measures of central tendency - icai knowledge …. measures of central tendency. the tendency of a...
TRANSCRIPT
Measures of Central Tendency
CPT Section D Quantitative Aptitude Chapter 11 Dr. R. B. Tiwari
Introduction
Central Behavior of the data.
Off Central Behavior of the data.
1. Measures of Central Tendency.
The tendency of a given set of observations to cluster around a single central or middle value and the single value that represents the given set of observations is described as a measure of central tendency or location or average.
e.g. a company is recognized by its high average profit and an educational institution is judged on the basis of the average marks obtained by the students.
Arithmetic Mean
Arithmetic Mean
Case (i) Arithmetic Mean (AM)
1 2 ...(i) nX X X
Xn
Xor X
n
+ + +=
=∑
Case (ii) For Discrete Series
i.e. For x1-f1, x2-f2,…xn-fn type data
If values of X are large enough
(ii)
1, 2,3....
.
ii
fdX A X C
NX A
where d i nC
A Assumed Mean
or C Common divisor for ungrouped data
= +
−= ∀ =
=
=
∑
Case (iii) For Continuous Series
i. e. If frequencies are given to corresponding class intervals
Example 1. Following are the daily wages in rupees of a sample of 9 workers: 58,62,48,53,70,52,60,84,75.Compute the mean wage.
Solution: Given that no. of workers, n = 9 ,and
58 62 48 53 70 52 60 84 75 562
562 62.449
Sum of the wages X
XTherefore the mean wage is givenby AM X
n
= + + + + + + + + =
= = = =
∑∑
Solution:
Example 2: Find the AM for the following data: X : 10 20 30 40 50 f : 2 5 6 3 4
Example 3.
Compute the mean weight of a group of BBA students of St. Xavier’s College from the following data:
• Weights in Kgs: 44-48 49-53 54-58 59-63 64-68 69-73
• No. of students. 3 4 5 7 9 8
Solution Weights in Kgs. No. of
Students (f)
Mid –Values (X) fX
44-48 3 46 138
49-53 4 51 204
54-58 5 56 280
59-63 7 61 427
64-68 9 66 594
69-73 8 71 568
Totals N=36 2211
Answer
2211 61.4236
fXX kgs
N= = =∑
Example 4.
Find the AM for the following frequency distribution.
• Class Intervals: 350-369 370-389 390-409 410-429 430-449 450-469 470-489
• Frequency : 23 38 58 82 65 31 11
Class intervals
Frequency (f)
Mid-Values
(X)
fd
350-369 23 359.50 -3 -69
370-389 38 379.50 -2 -76
390-409 58 399.50 -1 -58
410-429 82 419.50=A
0 0
430-449 65 439.50 1 65
450-469 31 459.50 2 62
470-489 11 479.50 3 33
Totals N=308 -43
419.5020
i ii
X A Xd
C− −
= =
Solution
Solution
( )43419.50 20
308416.71
fdX A X C
N
X
= +
−= +
=
∑
Note that here A is taken as guessed mid – value and C is a common divisor. Then AM is given as
Properties of AM
If all observations are equal to a constant, k say, Then AM is also equal to k.
The algebraic sum of deviations taken about AM is always zero. i.e. e.g. the AM of 3,5 and 7 is 5 and (3-5)+(5-5)+(7-5)=0
( )( )
0 .
0 .i
i i
X X for ungrouped data and
f X X for grouped frequancy distribution
− =
− =
∑∑
Properties of AM
AM is affected by change of origin and/or scale. i.e. if
.
fdX Ad and dC N
then X A C d
−= =
= +
∑
Properties of AM
If there are two groups having n1 and n2 observations and
as their respective AMs , then the AM of combined group is given as
1 2X and X
1 1 2 2
1 2
n X n XX
n n+
=+
Properties of AM
Example 5.
The mean salary for a group of 40 female workers is Rs. 5200 per month and that for a group of 60 male workers is Rs. 6800 per month. What is the combined salary?
Solution.
1 1
2 2
1 1 2 2
1 2
40, 520060, 6800
40 5200 60 680040 60
.6160
n Xn XThen combined mean salary is givenby
n X n XX
n nX X
Rs
= =
= =
+=
++
=+
=
Example 6.
