Measure of central tendency

Measures Of Central
Tendency
Quantitative Aptitude & Business Statistics

Quantitative aptitude & Business
Statistics: Measures Of Central
2
Statistics in Plural Sense as
Statistical data.
 Statistics in Plural Sense refers to
numerical data of any phenomena
placed in relation to each other.
 For example ,numerical data relating
to population ,production, price
level, national income, crimes,
literacy ,unemployment ,houses etc.,
 Statistical in Singular Scene as
Statistical method.

3
According to Prof.Horace
Secrist:
 “By Statistics we mean aggregate of
facts affected to marked extend by
multiplicity of causes numerically
expressed, enumerated or estimated
according to reasonable standard of
accuracy ,collected in a systematic
manner for a pre determined
purpose and placed in relation to
each other .”

4
Measures of Central Tendency

5
Def:Measures of Central Tendency
 A single expression
representing the whole
group,is selected which may
convey a fairly adequate idea
about the whole group.
 This single expression is
known as average.

6
Averages are central part of
distribution and, therefore ,they
are also called measures of
central tendency.

7
Types of Measures central
tendency:
There are five types ,namely
1.Arithmetic Mean (A.M)
2.Median
3.Mode
4.Geometric Mean (G.M)
5.Harmonic Mean (H.M)

8
Features of a good average
 1.It should be rigidly defined
 2.It should be easy to
understand and easy to
calculate
 3.It should be based on all the
observations of the data

9
 4.It should be easily
subjected to further
mathematical calculations
 5.It should be least affected
by fluctuations of sampling

10
Arithmetic Mean (A.M)
The most commonly used
measure of central tendency.
When people ask about the
“average" of a group of scores,
they usually are referring to
the mean.

11
 The arithmetic mean is
simply dividing the sum of
variables by the total
number of observations.

12
Arithmetic Mean for
raw data is given by
n
x
n
X
n
i
i
xxxx n
∑=++++
== 1......321

13
Find mean for the data
17,16,21,18,13,16,12 and 11

14
Arithmetic Mean for Discrete Series
∑
∑
=
=++++
=
++++
= n
i
i
n
i
ii
n
xfxfxfxf
f
xf
ffff
X nn
1
1
321
......
....
332211

15
Arithmetic Mean for
Continuous Series
C
N
fd
AX ×+=
∑

16
Calculation of Arithmetic mean
in case of Continuous Series
Marks 0-
10
10-
20
20-
30
30-
40
40-
50
50-
60
No. of
Students
10 20 30 50 40 30
From the following data calculate
Arithmetic mean

17
Marks Mid
values
(X)
No.of
Students
(f)
d= X-45
10
f.d
0-10 5 10 -4 -40
10-20 15 20 -3 -60
20-30 25 30 -2 -60
30-40 35 50 -1 -50

18
Marks Mid
values
(X)
No.of
Students
(f)
d= X-45
10
f.d
40-50 45 40 0 0
50-60 55 30 1 30
N=180 ∑fd=-
180

19
Solution
 Let us take assumed
mean =45
 Calculation from
assumed mean
 Mean =
35
180
10*180
45x
=
−
+=×+=
− ∑ C
N
fd
A

20
Calculation Of Arithmetic Mean
in case of Less than series
Marks
less
than /up
to
10 20 30 40 50 60
No. of
students
10 30 60 110 150 180

21
Solution:
Let us first convert Less than series
into continuous series as follows
Marks 0-10 10-
20
20-
30
30-
40
40-
50
50-60
No. of
students
10 20 30 50 40 30
180-
150=30

22
Calculation Of Arithmetic Mean
in case of more than series
Marks
more than
0 10 20 30 40 50 60
No. of
students
180 170 150 120 70 30 0

23
Solution:
Let us first convert More than series
into continuous series as follows
Marks 0-10 10-
20
20-
30
30-
40
40-50 50-
60
No. of
students
10 20 30 50 40 30
180-170=10 170-150=20
70-30=40
30-0=30

