.05 Median = 340.23 Mode = 342.72
Below 310310 –319320 – 329330 – 339340 – 349350 – 359360 – 369
370 & above
92024384847176
Total 209Problem = 14 From the following data relating to marks obtained by 675 students, in an examination. Calculate the Mean, Median, and Mode of the percentage marks obtained.Marks obtained less than 10% 20% 30% 40% 50% 60% 70% 80%No. of students 7 39 95 201 381 545 631 675
[Ans. Mean = 46.87, Median = 47.58, Mode = 49]Problem = 15 From the following data, calculate the value of the Mean, Median and Mode.Wages in Rs. above 10 20 30 40 50 60 70 80No. of workers 650 500 425 375 300 275 250 100[ Ans. Mean = 49.33, Median = 46.67, Z = 77.14] [Note: To find out modal class, use grouping Table & Analysis table]
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