Find the AM of the following data.
• Marks: less than 10 <20 <30 <40 <50
No. of students: 5 13 23 27 30
Solution
Marks No. of students
(f)
Mid-value (X)
fx
0-10 5 5 25
10-20 13-5=8 15 120
20-30 23-13=10 25 250
30-40 27-23=4 35 140
40-50 30-27=3 45 135
N=30 670
Solution: AM
670 22.3330
fXX
N= = =∑
Example.7
Find the missing value Y, if the following frequency distribution has AM as 5.
• X: 3 5 Y 6 • f: 1 3 2 4
Solution. AM is given by
1 3 3 5 2 6 4 42 21 3 2 4 10
42 2. . 510
50 42 24
fX Y YXN
Yi e
or Ytherefore Y
× + × + + × += = =
+ + ++
=
− ==
∑
MCQ’s on Arithmetic Mean Question Time
Question 1.
a) 2.5
b) 3.5
c) 4.5
d) 5.5
Answer: (b) 3.5
AM of 1,2,3,4,5,6 is
Question 2.
(a) 3
(b) 5
(c) 7
(d) 9
Answer: (b) 5
AM of 5,5,5,5,5, is
Question.3.
(a) 0
(b) 5
(c) -5
(d) None of these
Answer: (a) 0
The algebraic sum of deviations taken about the AM of 15, 20 and 25 is
Question 4.
(a) 11/3
(b) 5
(c) 4
(d) 4.5
Answer: (a) 11/3
If a variable assumes the values 1,2,3,4,5 with frequencies as 1,2,3,4,5 respectively, then AM is
(a) 40%
(b)50%
(c) 60%
(d) None of these
Answer: (a) 40%
The average salary of a group of unskilled workers is Rs. 10000 and that of a group of skilled workers is Rs. 15000.If
the combined salary is Rs. 12000, then what is the percentage of skilled workers?
Question.5
Question 6 Find the AM for the following frequency distribution
Marks : 5-14 15-24 25-34 35-44 45-54 55-64 No. Of students: 10 18 32 26 14 10
(a) 30
(b) 29
(c) 33.68
(d) 34.21
Answer: (c) 33.68
Question7
Find the AM of the following data.
• Marks: less than 20 <40 <60 <80 <100
• No. of students: 5 17 28 42 50
Question 7
(a) 51.2
(b) 52.2
(c) 53.2
(d) 54.2
Answer: (c) 53.2
Question 7
(a) 51.2
(b) 52.2
(c) 53.2
(d) 54.2
Answer: (c) 53.2
Question.8
Find the missing value Y, if the following frequency distribution has AM as 5.
• X: 6 4 Y 2 • f: 4 3 2 1
Question 8
(a) 5
(b) 6
(c) 7
(d) 8
Answer: (b) 6
Question.9
Find the AM of the following frequency distribution.
• X: 0-10 10-20 20-30 30-40 40-50 • f: 3 5 7 4 1
Question 9
(a) 20.5
(b) 21.5
(c) 22.5
(d) 23.5
Answer: (c) 22.5
Question.10
Find the AM of the following frequency distribution.
• X: 0-50 50-100 100-150 150-200 200-250 • f: 15 25 30 20 10
Question 10
(a) 115.5
(b) 116.5
(c) 117.5
(d) 118.5
Answer: (c) 117.5
Question.11
If the AM of the following frequency distribution is 14.7 for age in years of 20
individuals. Find the missing values. • X: 0-6 6-12 12-18 18-24 24-30 • f: 3 5 - - 4
Question 11
(a) 4, 4
(b) 6, 2
(c) 5, 3
(d) 7, 1
Answer: (b) 6, 2
Question.12 If the AM of the series 1,2,3,…n is 6. Find the value of n.
Question 12
(a) 7
(b) 9
(c) 11
(d) 12
Answer: (c) 11
Median
Median (Md) and other Partition Values
Median is a positional average, which divides the data into two equal parts.
