Statistics

Statistical-Measure-of-Central-Tendency

Posted by Asok K Nadhani on

Last Updated on: 11th July 2024, 02:41 pm

Statistical Measure of Central Tendency

Measures of Central Tendency

 In this part, we discus about various statistical measures of central tendency like

  • Frequency Distribution
  • Arithmetic, Geometric & Harmonic Mean
  • Mode, Median
  • Deviation

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Central Tendency Measures

A Measure of Central Tendency (also referred to as measures of centre, or central location) is a summary measure, describing a whole set of data with a single value, that represents the middle or centre of its distribution.

This single value is the point of location around which individual values cluster and is termed as the Measure of Central Tendency. Object of computing an average value for a set of observations is to obtain a single value which is representative of all the items.

There are three main measures of central tendency:  Mode, Median and Mean. Each of these measures describes a different indication of the typical or central value in the distribution.

Averaging

Average is the calculated “central” value of a set of numbers. The average depicts the characteristic of the whole group. An average represents the entire data. Its value lies somewhere in between the two extremes, i.e. the largest and the smallest items. For this reason an average is frequently referred to as a Measure of Central Tendency.

Objects of Averaging

  • Bird’s-eye view of the entire data : Measures of central value, by considering the mass of data in one single value, gives a bird’s-eye view of the entire data.

For example, average income  (obtained by dividing the total income by the number of companies) gives one single value that represents the entire industry, which is more meaningful than individual income value of each company.

  • Ease of comparisons : Reducing the mass of data to one single figure helps comparisons. Comparison can made either at a point of time or over a period of time.

For example, comparing average annual profits of different industries for a particular year helps us to know performance of various industries. Comparing for different time periods reveal who are  improving or who are deteriorating.

Types of Averages

  • Mathematical : Computed arithmetically from the set of values
  • Arithmetic Mean
  • Geometric Mean
  • Harmonic Mean
  • Positional : Indicates the position of the average in the set of numbers
  • Median
  • Mode
1 1

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Arithmetic Mean

An Arithmetic Mean or  Arithmetic Average may be defined as the quotient obtained by dividing the total of the values of a variate by the total number of their observations or items.

\displaystyle \overline{x}=\frac{{\left( {\sum{x}} \right)}}{n}, where \displaystyle \overline{x} = arithmetic average, x= value of a variable, \displaystyle \sum{X}= Sum of the values of a variable, n= Total number of observations or items.

Simple Arithmetic Mean

\displaystyle \overline{x} = (x1 + x2 + x3 + … + xn)  / n = \displaystyle \left[ {\frac{{\left( {\sum{X}} \right)}}{n}} \right]

Ex: Find the A. M. of the numbers 4, 10, 18, 22, 26.

Here n (the total number of items) = 5.

 \displaystyle \overline{x}= \displaystyle \frac{{\left( {4+\text{ }10\text{ }+\text{ }18\text{ }+\text{ }22\text{ }+\text{ }26} \right)}}{5}, \displaystyle \overline{x}= \displaystyle \frac{{80}}{5} = 16

Weighted Arithmetic Mean (Weighted Mean).

If the n values of a variable x1, x2, x3, …, xn are taken f1, f2, f3, …, fn times,  respectively (ί.e., if, f1, f2, f3, …, fn are the respective frequencies of x1, x2, x3, …, xn) then

Weighted Mean (X)  = (f1x1 + f2x2 + f3x3 + … + fn xn) / (f1 + f2 + f3 + … + fn) = \displaystyle \frac{{(\sum{{fx}})}}{{(\sum{f})}}

Ex. Find average income from the following table:

 Daily income (Variate = x)2591113Total 
 No. of employees (frequency  = f)4284220 
 Fx (f )x (x)810724426160 

Weighted Mean (or average income) = \displaystyle \frac{{\sum{{fx}}}}{{\sum{f}}}= \displaystyle \frac{{160}}{{20}} = 8

Properties of Arithmetic Mean

  • If the value of each observation is constant, say, k, then then Arithmetic Mean is also k. If the age of each baby in a crèche is 6 months, then the arithmetic mean of all the babies is also 6 months.
  • The algebraic sum of deviation of each value from the arithmetic mean is  zero., i.e \displaystyle \sum{{fi}}(xi-x)=0, where xi is the ith term and x is the AM.

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Arithmetic Mean : Short-cut method (Deviation Method)

Here, a value preferably from the middle, is first assumed to be the value of the arithmetic average. Then from the assumed average, the deviations of the different items of the series are found out. The average of such deviations are then added to the assumed average. The resultant figure comes out to be the value of the arithmetic average.

