Explain about DIAGRAMMATIC Representation. Bar Diagrams 1) Simple 'Bar diagram’: - It represents only one variable. For example sales, production, population figures etc. for various years may be shown by simple bar charts. Since these are of the same width and vary only in heights ( or lengths ), it becomes very easy for readers to study the relationship. Simple bar diagrams are very popular in practice. A bar chart can be either vertical or horizontal; vertical bars are more popular. Illustration:- The following table gives the birth rate per thousand of different countries over a certain period of time. Country Birth rate Country Birth rate India Germany U. K. 33 15 20 China New Zealand Sweden 40 30 15 Represent the above data by a suitable diagram. Comparing the size of bars, you can easily see that China's birth rate is the highest while Germany and Sweden equal in the lowest positions. Such diagrams are also known as component bar diagrams. 2) Sub - divided Bar Diagram:- While constructing such a diagram, the various components in each bar should be kept in the same order. A common and helpful arrangement is that of
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Explain about DIAGRAMMATIC Representation.
Bar Diagrams
1) Simple 'Bar diagram’: - It represents only one variable. For example sales, production,
population figures etc. for various years may be shown by simple bar charts. Since these are of the
same width and vary only in heights ( or lengths ), it becomes very easy for readers to study the
relationship. Simple bar diagrams are very popular in practice. A bar chart can be either vertical
or horizontal; vertical bars are more popular.
Illustration:- The following table gives the birth rate per thousand of different countries over a
certain period of time.
Country Birth rate Country Birth rate
India
Germany
U. K.
33
15
20
China
New
Zealand
Sweden
40
30
15
Represent the above data by a suitable diagram.
Comparing the size of bars, you can easily see that China's birth rate is the highest while Germany
and Sweden equal in the lowest positions. Such diagrams are also known as component bar
diagrams.
2) Sub - divided Bar Diagram:- While constructing such a diagram, the various components
in each bar should be kept in the same order. A common and helpful arrangement is that of
presenting each bar in the order of magnitude with the largest component at the bottom and the
smallest at the top. The components are shown with different shades or colors with a proper index.
Illustration:- During 1968 - 71, the number of students in University ' X ' are as follows. Represent
the data by a similar diagram.
Year Arts Science Law Total
1968 -
69
1969 -
70
1970 -
71
20,000
26,000
31,000
10,000
9,000
9,500
5,000
7,000
7,500
35,000
42,000
48,000
3) Multiple Bar Diagram:- This method can be used for data which is made up of two or
more components. In this method the components are shown as separate adjoining bars. The height
of each bar represents the actual value of the component. The components are shown by different
shades or colors. Where changes in actual values of component figures only are required, multiple
bar charts are used.
Illustration:- The table below gives data relating to the exports and imports of a certain country
X ( in thousands of dollars ) during the four years ending in 1930 - 31.
Year Export Import
1927 - 28 319 250
1928 - 29 339 263
1929 - 30 345 258
1930 - 31 308 206
Represent the data by a suitable diagram
4) Deviation Bar Charts:- Deviation bars are used to represent net quantities - excess or
deficit i.e. net profit, net loss, net exports or imports, swings in voting etc. Such bars have both
positive and negative values. Positive values lie above the base line and negative values lie below
it.
Illustration:-
Years Sales Net
profits
1985 - 86
1986 - 87
1987 - 88
10%
14%
12%
50%
-20
-10%
resent the above data by a suitable diagram showing the sales and net profits of
private industrial companies.
Pie Chart:
i) Geometrically it can be seen that the area of a sector of a circle taken radially, is proportional to
the angle at its center. It is therefore sufficient to draw angles at the center, proportional to the
original figures. This will make the areas of the sector proportional to the basic figures.
For example, let the total be 1000 and one of the component be 200, then the angle will be
In general, angle of sector at the center corresponding to a component
.
Tuition fees $ 6000
Books and lab. $ 2000
Clothes / cleaning $ 2000
Room and boarding $ 12000
Transportation $ 3000
Insurance $ 1000
Sundry expenses $ 4000
Total expenditure = $ 30000
Now as explained above, we calculate the angles corresponding to various items (components).
Tuition fees =
Book and lab =
Clothes / cleaning =
Room and boarding =
Transportation =
Insurance =
Sundry expenses =
Uses: - A pie diagram is useful when we want to show relative positions(proportions ) of the
figures which make the total. It is also useful when the components are many in number.
Explain about Graphical Representation with example.
A graph is a visual representation of data by a continuous curve on a squared ( graph ) paper. Like
diagrams, graphs are also attractive, and eye-catching, giving a bird's eye-view of data and
revealing their inner pattern.