24
Calculation of Arithmetic Mean in
case of Inclusive series
 From the following data ,calculate Arithmetic
Mean
Marks 1-10 11-20 21-
30
31-
40
41-
50
51-
60
No. of
Students
10 20 30 50 40 30

25
Solution
 Let us take assumed mean
=45.5
 Calculation from assumed
mean
 Mean =
35
180
10*180
45x
=
−
+=×+=
−
∑ C
N
fd
A

26
Marks Mid
values
No.of
Students
d=X-45.5
10
f.d
0.5-10.5 5.5 10 -4 -40
10.5-20.5 15.5 20 -3 -60
20.5-30.5 25.5 30 -2 -60
30.5-40.5 35.5 50 -1 -50
40.5-50.5 45.5 40 0 0
50.5-60.5 55.5 30 1 30
N=180 ∑fd=
-180

27
Calculation of Arithmetic Mean in
case of continuous exclusive series
when class intervals are unequal
 From the following data ,calculate
Arithmetic Mean
Marks 0-10 10-30 30-40 40-50 50-60
No. of
Students
10 60 50 40 20

28
 Since class intervals are unequal,
frequencies have been adjusted
to make the class intervals equal
on the assumption that they are
equally distributed throughout the
class
 Let us take assumed mean =45

29
 Calculation of Deviations from
assumed mean
 Mean=
778.32
180
10220
45x
=
−
+=×+=
−
∑ X
C
N
fd
A

30
Marks Mid
values
No. of
Students
d= X-45.5
10
f.d
0-10 5 10 -4 -40
10-20 15 30 -3 -90
20-30 25 30 -2 -60
30-40 35 50 -1 -50
40-50 45 40 0 0
50-60 55 20 1 30
N=180 ∑fd=-220

31
Combined Arithmetic Mean
(A.M)
 An average daily wages of 10
workers in a factory ‘A’ is
Rs.30 and an average daily
wages of 20 workers in a
factory B’ is Rs.15.Find the
average daily wages of all the
workers of both the factories.

32
Solution
 Step 1;N1=10 N2=20
 Step2:
 =20
15;30 21 == XX
21
2211
12
NN
XNXN
X
+
+
=

33
Weighted Arithmetic Mean
 The term ‘ weight’ stands for the
relative importance of the different
items of the series. Weighted
Arithmetic Mean refers to the
Arithmetic Mean calculated after
assigning weights to different values
of variable. It is suitable where the
relative importance of different items
of variable is not same

34
 Weighted Arithmetic Mean is
specially useful in problems relating
to
 1)Construction of Index numbers.
 2)Standardised birth and death rates

35
 Weighted Arithmetic Mean is
given by
∑
∑
∑ =
W
XW
X w
.

36
Mathematical Properties of
Arithmetic Mean
 1.The Sum of the deviations of
the items from arithmetic mean
is always Zero. i.e.
 2.The sum of squared
deviations of the items from
arithmetic mean is minimum or
the least
( ) 0=−∑ XX
( ) 0
2
≤−∑ XX

37
 3.The formula of Arithmetic
mean can be extended to
compute the combined
average of two or more
related series

38
 4.If each of the values of a
variable ‘X’ is increased or
decreased by some constant
C, the arithmetic mean also
increased or decreased by C .

39
 Similarly When the value of
the variable ‘X’ are multiplied
by constant say k,arithmetic
mean also multiplied the
same quantity k .

40
 When the values of variable
are divided by a constant say
‘d’ ,the arithmetic mean also
divided by same quantity

41
Merits Of Arithmetic Mean
 1.Its easy to understand and
easy to calculate.
 2.It is based on all the items of
the samples.
 3.It is rigidly defined by a
mathematical formula so that the
same answer is derived by every
one who computes it.

42
 4.It is capable for further
algebraic treatment so
that its utility is enhanced

43
 6.The formula of arithmetic
mean can be extended to
compute the combined
average of two or more
related series.