If Y= a +b X then Md(Y) = a + b Md(X)
Properties of Median
The sum of absolute deviations is minimum when it is measured about Md. i.e.
minimum if AX A is Md− =∑
Properties of Median
Determination of Median Case (i): When number of observations is odd, i.e. n is odd.
First arrange the observations in ascending or descending order . If no. of observations is n which is an odd number, then
12
nThen Md th term+ =
Example 1 : Find median of the following data 40, 29, 15, 58, 70
1 5 1 32 2
. . 40
rdnMd th term term
i e Md
+ + = = =
=
Solution: Let us arrange the observations in ascending order as 15, 29, 40, 58, 70 Here n= 5,Therefore
Case(ii)
1 ( ) ( 1)2 2 2
n nThen Md th term th term = + +
When number of observations is even, i.e. n is even. First arrange the observations in ascending or descending order.
Example 2. Find the Median of the following observations. 5, 25, 11, 14, 26, 30.
( )
1 ( ) ( 1)2 2 21 6 6 1( ) ( 1) 14 25 19.52 2 2 2
n nMd th term th term
Then Md th term th term
= + + = + + = + =
Solution: Let us arrange these observations into ascending order as 5, 11, 14, 25, 26, 30. Here n = 6 i.e. n is even, hence
Case (iii)
For x1-f1, x2-f2,…xn-fn type data, Form a column of less than type cumulative
frequencies.
Then Md = Value of the variable X corresponding to the cumulative frequency
just greater than N/2
Example 3.
Find the median of the following frequency distribution.
• X: 60 65 70 75 80 85 90 • f : 7 12 20 22 18 14 9
Solution:
X f Less than type c.f.
60 7 7
65 12 19
70 20 39
75 22 61
80 18 79
85 14 83
90 9 92
Total N=92
Solution
Here the c.f. just greater than N/2 i.e. 46, is 61. Therefore the corresponding value of X is Median, i.e. Md = 75
Case (iv)
For grouped data i.e. if frequencies are given corresponding to the class
intervals.
First find the Md class corresponding to the cumulative
frequency just greater than N/2.
2
limit
(besure that theclass intervals are , . . )
N cThen Md l h
fwhere l lower of the Md class
f frequency of the Md classc cumulative frequency preceeding the Md classh class width exclussivetype i e continuous classes
− = + ×
====
Median Formula for Continuous Series
Example. 4.
Find the Median of the following data.
• Marks: 0- 10 10-20 20-30 30-40 40-50 • No. of students: 5 8 10 7 6
Solution Marks
Class Intervals No. Of students
(f) c.f.
0-10 5 5
10-20 8 13=c
20-30 10=f 23
30-40 7 30
40-50 6 36
Totals N=36
Solution
2
36 132. . 20 10 25
10
N cThen Md l h
f
i e Md
− = + ×
− = + × =
The c.f. just greater than N/2 i.e. 18, is 36, which is the Md class. Hence, Note that if class intervals are not continuous then they are first made continuous by adding d/2 to the all upper limits and subtracting d/2 from all the lower limits, where d is the difference between lower limit of any class and upper limit of the previous class.
Quartiles There are three quartiles viz. Q1, Q2 and Q3,which divide the data into four equal parts.
Case (i) For x1-f1, x2-f2,…xn-fn type data
First form a column of cumulative frequencies then the value of the variable corresponding to the c.f. just greater than
iN/4 gives the ith quartile.
Then ith (i=1,2,3) Quartile (Qi) is given by
4
limit
(besure that theclass intervals are , . . )
i
i
i
i
iN cThen Q l h
fwhere l lower of the Q class
f frequency of the Q classc cumulative frequency preceeding the Q classh class width exclussivetype i e continuous classes
− = + ×
====
Quartile
Example 1 Find the third quartile for the following
frequency distribution.
X: 10 20 30 40 50 60 70 80 90 100 f: 3 5 10 20 30 40 45 35 7 5
Solution X f c.f. 10 3 3 20 5 8 30 10 18 40 20 38 50 30 68 60 40 108 70 45 153 80 35 188 90 7 195
100 5 200 Total N= 200
Here to calculate third quartile we see the value of X
corresponding to the
cummulative frequency just greater than
3N/4=3x200/4=150, which is 153. Hence third quartile
is given by
Q3 = 70
Solution
Case (ii)
If frequencies are given to corresponding class intervals.