Graphs of Frequency Distributions: -
The methods used to represent a grouped data are :-
1.Histogram
2.FrequencyPolygon
3.FrequencyCurve
4.Ogive or Cumulative Frequency Curve
1. Histogram :- It is defined as a pictorial representation of a grouped frequency
distribution by means of adjacent rectangles, whose areas are proportional to the
frequencies.
To construct a Histogram, the class intervals are plotted along the x-axis and corresponding
frequencies are plotted along the y - axis. The rectangles are constructed such that the
height of each rectangle is proportional to the frequency of the that class and width is equal
to the length of the class. If all the classes have equal width, then all the rectangles stand
on the equal width. In case of classes having unequal widths, rectangles too stand on
unequal widths (bases). For open-classes, Histogram is constructed after making certain
assumptions. As the rectangles are adjacent leaving no gaps, the class-intervals become of
the inclusive type, adjustment is necessary for end points only.
For example, in a book sale, you want to determine which books were most popular, the high-
priced books, the low-priced books, books most neglected etc. Let us say you sold a total 31 books
Calculate the arithmetic mean of incomes. Let income be denoted by the symbol X.
Solution.
CALCULATION OF ARITHMETIC MEAN
Monthly Income (Rs.) 1,920
Employee Employee Monthly Income
(Rs.) 1,780 1,760 ,690
1, ,100 ,810
1,050 1,950
EX = 16,650
2
3 4 1750
10 1,840 N= 10
X = EX = 16,650, N= 10
= 16,65 = 1,665. 10
Hence ne average income is Rs. 1,665.
Short-cut method. The arithmetic mean can be calculated by using 1s known as an arbftrary m as an arbitrary orlgin. When deviations are taken from an
Origin, the formula for calculating arithmetic mean is
X=A+ N
metiEuc mean of a sample is designated by the symbol which X is read X- and the arithmetic Sman of a population is designated by the Greek Letter u pronounced as 'mu.
Tmathema is the letter capital sigma of the Greek alphabet and is used in
to denote the sum of the values. ymbol E
NETHOTSR STATSTICAL MET 182
where A is the assumed mean and d is the deviation of iten assumed mean, ie., d = (X - A).
Steps.
items iron
(1) Take an assumed mean.
(2) Take the deviations of items from the assumed mean and .
these deviations by d.
(3) Obtain the sum of these deviations, i.e., Zd.
denor
(4) Apply the formula : X=A+
From Ilustration 1 calculate arithmetic mean by taking 1,800 as t N
assumed mean.
Solution.
CALCULATION OF ARITHMETIC MEAN
(X-1800) -20
Employee Income
1,780 2 1,760 -40
1,690
1,750 3 -110
4 -50
5 1,840 +40
1,920 +120
7 1,100 -700
8 1,810 10
1,050 -750 10 1,950 +150
N= 10 2d =- 1350
= A N
A 1800, 2d= - 1350, N = 10
X = 180o - 1,350 10
= 1,800 135 1,665.
Hence the average income is Rs. 1,665. Note. The reader will find that the calculations here are more than s
we used the formula
We had
N This is true for ungrouped data. But for grouped data consideraC possible by adoping the short-cut method.
saving in
Calculation of Arlthmetic Mean-Discrete Serles
In discrete serles arlthmetic mean may be computed by app (1) Direct method, or
(il) Short-cut method.
ing
Any value whether existing in the data or not can be taken as esU and the final anSwer would be the same. However, the nearer the to the actual mean, ue lesser are the calculations.
Ie asstumed DSUmed e
183 MEASURES OF CENTRAL VALUE
Direct Method
rhe formula for computing mean is
N
where, f= Frequency: X = The variable in question; N* = Total number of observations, ie., f. Steps: ( Multiply the frequency of each row with the variable and obtain the total y.
(ti) Divide the total obtained by step () by the number of observations, Le., total frequency.
lustration 2. From the following data of the marks obtained by 60 students of a class, calculate the arithmetic mean:
Marks No. of Students Marks No. of Students 20 8 50 10 30 12 60 6 40 20 70
Let the marks be denoted by X and the number of students by Solution.
CALCULATION OF ARITHMETIC MEAN
Marks No. of students fX X
8 160 20
12 360 30
0 800 40
10 500 50
360 60
4 280 70
N = 60 fX = 2,460
-22460 60
41
Short-cut Method According to this method.
N Henc the average marks 41.
=A+ 2 A = Assumed mean; d = (X - A}; N = Total number of observations.
e., 2f. where A =
total der should note carefully that n diserete and continuous frequency distributions the
number Der of observatlons, Le. N= the sum ot Irequeincy or N= X
STATSTICAL METH 184
Steps:
(i Take the deviations of the variable X from the assumed
denote the deviations by d.