44
 7.It has sampling stability .It
is least affected by sampling
fluctuations

45
Limitations of Arithmetic Mean
 1.Affected by extreme values
i.e . Very small or very big
values in the data unduly
affect the value of mean
because it is based on all the
items of the series.

46
 2.Mean is not useful for
studying the qualitative
phenomenon.

47
Median
 The middle score of the
distribution when all the scores
have been ranked.
 If there are an even number of
scores, the median is the
average of the two middle
scores.

48
 In an ordered array, the median is
the “middle” number
If n or N is odd, the median is the
middle number
If n or N is even, the median is the
average of the two middle
numbers

49
Potential Problem with Means
Mean
Mean
Median
Median

50
Median
0 1 2 3 4 5 6 7 8 9 10 0 1 2 3 4 5 6 7 8 9 10 12 14
Median = 5 Median = 5

51
Median for raw data
 When given observation are even
 First arrange the items in ascending
order then
 Median (M)=Average of
Item
2
1
2
+
+=
NN

52
 Find the Median for the raw data
 25,55,5,45,15 and 35
 Solution ;Arrange the items
 5,15,25,35,45,55,here N=6
 Median =Average of 3rd and 4th
item=30

53
Median for raw data
 When given observation are odd
 First arrange the items in ascending
order then
 Median (M)=Size of
Item 2
1+
=
N

54
Median for continuous series
c
f
m
N
LM ×












−
+= 2
Where M= Median; L=Lower limit of
the Median Class,m=Cumulative
frequency above median class
f=Frequency of the median class
N=Sum of frequencies

55
Quartiles
 The values of variate that
divides the series or the
series or the distribution into
four equal parts are known as
Quartiles .

56
 The first Quartile (Q1),known
as a lower Quartile is the
value of variate below which
25% of the observations.

57
 The Second Quartile known as
middle Quartile(Q2)known as
middle Quartile or median ,the
value of variates below which
50% of the observations

58
 The Third Quartile known as
Upper Quartile(Q3)known as
middle Quartile or median ,the
value of variates below which 75
% of the observations.

59

th
N
SizeQ
4
1
1
+
= Item
th
N
SizeQ
4
)1(3
3
+
= Item

60
Octiles
distribution into eight equal
parts are known as Octiles .
 Each octile contains 12.5% of
the total number of
observations .

61
 Since seven points are
required to divide the data
into 8 equal parts ,we have
7 octiles.

62

th
Nj
SizeOj
8
)1( +
= Item
th
N
SizeO
8
)1(4
4
+
= Item

63
Deciles
distribution into Ten equal
parts are known as Deciles .
 Each Decile contains 10% of
the total number of
observations .

64
 Since 9 points are required to divide
the data into 10 equal parts ,we
have 9 deciles(D1 to D9)

65

th
Nj
SizeDj
10
)1( +
= Item
th
N
SizeD
10
)1(5
5
+
= Item

66
Percentiles
 The values of variate that divides
the series or the distribution into
hundred equal parts are known as
Percentiles .
 Each percentile contains 10% of
the total number of observations .
 Since 99 points are required to
divide the data into 10 equal parts
,we have 99 deciles(p1 to p99)

67

th
Nj
SizePj
100
)1( +
= Item
th
N
Sizep
100
)1(50
50
+
= Item

68
Relation Ship Between Partition
Values
1.Q1=O2=P25 value of variate which
exactly 25% of the total number of
observations
2.Q2=D5=P50,value of variate which
observations.
3. Q3=O6=P75,value of variate which
observations

69
Calculation of Median in case of
Continuous Series
Marks 0-10 10-20 20-30 30-40 40-50 50-
60
No. of
Students
10 20 30 50 40 30
From the following data
calculate Median

70
Marks No. of
Students
(f)
Cumulative
Frequencies
(c.f.)
0-10 10 10
10-20 20 30
20-30 30 60
30-40 50 110
40-50 40 150
50-60 30 180
N=180

71
 Calculate size of N/2
90
2
180
2
==
N

72
10
50
60
2
180
30 ×












−
+=M
36630 =+=M

73
Merits of Median
 1.Median is not affected by
extreme values .
 2.It is more suitable average
for dealing with qualitative
data ie.where ranks are given.
 3.It can be determined by
graphically.