  • Individual Series: \displaystyle \bar{x}=\text{ A+}\frac{{(\sum{d})}}{N}
  • Discrete Series :  \displaystyle \bar{x}=\text{ A+}\frac{{(\sum{{fd}})}}{{\sum{f}}}
  • Continuous Series : \displaystyle \bar{x}=\text{ A+}\frac{{(\sum{{fd}})}}{{\sum{f}}}

Where A = Assumed mean, \displaystyle \sum{{}}d = sum of deviation from assumed mean,  \displaystyle \sum{{}}fd = sum of products of deviations from assumed mean and their corresponding frequencies

Ex. Find Arithmetic Mean

Variate (x)Frequency (f)d=x–Afx= (x)X(d)
24-7-28
54-4-16
9600
11428
13248
Total20¾-28

Let A (assumed mean) = 9. Now, A.M.=A+\displaystyle \frac{{\sum{{fd}}}}{{\sum{f}}} = 9+ (-28) / 20 = 9 – 1.4 = 7.6

Note: If, instead, we take A = 11 or 5, we would get the same result. So if the value of origin (A) is changed, Mean is unchanged.

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Arithmetic mean : Step Deviation Method

Under this method, the figures of deviations are reduced by dividing them all by a common factor.

  • Individual Series: \displaystyle \bar{x}=\text{ }A+\left[ {\left( {\frac{{\sum{{d'}}}}{N}} \right)} \right]\text{ }\times \text{ }c
  • Discrete Series :  \displaystyle \bar{x}=\text{ }A+\left[ {\left( {\frac{{\sum{{d'}}}}{{\sum{f}}}} \right)} \right]\text{ }\times \text{ }c
  • Continuous Series : \displaystyle \bar{x}=\text{ }A+\left[ {\left( {\frac{{\sum{{d'}}}}{{\sum{f}}}} \right)} \right]\text{ }\times \text{ }c

Where c = Common factor by which each of the deviations is divided, d’ = the deviations from the assumed average divided by the common factor, i.e \displaystyle \frac{{\left( {X-A} \right)}}{c}

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Step Deviation Method- Problem

Ex. Find A. M. from the following Table

x:102030405060
f:6461284

Calculation of A. M.

x(f)D=x-Ad’ =\displaystyle \frac{{\left( \mathbf{d} \right)}}{{10}}f d‘d1‘ =\displaystyle \frac{{\left( \mathbf{d} \right)}}{5}f d1‘
(1)(2)(3)(4) = (3) \displaystyle \div 10(5) = (2) x (4)(6) = (3) \displaystyle \div 5(7) = (2) x (6)
106-30-3-18-6-36
204-20-2-8-4-16
306-10-1-6-2-12
401200000
5081018216
6042028416
Total40–––16––32

Let A (assumed mean) = 40.

In the above Table two common factors (ί.e., two scales) 10 and 5 have been taken to scale down the data and shown separately in the Table.

For the scale i = 10, A.M.\displaystyle =A+\left( {\frac{{\sum{{fd'}}}}{{\sum{f}}}} \right)\times I= 40+ [{\displaystyle \frac{{-32}}{{40}}} x 5] = 40-4 = 36

We get the same result for different scales (whether we take the scale as 10 or 5, the result would be same)

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Deviation Method – Problem

Ex. From the following data relating to the marks secured in English paper by the College students, calculate the average of marks under all the possible methods:

Class of marks :    0-1010-2020-3030-4040-50
No. of students  :108172520

Computation of the arithmetic average of marks

Direct method :

Marks  ClassNo. of Students fMid values xfx
0-10
10-20
20-30
30-40
40-50		10
8
17
25
20		5
15
25
35
45		50
120
425
875
900
Total\displaystyle \sum{f} = 80―\displaystyle \sum{{fx}} = 2370

We have , \displaystyle \overline{x}=\frac{{\left( {\sum{{fx}}} \right)}}{{\sum{f}}}=\frac{{2370}}{{80}}=29.625 (or 30 appx)

Short-cut Method (at A = 25)

Marks ClassNo. of Students  fMid-values xd = (x − A)
A = 25		  fd
0-10
10-20
20-30
30-40
40-50		10
8
17
25
20		5
15
25
35
45		−20
−10
0
10
20		−200
−80
00
250
400
Total\displaystyle \sum{f}=80――\displaystyle \sum{{fd}} = 370

\displaystyle \overline{x}=A+\frac{{\left( {\sum{{fd}}} \right)}}{{\sum{f}}} = \displaystyle 25+\frac{{370}}{{80}} = 25+4.625 = 29.625 (or 30 Appx)