(i) Multiply these deviations with the respective frequency and tal
total fd
(iw) Divide the total obtained in third step by the total frequencu
(9 Take an assumed mean.
Tmean and e te
lustration 3. Calculate arithmetic mean by the short-cut method using fr distribution of illustration 2.
frequecy
Solution.
CALCULATION OF ARITHMETIC MEAN
fd No. of Students f
(X 40) Marks X
20 8 -20 -160
30 12 -10 -120
40 20 0
50 10 +10 +100
60 6 +20 +120
70 +30 +120
N 60 2fd = 60
60 X= A+= 40 + 40+1 = 41 60
Calculation of Arlthmetic Mean-Continuous Series
a In continuous series, arithmetic mean may be computed by appy of the following methods: (0 Direct method, (0 Short-cut method.
Direct Method When direct method is used
N where m = mid-point of varlous classes; f = the frequeney N the total frequency. ency
of each e
the reader should be famillar with the terms 'groupe erles
ereas
USced
the
data also. Ungrouped dala refer to the Individual observations W*
tta' and
Note. For the sake ol clarity and ease of understanding we naghoul individua observatlons, discrete serles and continuous series
the
However, foupe.
refer to coninuous SCies and the discrete serles.
Mid polnt=Ower limit + Upper limit 2
185 MEASU
OF CENTRAL VALUE
Steps:
Obtain the mid-point of each class and denote it by m. these mid-points by the respective frequency of each class
tMultiply
and obtain the total Yfm.
aDivide the total obtained in step (0 by the sum of the frequency, ie., N.
lustratio ehration 4. From the following data compute arithmetic mean by direct method:
0-10 10-20 20-30 30-40 40-50 50-60 Marks
10 5 10 25 30 20 No. of students
Solution.
CALCULATION OF ARITHMETIC MEAN BY DIRECT METHOD
Marks Mid-point No. of Students tm m
5 25 0-10 5
15 10 150 10-20
25 625 20-30 25
35 30 1,050 30-40
20 900 40-50 45
10 550 50-60 55
N 100 fm = 3,300
Xim 3,300 33. N 100
omput by applying the following formula: Short-cut Method When short-cut method is used, arithmetiç mean is
K=A+ N
assumed mean; d= deviatlons of mid-points from assumed Stee, (m - A); N = total number of observatlons.
Steps: Take an assumed mean.
ig Tom the mid-point of each class deduct the assumed mean. ti Multtply Pro
any the respective frequencles of each class by these devlations and obtain the total 2jd.
Apply the formula: X= A+ fd N Calct
ationarithmetic mean by the short-cut method from the data of
4.
MEASURES
OF CENTRAL VALUE
ulation of Median-Individual Observations
197
Sreps
Arrange the data in ascending or descending order of magnitude. (Both arrangements would give the same answer.)
h In a group composed of an odd number of values such as 7, add l1 to the total number of values and divide by 2. Thus, 7 + 1 would be 8 which divided by 2 gives 4-the number of the value starting
at either end of the numerically arranged groups will be the median value. In a large group the same method may be followed. In a group of 199 items the middle value would be 100th value.
199+ This would be determined by In the form of formula 2
Med. Size of h item.
lustration 11. From the following data of the wages of 7 workers compute the median
wage
1200 1160 1400 Wages (in Rs.) 1100 1150 1080 1120
Solution:
CALCULATION OF MEDIAN
Wages arranged inn ascending order
SI. No. Wages arranged in ascending order
SI. No.
1160 1080
6 1200 1100 7 1400 3 1120
1150
Size ofth item = = 4th tem. of 4th item =
150 S find that median is the middlemost item : 3 persons get a wage less than Rs. We t qual number, i.e., 3, get more than Rs. 1150.
1150. Hence the median wage = Rs. 1150
The Iterocedure for determining the median of an even-numbered group
ues1s not as obvious as above. If there were, for instance, different th a and group, the median is really not determinable since both the
bth values are in the centre. In practice, the median value for a hetposed of an even number of items is estimated by fînding the
YOup arthme he mean of the two middle values-that is, adding the two values ontdle and dividing by two. Expressed in the form of formula, it nOunts to
Median = Size of th item
The bbreviation Med. represents median.