74
Limitations of Median
1.It is not based all the items of
the series .
2.It is not capable of algebraic
treatment .Its formula can not
be extended to calculate
combined median of two or
more related groups.

75
0
X
Y
M
Less than
Cumulative
curve
More than
Cumulative Curve
Median By Graph
Q3Q1 CI
Frequency
N/2
3N/4
N/4

76
Mode
 A measure of central tendency
 Value that occurs most often
 Not affected by extreme values
 Used for either numerical or
categorical data
 There may be no mode or several
modes
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Mode = 9
0 1 2 3 4 5 6
No Mode

77
Mode
 The most frequent score in the
distribution.
 A distribution where a single
score is most frequent has one
mode and is called unimodal.

78
 A distribution that consists
of only one of each score has
n modes.
 When there are ties for the
most frequent score, the
distribution is bimodal if two
scores tie or multimodal if
more than two scores tie.

79
 Calculate the mode from the following
data of marks obtained by 10 students.
 20,30,31,32,25,25,30,31,30,32
 Mode (Z)=30

80
Mode for Continuous Series
c
fff
ff
LZ ×





−−
−
+=
201
01
2
Where Z= Mode ;L=Lower limit of the Mode Class
f0 =frequency of the pre modal class
f1=frequency of the modal class
f2=frequency of the post modal class
C=Class interval of Modal Class

81
Calculation of Mode :Continuous
Series
Marks 0-
10
10-
20
20-
30
30-
40
40-
50
50-
60
No. of
Students
10 20 30 50 40 30
From the following data calculate
Mode

82
Marks No. of
Students
(f)
0-10 10
10-20 20
20-30 30
30-40 50 f1
40-50 40
50-60 30
N=180
f0
f2

83
667.36667.630
10
4030502
6050
30
2 201
01
=+=
×





−−×
−
+=
×





−−
−
+=
Z
c
fff
ff
LZ

84
x0
Y
Z
10 20 30 40 50 60
10
20
30
40
50
Calculation Mode Graphically

85
Relationship between Mean,
Median and Mode
 The distance between Mean
and Median is about one
third of distance between the
mean and the mode.

86
Karl Pearson has expressed the
relationship as follows.
Mean –Mode=(Mean-Median)/3
Mean-Median=3(Mean-Mode)
Mode =3Median-2Mean
Mean=(3Median-Mode)/2

87
Example
 For a moderately skewed
distribution of marks in statistics for
a group of 200 students ,the mean
mark and median mark were found
to be 55.60 and 52.40.what is the
modal mark?

88
Solution
 Since in this case mean=55.60and
median =52.40 applying ,we get
 Mode=3median -2Mean
 =3(52.40)-2(55.60)
 Mode =46

89
Example
 If Y=2+1.50X and mode of X is 15 ,What
is mode of Y
 Solution
 Y m=2+1.50*15=24.50

90
Merits of Mode
 1.Mode is the only suitable
average e.g. ,modal size of
garments, shoes.,etc
 2.It is not affected by extreme
values.
 3.Its value can be determined
graphically.

91
Limitations of Mode
 1.In case of bimodal /multi
modal series ,mode cannot be
determined.
 2.It is not capable for further
algebraic treatment, combined
mode of two or more series
cannot be determined.

92
 3.It is not based on all the items
of the series
 4.Its value is significantly
affected by the size of the class
intervals

93
Geometric mean
nn
i
i
n
niG
x
xxxxx
/1
1
21






=
=
∏=


94
 Take the logarithms of each item of
variable and obtain their total i.e ∑ log
X
 Calculate G M as follows








=
∑
n
X
AntiMG
log
log.

95
Computation of G.M -Discrete
Series
 Take the logarithms of each item of
variable and multiply with the
respective frequencies obtain their
total
i.e ∑ f .log X
 Calculate G M as follows








=
∑
N
Xf
AntiMG
log.
log.