Step Deviation Method

Marks ClassNo. of Students (f)Mid Values (x)d = (x − A)
A = 25		d’ =d/c
(C = 10)		Fd‘
0-10
10-20
20-30
30-40 40-50		10
8
17
25
20		5
15
25
35
45		−20
−10
0
10
20		−2
−1
0
1
2		−20
−8
0
25
40
Total\displaystyle \sum{f}=80―――\displaystyle \sum{{Fd'}}=37

\displaystyle \overline{x}=A+\left[ {\frac{{\left( {\sum{{Fd'}}} \right)}}{{\sum{f}}}} \right]\times c = 25 + \displaystyle \left( {\frac{{37}}{{80}}} \right) x 10 = 25+4.625 = 29.625 (or 30 Appx)

Weighted Arithmetic Average

Weight in relation to the statistical data means the relative importance of the data. All the items of a series may not be equally important for the purpose of study. Different weights are given to the different items in accordance with the nature and purpose of the study.

Direct Method

\displaystyle \overline{X}=A+\left( {\frac{{\sum{{dw}}}}{{\sum{w}}}} \right) , where \displaystyle \overline{x}= Weighted Arithmetic Mean, A = Assumed Mean,
\displaystyle \sum{{dw}}= Sum of the product of deviation of variable x and weights, 
\displaystyle \sum{w} = Sum of the weights.

Short Cut Method

\displaystyle \overline{x}=\left( {\frac{{\sum{{wx}}}}{{\sum{w}}}} \right)

Weighted Arithmetic Average – Problem

Ex. From the following data find out the weighted Mean of the pass percentage of students of Calcutta University:

CoursesPass % XNo. of Students (W)WXd =X – A
A = 70		dW
M.A.
M.Sc. M.Com.		60
70
75		10
15
20		600
1050
1500		-10
0
+5		-100
0
100
Total \displaystyle \sum{w} = 45\displaystyle \sum{{wx}} = 3150 \displaystyle \sum{w}= 0

Direct Method :

Weighted Mean = \displaystyle \left( {\frac{{\sum{{wx}}}}{{\sum{w}}}} \right)=\frac{{3150}}{{45}} =70

Short Cut Method :

Weighted Mean =\displaystyle A+\left( {\frac{{\sum{{dw}}}}{{\sum{w}}}} \right)=70+\frac{0}{{45}}=70

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Combined Arithmetic Mean for a group

If there are two groups containing n1 and n2 observations and \displaystyle \overline{{{{x}_{1}}}} and \displaystyle \overline{{{{x}_{2}}}} as the respective arithmetic means, then the combined AM is given by

x = [(n1 \displaystyle {{{{\overline{x}}}_{1}}}) + (n2\displaystyle {{{{\overline{x}}}_{2}}})]  / (n1 + n2)

Ex: The mean salary for a group of 60 female workers is Rs.7200 per month and that for a group of 80 male workers is Rs.9800 per month. Compute the combined salary

Here, \displaystyle {{{n}_{1}}}= 60, \displaystyle {{{n}_{2}}}= 80, \displaystyle {{{{\overline{x}}}_{1}}} = Rs.7200 and  \displaystyle {{{{\overline{x}}}_{2}}}= Rs.9800 hence, the combined mean salary per month = \displaystyle \overline{x}= = (60 x 7200 + 80 x 9800) / (60+80) = (4,32,000 + 784000) / 140 = 8686

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Arithmetic Mean – Merits and demerits

Merits:

  • It is easy to compute and simple to understand.
  • For counting mean, all the data are utilized. It can be ascertained even when only the number of items and their aggregate are known.
  • It is capable of further mathematical treatment.
  • It provides a good basis to compare two or more frequency distribution.
  • Mean does not need the arrangement of data.

Demerits:

  • It may give considerable weight to extreme items. Mean of 2, 6, 301 is 103 and none of the value is sufficiently represented by the mean 103.
  • In some cases, arithmetic mean may give misleading impressions. For example , average number of patients admitted in a hospital is 10.7 per day. Here mean is a useful information, but does not represent the actual item.
  • It can hardly be identified by inspection.

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Geometric Mean (G.M.)

The geometric mean (G) of the n positive values of a variate x1, x2, x3, …., xn is  the n root of the product of the values

 G =\displaystyle \sqrt[n]{{{{x}_{1}}}}. x2. x3. …., xn= (x1. x2. x3. …., xn)1/n.

Now taking logarithms on both sides

log G = (\displaystyle \frac{1}{n}) log (x1. x2. …. . xn) = (\displaystyle \frac{1}{n}) ((log x1 + log x2 + …. + log xn) =  (\displaystyle \frac{1}{n}) \displaystyle \sum{{\log x}}. 