Then ith (i=1,2,3) Quartile (Qi) is given by
4
limit
(besure that theclass intervals are , . . )
i
i
i
i
iN cThen Q l h
fwhere l lower of the Q class
f frequency of the Q classc cumulative frequency preceeding the Q classh class width exclussivetype i e continuous classes
− = + ×
====
Quartile
Example 2
• Marks: 0- 10 10-20 20-30 30-40 40-50 50-60 • No. of students: 5 8 10 7 6 4
Find the First and Third Quartiles for the following frequency distribution.
Solution Marks
Class Intervals No. Of students
(f) c.f.
0-10 5 5
10-20 8 13
20-30 10 23
30-40 7 30
40-50 6 36
50-60 4 40
Totals N=40
1
1
4
40 54. . 10 10 16.25
8
N cThen Q l h
f
i e Q
− = + ×
− = + × =
And
3
3
34
3 40 304. . 40 10 40
6
N cThen Q l h
f
i e Q
− = + ×
× − = + × =
Solution
Deciles
There are nine deciles viz. D1, D2,…D9, which divide
the data into 10 equal parts.
Case (i) For x1-f1, x2-f2,…xn-fn type data, Form a column of less than type cumulative frequencies.
Then ith decile is the value of the variable X corresponding to cumulative frequency just
greater than iN/10.
Example 1
Find the 5th decile for the following frequency distribution.
X: 10 20 30 40 50 60 70 80 90 100 f: 3 5 10 20 30 40 45 35 7 5
Solution X f c.f. 10 3 3 20 5 8 30 10 18 40 20 38 50 30 68 60 40 108 70 45 153 80 35 188 90 7 195
100 5 200 Total N= 200
Solution
Here we have formed the column of cumulative frequencies, now see the
c.f. just greater than 5N/10=5x200/10=100, which is 108,
hence the corresponding value of X is fifth decile, i.e.
D5 = 60
Case (ii) For grouped data, i.e. frequencies against class intervals are given.
Then ith decile class is obtained corresponding to
the cumulative frequency just greater than iN/10.
Then ith (i-1,2,…9) decile (Di) is given by
10
limit
(besure that theclass intervals are , . . )
i
i
i
i
iN cD l h
fwhere l lower of the D class
f frequency of the D classc cumulative frequency preceeding the D classh class width exclussivetype i e continuous classes
− = + ×
====
Example 2
Find the seventh decile for the following frequency distribution.
Marks : 0- 10 10-20 20-30 30-40 40-50 50-60 No. of students: 5 8 10 7 6 4
Solution
Marks Class
Intervals
No. of students (f)
c.f.
0-10 5 5
10-20 8 13
20-30 10 23
30-40 7 30
40-50 6 36
50-60 4 40
Totals N=40
First we find the seventh decile class
corresponding to the c.f. just greater than 7N/10=28, i.e. 30. Hence
7
7
7
710
7 40 2310. . 30 10
7i. . 30 7.14 37.14
N cThen D l h
f
i e D
e D
− = + ×
× − = + ×
= + =
Percentiles:
Percentiles divide the whole data into 100 equal
parts. They are 99 in number and denoted by
P1,P2,…P99
Case (i) For x1-f1, x2-f2,…xn-fn type data.
First, form a column of less than type cumulative frequencies. Identify the ith
percentile class for which the cumulative frequency is just greater than iN/100.
Example 1
Find the 65th percentile for the following frequency distribution.
X: 10 20 30 40 50 60 70 80 90 100 f: 3 5 10 20 30 40 45 35 7 5
Solution X f c.f. 10 3 3 20 5 8 30 10 18 40 20 38 50 30 68 60 40 108 70 45 153 80 35 188 90 7 195
100 5 200 Total N= 200
The 65th percentile is the value of X for which corresponding cumulative frequency is just greater
than 65N/100=65x200/100=130, which is 153 .
