Unit - 4 MEASURES OF CENTRAL TENDENCY AVERAGES Mass 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 sample 4. To establish relationship between the two series 5. To provide basis for decision making 6. To calculate the representative single value from the given data. 49
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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.
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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.
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
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
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.
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.
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.
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
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
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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
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.
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
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
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)
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)
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)
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.
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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)
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
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.
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
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
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
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.
79
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
= 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
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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
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
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.
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.
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]