G= antilog (\displaystyle \frac{1}{n}) \displaystyle \sum{{\log x}}

So, we find that the logarithm of the G.M. of x1, x2, …., xn = A.M. of logarithms of x1, x2, …., xn .

Uses of Geometric Mean.

  • It is used to find average of the rates of continuously increasing changes (like population growth etc)
  • It is considered to be the best average for the construction of index numbers.

Geometric Mean – Merits and demerits

Merits:

  • It is not influenced by the extreme items to the same extent as mean.
  • It is rigidly defined and its value is a precise figure.
  • It is based on all observations and capable of further algebraic treatment.
  • It is useful in calculating index numbers.

Demerits:

  • It is neither easy to calculate nor it is simple to under stand.
  • If any value of a set of observations is zero, the geometric mean would be zero, and it cannot be determined.
  • If again any value becomes negative, geometric mean becomes imaginary.

Geometric Mean – Properties

  • The Product of n values of a variate of a variate is equal to the n-th power of their G.M. ί.e., x1 . x2 .…. . xn = Gn (it is clear from the definition).
  • Taking G as geometric mean of n observations x1, x2, …, xn the ratios of each observation  to the geometric mean are \displaystyle \left( {\frac{{{{x}_{1}}}}{G}} \right),\displaystyle \left( {\frac{{{{x}_{2}}}}{G}} \right)….\displaystyle \left( {\frac{{{{x}_{n}}}}{G}} \right)
  • If G1, G2,  … are the geometric means of different groups having observations n1, n2, …. respectively, then G.M. (G) of composite group is given by

 G = (G1n1 . G2 n2 . ….) 1/N, N=n1 + n2 +….,

i.e Log G = \displaystyle \frac{1}{n} [n1 log G1 + n2 log G2 + …..]

  • The logarithm of G.M. of n observations is equal to the A.M. of logarithms of n observations.
  • The product of the ratios of each of the n observations to the G.M. is always unity.

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Geometric Mean : Problem

Ex. Find the G.M. of 111, 171, 191, 212.

Let G indicate the G.M. of the numbers

G=\displaystyle \sqrt[4]{{111\times 171\times 191\times 212}} , here n =4

Taking logarithm of both sides, log G = [1/4 (log 111 + log 171 + log 191  + log212)]

[\displaystyle \frac{1}{4}(2.0453 + 2.2330 + 2.2810 + 2.3263)] = \displaystyle \frac{1}{4}(8.8856) = 2.2214

G=Antilog (2.2214) = 166.5

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Weighted Geometric Mean : Problem

Ex. Find the G.M. of 111, 171, 191, 212 having weighted by 3, 2, 4, 5 respectively.

Xflog xf log x
11132.04536.1359
17122.23304.4660
19142.28109.1240
21252.326311.6315
Total14–31.3574

Log G = (\displaystyle \sum{f} log x)  / (\displaystyle \sum{f}) = \displaystyle \frac{{31.3574}}{{14}} = 2.2391

G = antilog 2.2391 = 173.4

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Weighted Geometric Mean : Problem

The weighted geometric mean of the four numbers 8, 25, 17 and 30 is 15.3. If the weights of the first three numbers are 5, 3 and 4 respectively, find the weight of the fourth number.

Let  f4as be the  weight of the fourth number 30, we get the following figures

xflog xf log x
850.90314.5155
2531.39794.1937
1741.23044.9216
30f41.47711.4771 f4
Total 12 + f413.6308 + 1.4771 f4

log G = (\displaystyle \sum{f} log x) / (\displaystyle \sum{f}). So, log 15.3 = [(13.6308 + 1.4771 f4)/ (12 +f4)]

or 1.1847 = [(13.6308 + 1.4771 f4)/ (12 +f4)]

or 14.2164 – 13.6308 = (1.4771 f4 ) – (1.1847 f4)

or .5856 = .2924 f4 . or f4 = \displaystyle \frac{{.5856}}{{.2924}} = 2

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Harmonic Mean (HM)

The Harmonic Mean for n observations x1, x2, …, xn is the total number  divided by the sum of the reciprocals of the numbers.

H = [(n) / { (\displaystyle \frac{1}{{{{x}_{1}}}}) + (\displaystyle \frac{1}{{{{x}_{1}}}}) + …. (\displaystyle \frac{1}{{{{x}_{n}}}})} ] = \displaystyle \frac{n}{{\sum{{}}}} (\displaystyle \frac{1}{x})

So, \displaystyle \frac{1}{H} = [(\displaystyle \sum{{\frac{1}{x}}}) / n ]

So, reciprocal of H.M. = A.M. of reciprocals of the numbers. 