96
Merits of Geometric Mean
 1.It is based on all items of
the series .
 2 It is rigidly defined
 3.It is capable for algebraic
treatment.

97
 4.It is useful for averaging
ratios and percentages rates
are increase or decrease

98
Limitations of Geometric
Mean
 1.Its difficult to understand
and calculate.
 2.It cannot be computed
when there are both negative
and positive values in a
series

99
 3.It is biased for small values
as it gives more weight to
small values .

100
Calculation of G.M
:Individual Series
 From the following data
calculate Geometric Mean
Roll No 1 2 3 4 5 6
Marks 5 15 25 35 45 55

101
Computation of G.M :Individual
Series
X log X
5 0.6990
15 1.1761
25 1.3979
35 1.5441
45 1.6532
55 1.7404
∑log X=8.2107

102
36.23
)3685.1log(
6
2107.8
log
log.
=
=






=








=
∑
Anti
Al
n
X
AntiMG

103
 Find the average rate of increase
population which in the first decade
has increased by 10% ,in the second
decade by 20% and third by 30%

104
Decade % rise Population at
the end of the
decade
logx
1
2
3
10
20
30
110
120
130
2.0414
2.0792
2.1139
∑log
X=6.2345

105
8.119
)0782.2log(
2345.6
log
log.
=
=






=








= ∑
Anti
Al
n
X
AntiMG
Average Rate of increase in Population
is 19.8%

106
Weighted Geometric Mean








=
∑
∑
w
Xw
AntiMG
log.
log.

107
Harmonic Mean (H.M)
 Harmonic Mean of various items of a
series is the reciprocal of the
arithmetic mean of their reciprocal
.Symbolically,
nXXXX
N
MH
1
.......
111
.
321
++++
=

108
 Where X1,X2,X3…….X n refer to the
value of various series.
 N= total no. of series

109
Merits of Harmonic Mean
 1.It is based on all items of
the series .
 2 It is rigidly defined
 3.It is capable for algebraic
treatment.

110
 4.It is useful for averaging
measuring the time ,Speed
etc

111
Limitations of Harmonic Mean
 1.Its difficult to understand
and calculate.
 2.It cannot be computed
when one or more items are
zero

112
 3.It gives more weight to
smallest values . Hence it is
not suitable for analyzing
economic data .

113
Calculation of H.M :Individual
Series
 From the following data
calculate Harmonic Mean
Roll
No
1 2 3 4 5 6
Mark
s
5 15 25 35 45 55

114
Computation of H.M :Individual
Series
X l/x
5 0.2000
15 0.0666
25 0.0400
35 0.0286
45 0.0222
55 0.0182
∑(1/x)=0.3756

115
9744.15
3576.0
6
1
1
=
=
=
∑ =
n
i
i
H
x
n
x

116
 Compute AM ,GM and HM for the
numbers 6,8,12,36
 AM=(6+81+12++36)/4=15.50
 GM=(6.8.12.36)1/4=12
 H.M=9.93
36
1
12
1
8
1
6
1
4
.
+++
=MH

117
Weighted Harmonic Mean
∑
∑=
)(
i
i
i
X
w
w
HM

118
 Find the weighted AM and HM of first n natural
numbers ,the weights being equal to the
squares of the Corresponding numbers.
X 1 2 3 …n
W 12 22 32 ..n2

119
 Weighted
∑
∑=
Wi
XiWi
AM
.
)12(2
)1(3
+
+
=
n
nn





 ++





 +
=
++++
++++
6
)12)(1(
4
)1(
.....321
.....321
22
2222
3333
nnn
nn
n
n

120
∑
∑=
)(
i
i
i
X
w
w
HM
3
12
2
)1(
6
)12)(1(
.....321
.....321
23222
+
=



 +





 ++
=
++++
++++
n
nn
nnn
n
n

121
 The AM and GM of two observations are 5
and 4 respectively ,Find the two
observations.
 Solution : Let the Two numbers are a and
b given
 ( a+b)/2=10 ;a + b=10
 GM=4 ab=16
 (a-b)2=(a+b)2-4ab=100-64=36
 a-b=6 a=8 and b=2