TSTICAL ME 198
Thus we find that it is both when N is odd as well
(one) has to be added to determine median value. as even tha
Illustration 12. Obtain the value of median from the following data:
Geometric mean is specially useful in the following cases:
increase ser sales, production, population or other economic or busines
For example, from 1996 to 1998 prices increased by 5, 10 e .The geometric mean is used to lind the average per cent incre.
per cent respectively. The average annual increase is not 11 Der.. Cent
5+10+ 18- 11| as given by the arithmetic average but 10.9 per 3
cent as obtained'by the geometric mean.
.Geometric mean is theoretically considered t be the best average the construction of index numbers. t satisties the time reversal test
and gives equal weight to equal ratio of change.
This average is most suitable when large weights have to be given t small items and small weights to large items, situations which we usually come across in social and economic fields.
The following examples illustrate the use of geometric mean
n
llustration 33. The price of a commodity increased by 5% from 1995 to 1996, 8% from 1996 to 1997 and 77% from 1997 to 1998. The average increase from 1996 to 1998 is quoted as 26% and not 30%. Explain and verify the result.
Solution. The appropriate average here is the geometric mean and not the aithmete mean. The arithmetic mean of 5, 8, 77 is 30 but this is not the correct answer. Corect answer shall be obtained if we calculate geometric mean.
% Rise2 X Price at the end of the year
taking preceding year as 100
logX
5 105 2.0212 8 108 2.0334
77 177 2.2480
2 log X = 6.3026
G.M. A.LogX=AL = A.L. (2.1009) 126.2
The average increase from 1996 to 1998 = 126.2 100 = 26.2% or approx. 2070. Verification. When the average rise is 30%. Year
Rate of change Total change Price at the end
of each year
130.0 year 30% on 100
30 year 30% on 130
39 169.0 year 30% on 169
When the average rise is 26%: 50.7 219.7
CENTRAL VALUE 233 4SURES OF
N H.M. = In discrete seriesS,
N Ef/m)
N H.M. = In continuous series,
colculation of Harmonic Mean--Individual Observatlons
individual serles harmonic mean is computed by applylng the following
formula:
N H.M. =
X, Xo, Xg. etc. refer to the various items of the variable.
lustration 41. (a) Find the harmonic mean from the following:
In discrete series, harmonic mean is computed by applying tho
formula e folowing
N N H.M. = 1f/0
fx
Steps: Take the reciprocal of the various items of the variable X.
(0 Multiply the reciprocal by frequencies and obtain the total zf
(ii Substitute the values of N and 2fx in the above formula.
Note. Instead of first finding out the reciprocals and then multiplvina them by frequencies it will be far more easier to divide each frequency b the respective value of the variable.
Mlustration 42. From the following data compute the value of harmonic mean
Marks 10 20 25 40 50 No. of students 20 30 50 15 5
For calculating harmonic mean in continuous series the procchere same as applied to discrete serles. The only difference 1s
take the reciprocal of the mid-points.
cedure is the here we
llustration 43. From the following data compute the value of harmonic me Class interval
50-60
10-20 20-30 30-40 40-50
3 Frequency 6 10 7
235 SURES OF CENTRAL VALUE
Solution.
CALCULATION OF HARMONIC MEAN
f/m Mid-points Frequency f
Class interval
m
0.267 15 4 10-20
0.240 25 20-30
0.286 35 10 30-40
0.156 45 40-50
3 0.055 55 50-60
N = 30 Z(fm) = 1.004
H.M. 2(f/m)1.00429.88. N 30 29.88.
Uses of Harmonic Mean'
The harmonic mean is restricted in its field of usefulness. It is useful for computing the average rate of increase in profits of a concern or average speed at which a journey has been performed or the average price at which an article has been sold. The rate usually indicates the relation Detween two different types of measuring units that can be expressed reciprocally. For example if a man walked 20 km. in 5 hours the rate of is walking speed be expressed as
20 km. 4 km. per hour
5 hours
re the unit of the first term is a km. and the unit of the second term san hour. Or reciprocally,
5 km. hours per km. 20 hours
T ne unit of the first term is an hour and the unit of the secondH Term is km. where
VGraeOn 44. An automobile driver travels from plain to hill station 100 km. distance at an speed of
Mustrati Kn. per ur. What is
30 km. per hour. He then makes the return trip at average speed of 20 his average speed over the entire distance (200 km.)?
Solution. an of the problem is given to a layman he is most likely to compute the arithmetic *anl of two speeds, 1.e, 30 km. + 20 Kl = 25 km. p.h. X =
2 But this is not ation. Ha not the correct average. Hamonic mean would be more suitable in this *narmonic mean of 30 and 20 IS
2x 120 10
2 2 24 km. p.h. 1 10 20 30
H.M. =
120
onic mean is a measure of central terndeney lor data expressed as rates Kms. per hour, kms. per itre, hours per semester tonnes per month ete.