Harmonic Mean – Merits and Demerits

Merits :

  • Like A.M. and G.M, HM is also dependent on all observations.
  • HM is Capable of further algebraic treatment.
  • HM is extremely helpful while averaging certain type of rates and ratios.

Demerits:

  • HM is not readily understood nor can it be computed easily.
  • HM value  may not be a member of the given set of numbers.
  • HM cannot be computed when there are both negative and positive values in a series or one or more values is zero.

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Harmonic Mean : Problem

Ex. Find the H.M. of 6, 12, 24 and 30

H.M.= [4/ {(\displaystyle \frac{1}{6}) + (\displaystyle \frac{1}{{12}}) + (\displaystyle \frac{1}{{24}}) + (\displaystyle \frac{1}{{30}})}] = [4/ {(20 + 10 + 5 + 4) / 120}]

= [4/ {(39) / 120}] = [(4x 120) / 39] = \displaystyle \frac{{480}}{{39}}

HM = \displaystyle \frac{{480}}{{39}} = 12.31 appx

Ex. Find the H.M. of 1, \displaystyle \frac{1}{2}, \displaystyle \frac{1}{3}, ,,,, \displaystyle \frac{1}{n}

H.M = [n / {1+2+ 3+…n}] = n /  [{(n( n+1)} / 2] = [2n /  {n(n+1)}] = 2 / (n+1)

[Note : Sum of numbers in AP  in the denominator  {1+2+ 3+…n} = (n(n+1) /2]

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Harmonic Mean : Problem

A car covered a distance of 50 miles four times. The first time at 50 m.p.h. the second at 20 m.p.h. the third at 40 m.p.h, and fourth at 25 m.p.h. Calculate the average speed and explain the choice of the average.

Note : For the statement x units per hour, when the different values of x (ί.e., distances) are given,  to find average, we should use H.M.

If hours ί.e., (time of journey) are given, to find average, we should use A.M.

In this problem, miles (distances) are given, so we use H.M.

Average Speed (H.M.) = 4/ {(1/50) + (1/20) + (1/40) + (1/25)}

= [4/ {(20 + 50 +25 + 40) / 1000}] = [4/ (1000 / 35)] = [(4 x 1000) / 35] = 800 /27 = 29.33 mph (appx)

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Weighted Harmonic Mean

Weighted HM = N / { (f1/x1) + (f2/x2) + … (fn/xn)}], where \displaystyle \sum{f} = N

Ex: A person traveled 20 k.m. at 5 k.m.p.h. and again 24 k.m. at 4 k.m.p.h, to find average speed.

Since distances are given. So, we should apply H.M.(weighted) to get Average Speed

Average Speed (A) = [{(20+24)} / {(20/5)+ (24/4)}] = 44 / (4+6) = 44/10 = 4.4 kmph

Ex : A person traveled 20 hours at 5 k.m.p.h. and again 24 hours at 4 k.m.p.h, to find average speed.

Here, times of journey are given. So, we should apply A.M.(weighted) to get Average Speed

Average Speed (A) = (20 x 5 + 24 x 4) / (20+24) = (100 + 96) / 44 = 4.45 kmph appx

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Median

Median is an average of position or a positional average. This is called so, because its value is determined with reference to its position in the value column of a series.

The median is that value of the variable, which divides the group into two equal parts. One  part comprising all value greater and the others all values less than the median.

Advantages of Median : Median is rigidly defined. Median is not affected by the values of extreme items. Median is very easy to calculate. Median can be calculated even if data is incomplete.

Disadvantages of Median: Median is not based on all the observations of the series. Median is not capable of further algebraic treatment like mean, geometric mean and harmonic mean. If the number of items is very small, Median may give erroneous result. Median is very much affected by fluctuation in sampling. At times, Median produces a value which is never found in the series.

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Median – Computation of

Median of Individual Series

The items of the series are arranged in ascending or descending order.

Median = value of    [(n+1)/ 2 ] th term.,

i.e the item  corresponding to [(n+1)/ 2 ] th term,  where, n = number of items. 

Ex. Find the median of marks:

3, 11, 6, 8, 13, 16, 15, 20

Arrangement  in ascending order : 3, 6, 8, 11, 13, 15, 16, 20. Here n = 8 (even number).

Median = Average value of (n/2) the term and the next item., i.e average of  (8/2) th term and next term, i.e average of   4th & 5th term = (11 + 13) / 2 = 12

Median of Discrete Series

Discrete Series contains discrete variable. Discrete variable refers to characteristic which cannot be expressed in fractions. For example number of person in a room (as number of persons cannot be fraction), while continuous Variable is like Weight or Height of Person

Median Computation – Simple Frequency Distribution

First, Cumulative frequency is calculated. Now the value of the variable corresponding to the cumulative frequency [(N+1) / 2]gives the median, when N is the total frequency.