122
 The relationship between AM ,GM
and HM
 G2=A.H

123
 1.The empirical relationship among
mean, median and mode is ______
 (a) mode=2median–3mean
 (b) mode=3median-2mean
 (c) mode=3mean-2median
 (d) mode=2mean-3median

124
 1. The empirical relationship among
mean, median and mode is ______
 (a) mode=2median–3mean
 (b) mode=3median-2mean
 (c) mode=3mean-2median
 (d) mode=2mean-3median

125
2. In a asymmetrical
distribution ____
 (a) AM = GM = HM
 (b) AM<GM<AM
 (c) AM<GM>HM
 (d) HMGMAM ≠≠

126
2. In a asymmetrical
distribution ____
(a) AM = GM = HM
(b) AM<GM<AM
(c) AM<GM>HM
(d) HMGMAM ≠≠

127
 3. The points of intersection of
the “less than and more than”
ogive corresponds to ___
 (a) mean
 (b) mode
 (c) median
 (d) all of above

128
 .3.The points of intersection of
the “less than and more than”
ogive corresponds to ___
 (a) mean
 (b) mode
 (c) median

129
•4. Pooled mean is also called
 (a) mean
 (b) geometric mean
 (c) grouped mean
 (d) none of these

130
 4. Pooled mean is also called
 (a) mean
 (b) geometric mean
 (c) grouped mean

131
 5. Relation between mean,
median and mode is
 (a)mean–mode=2(mean-median)
 (b)mean–median=3(mean–mode)
 (c) mean–median=2(mean–
mode
 (d)mean–mode=3(mean–median)

132
 5. Relation between mean, median
and mode is
 (a)mean–mode=2(mean-median)
 (b)mean–median=3(mean–mode)
 (c) mean–median=2(mean–
mode
 (d)mean–mode=3(mean–median)

133
 6. The geometric mean of 9, 81, 729
is _____
 (a) 9
 (b) 27
 (c) 81

134
 6. The geometric mean of 9, 81,
729 is _____
 (a) 9
 (b) 27
 (c) 81

135
 7. The mean of the data set of 1000
items is 5. From each item 3 is
subtracted and then each number is
multiplied by 2. The new mean will be
_____
 (a) 4
 (b) 5
 (c) 6
 (d) 7

136
 7. The mean of the data set of 1000
items is 5. From each item 3 is
subtracted and then each number is
multiplied by 2. The new mean will be
 (a) 4
 (b) 5
 (c) 6
 (d) 7

137
 8. If each item is reduced by
15, AM is ____
 (a) reduced by 15
 (b) increased by 15
 (c) reduced by 10

138
 8. If each item is reduced by
15, AM is ____
 (a) reduced by 15
 (b) increased by 15
 (c) reduced by 10

139
 9. In a series of values if one value is
0 ____
 (a)both GM and HM are zero
 (b)both GM and HM are intermediate
 (c) GM is intermediate and HM is zero
(d)GM is zero and HM is intermediate

140
 9. In a series of values if one value is
0 ____
 (a) both GM and HM are zero
 (b)both GM and HM are intermediate
 (c) GM is intermediate and HM is zero
 (d)GM is zero and HM is intermediate

141
 10.Histogram is useful to determine
graphically the value of
 (a) Mean
 (b) Mode
 (c) Median

142
 10.Histogram is useful to
determine graphically the value of
 (a) Mean
 (b) Mode
 (c) Median

143
 11.The positional measure of
central Tendency
 (a) Arithmetic Mean
 (b) Geometric Mean
 (c) Harmonic Mean
 (d) Median

144
 11.The positional measure of
central Tendency
 (a) Arithmetic Mean
 (b) Geometric Mean
 (c) Harmonic Mean
 (d) Median

145
 12.The average has relevance for
 (a) Homogeneous population
 (b) Heterogeneous population
 (c) Both
 (d) none