Ex. Find the median of the following frequency distribution:

x :123456
f :71217193134

Median = Value of {(n+1) /2} th term = (120+1) /2 th term = 60.5th term

From the last column, it is found 60.5 is greater than the cumulative frequency 55, but less than the next cumulative frequency 86, corresponding to x= 5. All the 31 items (from 56 to 86) have the same variate 5. And 60.5 th item is also one of these 31 item.  So, Median is 5

xfCumulative frequency
177
21219
31736
41955
53186
634120= (N)

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Continuous Series

Grouped Frequency Distribution

We are to determine the particular class in which the value of the median lies, by \displaystyle \frac{N}{2} (and not by \displaystyle \frac{{(N+1)}}{2}, as in continuous series \displaystyle \frac{N}{2} divides the area of the curve into two equal parts). After locating median, its magnitude is measured by applying the formula of interpolation given below:

Median = l1 + [(l2 – l1 ) / f] (m-c),

where m=\displaystyle \frac{N}{2}, i1 = lower limit of the class in which median lies, i2 = upper limit of the class in which median lies, m = middle item (i.e. item at which median is located or (\displaystyle \frac{N}{2}) th term, c = cumulative frequency of the class preceding the median class

Note. The above formula is based on the assumption that the frequencies of the class-interval in which median lies are uniformly distributed over the entire class-interval.

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Median & Median Class – Problems

Ex. Find the median and median-class of the data given below:

Class- boundaries  FrequencyCumulative frequency
15-25
25-35
35-45
45-55
55-65
65-75		4
11
19
14
0
12		4
15
34
48
48
60 (= N)

Median = Value of \displaystyle \frac{N}{2} th term = value of (\displaystyle \frac{{60}}{2}), .e 30th term, which is greater than cumulative frequency 34. So, median lies in the class 35-45

Now, median = l1 +[ (I2 – l1) ] / f (m-c), where l1 = 35 2 = 45, f = 19, m = 30, c = 15

= 35+ [{(45-35) / 19} (30-45)] = 35 + (10/19) x 15 = 35 + 7.89 = 42.89

Hence the required median class is (35 – 45).

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Median & Median Class – Problems

Ex. Compute the median from the following data:

Mid-valueFrequencyMid-valueFrequencyMid-valueFrequency
11561457217538
1252515511618522
135481656019513

First find the class-boundaries from the mid-values given.

Class boundariesFrequencyCumulative frequencyClass boundariesFrequencyCumulative frequency
110-12066150-160116267
120-1302531160-17060327
130-1404879170-18038365
140-15072151180-19022387
   190-20013400

Median = value of \displaystyle \frac{N}{2} th term = value of \displaystyle \frac{{400}}{2} the term or 200th term. So, median lies in the class (150-160)

Median = l1 +[ (l2 – l1) ] / f (m-c), where l1 = 150, l2 = 160, f = 116, m = 200, c = 151

= 150 + [( 160-150) / 116 (200-151)] = 150 + (10/116) x (49) = 150 + 4.22 = 154.22

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Median & Median Class – Problems

Ex. Compute Median of the following data

Marks above :    020406080
No. of students  :7450403512

First rearrange the series, in order of the specific class intervals alongwith their corresponding frequencies, in as much as the cumulative frequencies are in descending order:

Computation Table

MarksNo. of Students FrequencyCumulative frequency
0-20
20-40
40-60
60-80
80-100		24
10
5
23
12		24
34
39
62
74
TotalN = 74―

Median = Value of m th Item = value of (\displaystyle \frac{N}{2}) the Item or \displaystyle \frac{{74}}{2}., ie. 37th Item. This lies in the class (40-60).

By interpolation, we have l1 +[ (l2 – l1) ] / f (m-c) = 40+ {(60-40) / 10} (37-34) = 40+[(\displaystyle \frac{{20}}{{10}}) x3] = 46. So, the value of the median is 46

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Quartiles, Deciles & Percentiles

There are other positional averages which are determined just in the similar manner as that of median.

  • Quartiles, (2) Deciles, (3) Percentiles, (4) Octiles, (5) Septiles, (6) Quintles, (7) Hexiles
  • Quartiles divide a series into 4 equal parts and as such there can be 3 quartiles and are denoted as Q1, Q2 & Q3.
  • Deciles divide a series into 10 equal parts and as such there can 9 Deciles and are denoted as D1, D2…, D9.
  • Percentiles divide a series into 100 equal parts and as such there can be 99 percentiles P1, P2…, P99.
  • Octiles divide a series into 8 equal parts and as such there can 7 octiles and are denoted as O1, O2…, O7.
  • Septiles divide a series into 7 equal parts as such there can 6 Septiles and are denoted as S1, S2, ……. , S6
  • Quintiles (or pentile) divide a series into 5 equal parts and as such there can 4 quintiles and are denoted as qt1, qt2……., qt4.
  • Hexiles divide a series into 6 equal parts and as such there can 5 hexiles and are denoted as H1, H2… H5.