146
 12.The average has relevance for
 (a) Homogeneous population
 (b) Heterogeneous population
 (c) Both
 (d) none


147
 13.The sum of individual
observations is Zero When taken
from
 (a) Mean
 (b) Mode
 (C) Median
 (d) All the above


148
 13.The sum of individual
observations is Zero When taken
from
 (a) Mean
 (b) Mode
 (C) Median
 (d) All the above

149
 14.The sum of absolute
deviations from median is
 (a) Minimum
 (b) Zero
 (C) Maximum
 (d) A negative figure

150
 14.The sum of absolute
deviations from median is
 (a) Minimum
 (b) Zero
 (C) Maximum
 (d) A negative figure

151
 15.The mean of first natural
numbers
(a)n/2
(b)n-1/2
(c)(n+1)/2
(d) none

152
 15.The mean of first natural
numbers
(a)n/2
(b)n-1/2
(c)(n+1)/2
(d) none

153
 16.The calculation of Speed
and velocity
(a) G.M
(b) A.M
(c) H.M
(d) none is used

154
 16.The calculation of Speed and
velocity
(a)G.M
(b)A.M
(c)H.M
(d)none is used

155
 17. The class having maximum
frequency is called
 A) Modal class
 B) Median class
 C) Mean Class
 D) None of these

156
 17. The class having maximum
frequency is called
 A) Modal class
 B) Median class
 C) Mean Class
 D) None of these

157
 18. The mode of the numbers 7, 7, 9,
7, 10, 15, 15, 15, 10 is
 A) 7
 B) 10
 C) 15
 D) 7 and 15

158
 18. The mode of the numbers 7, 7, 9,
7, 10, 15, 15, 15, 10 is
 A) 7
 B) 10
 C) 15
 D) 7 and 15

159
 19. Which of the following measures
of central tendency is based on only
50% of the central values?
 A) Mean
 B) Mode
 C) Median
 D) Both (a) and (b)

160
 19. Which of the following measures
of central tendency is based on only
50% of the central values?
 A) Mean
 B) Mode
 C) Median
 D) Both (a) and (b)

161
 20. What is the value of the first
quartile for observations 15, 18, 10,
20, 23, 28, 12, 16?
 A) 17
 B) 16
 C) 15.75
 D) 12

162
 20. What is the value of the first
quartile for observations 15, 18, 10,
20, 23, 28, 12, 16?
 A) 17
 B) 16
 C) 15.75
 D) 12

163
 21. The third decile for the numbers
15, 10, 20, 25, 18, 11, 9, 12 is
 A) 13
 B) 10.70
 C) 11
 D) 11.50

164
 21. The third decile for the numbers
15, 10, 20, 25, 18, 11, 9, 12 is
 A) 13
 B) 10.70
 C) 11
 D) 11.50

165
 22. In case of an even number of
observations which of the following is
median?
 A) Any of the two middle-most value..
 B) The simple average of these two
middle values
 C) The weighted average of these two
middle values.
 D) Any of these

166
 22. In case of an even number of observations
which of the following is median?
 A) Any of the two middle-most value..
 B) The simple average of these two middle
values
 C) The weighted average of these two middle
values.
 D) Any of these

167
 23. A variable is known to be
_______ if it can assume any value
from a given interval.
 A) Discrete
 B) Continuous
 C) Attribute
 D) Characteristic

168
 23. A variable is known to be
_______ if it can assume any value
from a given interval.
 A) Discrete
 B) Continuous
 C) Attribute
 D) Characteristic

169
 24. Ogive is used to obtain.
 A) Mean
 B) Mode
 C) Quartiles
 D) All of these

170
 24. Ogive is used to obtain.
 A) Mean
 B) Mode
 C) Quartiles
 D) All of these

171
 25. The presence of extreme
observations does not affect
 A) A.M.
 B) Median
 C) Mode
 D) Any of these

172
 25. The presence of extreme
observations does not affect
 A) A.M.
 B) Median
 C) Mode
 D) Any of these

THE END
Measures Of Central Tendency

Measure of central tendency

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