Basic Formula

Individual & Discreet SeriesContinuous  Series
Quartiles
Q1= Value of [(N+1) /4] th term
Q2= Value of [2(N+1) /4] th term
Q3= Value of [3(N+1) /4] th term		Q1= Value of N /4 th term
Q2= Value of 2N /4 th term
Q3= Value of 3N /4 th term
Deciles
D1= Value of [(N+1) /10] th term ……
D9= Value of [9(N+1) /10] th term		D1= Value of (N /10) th term ……
D9= Value of 9(N/10) th term
Percentiles
P1= Value of [(N+1) /100] th term …..
P99= Value of [99(N+1) /100] th ter		P1= Value of [(N/100] th term ….
P99= Value of [(99N/100] th term

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Quartiles, Deciles & Percentiles : Individual Series

Ex. Find Q1, Q3, D4 and P 60 from the series (Kg):

19,27,24, 39,57,44, 56,50,59, 67,62,42, 47,60,26, 34,57,51, 59,45.

Arranging in ascending order, we have the data (Here n=20):

Sl#WtSl#WtSl#WtSl#WtSl#Wt
11953494513561759
224639104714571860
326742115015571962
427844125116592067

1. Q1 (first quartile) = size of [(n+1) / 4] th term = size of (20+1)/ 4= 5.25 th term = size of 5th term + \displaystyle \frac{1}{4} (size of 6th term – 5th term) = 34+ {(\displaystyle \frac{1}{4}) (39-34) = 34 + (\displaystyle \frac{5}{4}) = 34+1.25 = 35.25 kg

2. Q3 (Third  quartile) = size of [3(n+1) / 4] th term = size of 3(20+1)/ 4= \displaystyle \frac{{63}}{4} = 15.75 th term = size of 15th term + 3/4 (size of 16th term – 15th term) = 57+ (\displaystyle \frac{3}{4})x (59-57) = 57 + 3/2 = 58.5 kg

3. D4 (fourth decile) = size of  4 (n + 1) / 10 = size of [4 (20 + 1) / 10 ] th term = 8.4 th term = size of 8th item + .4 x (size of 9th item – size of 8th item) = 44+ .4 x (45-44) = 44.4 kg

4. P60 (sixty-th percentile) = size of  [ 60 (20 + 1) / 100] th term = 12.6 th term = 12th term + .60 (size of 13th term – size of 12th term) = 51 + .6 (56-51) = 51 + 3 = 54 kg

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Quartiles, Deciles & Percentiles : Discrete Series

Ex.

Weight (Kg.)FrequencyCumulative  frequency
4022
4268
45816
501026
51632
541446
561258
59866
601480
621292
64698 (= N)

Q1 = size of (N+1)/4 th term
= size of (98+1) / 4th term  
= 24.75th term = 50 kg  

Q2 = size of (3N+1)/4 th term
= 3(98+1) / 4th term
= 74.25th term = 60kg  

P60 = size of 60 (N+1) / 4th term
= {60x (98 + 1)} /4 th term
= 59.4th term = 59 kg  
(N is the total frequency = 98)

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Quartiles, Deciles & Percentiles : Continuous Series :

Ex.

Weight (Kg.)FrequencyCumulative  frequency
20-2422
24-2835
28-32510
32-361020
36-40828
40-44634
44-481650
48-521262
52-561072
56-60779
60-64584

Like median, the value of quartiles, deciles and percentiles lie in various class-intervals and the actual values to be calculated by applying interpolation formulae.

Q1 = size of (N/4) th term = size of (\displaystyle \frac{{84}}{4}) th term = size of 21st term, which lies in the class (36 – 40). Now, l1 = 36, l2 = 40, f = 8, q = 21, c = 20

So, Q1 = l1 + {(l2 – l1) / f} ( q-c) = 36 + {(40-36) /8} (21-10) = 36+(4/8) = 36.5 kg

Q3 = size of (3N/4) th term = size of (3X84)/4 th term = size of 63rd, which lies in the class (52 – 56), Now, l1 = 52, l2 = 56, f = 10, q = 63, c = 62

So, Q3 = l1 + {(l2 – l1) / f} ( q-c) = 50+ {(56-52) /10} (63-62) = 52+.4 = 52.4 kg

D4 = size of (\displaystyle \frac{{4N}}{{10}}) th term = size of (4X84) /10th term = 33.6th item. So, D4 lies in the class (40 – 44) , So, D4 = 40+ {(44-40) / 6} (33.6-28) = 40+ (4/6) x 5.6 = 40 + 3.7 = 43.7 kg

P60 = Size of (\displaystyle \frac{{60N}}{{100}}) the term = (60×84) /100 the term = 50.4th term. So, P60 lies in the class (48 – 52), So, P60 = 48 + {(52-48) / 12} (50.4 – 50) = 48 + {(4/12) x (.4)} = 48 + 1.3 = 49.3 kg

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Mode

Mode is the value of the variate which occurs most frequently. Mode represents the most frequent value of a series.

When one speaks of the ‘average salary’, ‘average student’, etc., we often mean the modal salary, the modal student. It we say that the modal Salary obtained by employees in an office are Rs.9000, we mean that the largest number of employees got the similar amount. High & Low Salaries which are not frequent (like Rs.1 lac and as Rs.600) are non-modal.

Calculation of Mode

Mode can be determined from a series of individual observations when it is converted to a discrete series (or continuous series).

-In a discrete series, the value of the variant having the maximum frequency is the mode.

-In continuous series, the class-interval, having the maximum frequency is the modal class. However the exact location of mode is done by interpolation formula like median.

Location of modal value in case of discrete series is possible if there is concentration of items at one point. If again there are two or more values having same maximum frequencies (i.e., more concentrations), it becomes difficult to determine mode. Such items are known as bi-modal, tri-modal or multi-modal according as the items concentrate at 2, 3 or more values.

Mode – Merits and demerits

Merits: Mode can often be located by inspection. Mode is not effected by extreme values. It is often a really typical value. Mode is simple and precise. Mode is an actual item of the series except in a continuous series. Mode can be determined graphically, unlike Mean.

Demerits: Mode is unsuitable for algebraic treatment. When the number of observations is small, the Mode may not exist, while the Mean and Median can be calculated. The value of Mode is not based on each and every item of series. Mode does not lead to the aggregate, if the Mode and the total number of items are given.

Relationship between Mean, Median and Mode

A distribution in which the values of Mean, Median and Mode coincide, is known symmetrical. If these values are not equal, then the distribution is said asymmetrical or skewed.

In a moderately skewed distribution

Mean – Mode = 3 (Mean – Median).

So, if any two values are known, we can find the other.

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Computation of Mode by Individual Observations

The individual observations are to be first converted to discrete series (if possible). The variate having the maximum will be the mode.

Calculate mode from the Mark data : 9, 13, 23, 26, 23, 11, 10, 16.

The following Table is created showing the frequency of each occurrence.  Individual observations are converted into a discrete series

MarksFrequency
91
101
111
131
161
232
261

Here marks 23 occurs maximum number of times, i.e., 2. Hence, the modal marks are 23, or, mode = 23 marks.
Alternatively: Grouping the numbers we get : 9, 10, 11, 13, 16, (23, 23), 26
Now 23 occurs maximum number, i.e., 2. So, mode = 23 marks.  

Multi Modal

When there are two or more values having the same maximum frequency, then mode is ill-defined. Such a series is known as bi-modal or multi-modal as the case may be.

Marks obtained : 22, 12, 18, 15, 18, 12.

Marks Obtained : 22, (12, 12), 15, (18 , 18)

Here 12 occurs 2 times (max.) and 18 occurs 2 times (max). It is bi-modal. Here, Mode is ill-defined.

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Computation of Mode by Continuous Series

By inspections or by preparing Grouping Table and Analysis table, ascertain the modal class. find the exact value of mode :

Mode = l+ [{(f1 – f0) / (2f1 – f0 – f2)} x i],

Where  l = lower class-boundary of modal class, f1 = frequency of modal class, f0 = frequency of the class preceding the modal class, f2 = frequency of the class succeeding the modal class, I =  size of class-interval of modal class.

Compute mode from the following Cumulative Frequency data:

MarksNo. of  ExamineesCumulative Frequency converted into Simple Frequency distribution
above 105910-205
 “        20   5420-308
 “        30   4630-4012
 “        40   3440-5016
‘’       501850-608
 “        60   1060-7010
 “        70   0  

The modal class is (40-50), since the max, frequency is 16. Here, l = 40, f1 = 16, f2 = 8, l = 10, f0 = 12.

Mode = l+ [{(f1 – f0) / (2f1 – f0 – f2)} x l]

Mode = [40+ (16 – 12) / {(32 – 12 – 8) x10}] = [40+ {(\displaystyle \frac{4}{{12}}) x 10}] = 40+ 3.33 = 43.33 marks

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