Hence P65 = 70
The 65th Percentile
Case (ii) For grouped data, i.e. frequencies are given corresponding to class intervals.
First obtain the ith percentile class which is corresponding to the cumulative frequency just
greater than iN/100.
Then ith (i=1,2,…99) percentile (Pi) is given by
100
limit
(besure that theclass intervals are , . . )
i
i
i
i
iN cP l h
fwhere l lower of the P class
f frequency of the P classc cumulative frequency preceeding the P classh class width exclussivetype i e continuous classes
− = + ×
=
=
=
=
Percentile
Example 2.
• Marks: 0- 10 10-20 20-30 30-40 40-50 50-60 • No. of students: 5 8 10 7 6 4
Find the 52nd percentile for the following frequency distribution.
Solution Marks
Class Intervals No. Of students
(f) c.f.
0-10 5 5
10-20 8 13
20-30 10 23
30-40 7 30
40-50 6 36
50-60 4 40
Totals N=40
The 52nd percentile
52
52
52
52100
52 40 1310020 10
1020 7.8 27.8
N cP l h
f
or P
P
× − = + ×
× − = + ×
= + =
First we obtain the 52nd Percentile class which is corresponding to the c.f. just greater than 52N/100 i.e. 52x40/100=20.8, which is 23, shown as red.
Partition Values for Ungrouped Data:
For unclassified data the pth Quartile is given by (n+1)pth value.
Take p=1/4 for first quartile p=2/4 for second quartile
p=3/4 for third quartile
Example
Find Q3, for the wages of labourers: • Rs. 82, 56, 90, 50, 120, 75,
79, 80, 130, and 65.
Solution
( )
( )
33143i. . 10 1 8.254
. . 8 0.25(9 value 8 value)i.e. 90 0.25(120 90)i.e. 90.5
th
th th th
Q n th value
e th value value
i e value
= + ×
= + × =
= + −= + −=
Arrange the wages in ascending order as : 50, 56, 65, 75, 79, 80, 82, 90, 120, 130
Partition Values for Ungrouped Data:
For unclassified data the pth Decile is given by (n+1)pth value.
Take p=1/10 for first decile, p=2/10 for second decile, . . . p=9/10 for ninth decile.
Example
Find D7, for the wages of labourers:
• Rs. 82, 56, 90, 50, 120, 75, 79, 80, 130, and 65.
Solution
( )
( )
7
7
7110
7i. . 10 1 7.710
. . 7 0.7 (8 value 7 value)i.e. 82 0.7(90 82)i.e. D 87.6
th
th th th
D n th value
e th value value
i e value
= + ×
= + × =
= + −= + −
=
Arrange the wages in ascending order as : 50, 56, 65, 75, 79, 80, 82, 90, 120, 130
Partition Values for Ungrouped Data:
For unclassified data the pth Percentile is given by (n+1)pth value.
Take p=1/100 for first percentile, p=2/100 for second percentile, . . . p=99/100 for ninety ninth percentile.
Example
Find P73, for the wages of labourers: • Rs. 82, 56, 90, 50, 120, 75,
79, 80, 130, and 65.
Solution
( )
( )
73
73
731100
73i. . 10 1 8.03100
. . 8 0.03(9 value 8 value)i.e. 90 0.03(120 90)i.e. 90.9
rd
th th th
P n th value
e th value value
i e value
P
= + ×
= + × =
= + −= + −
=
Arrange the wages in ascending order as : 50, 56, 65, 75, 79, 80, 82, 90, 120, 130
Question Time M.C.Q's. on Median and other
Partition Values
Question 1
What is the Median of 5, 8, 6, 9, 11 and 4.
a) 6
b) 7
c) 8
d) None of these
Answer: (b) 7
Question 2
What is the value of the first Quartile for the observations 15, 18,10, 20, 23, 28, 12, 16?
(a) 17
(b) 15.75
(c) 12
(d) 12.75
Answer: (d) 12.75
Question 3.
(a) 13
(b) 10.70
(c) 11
(d)11.50
Answer: (b) 10.70
The third decile for the numbers 15, 10, 20, 25, 18, 11, 9, 12 is
Question 4
Find the median of 12, 4, 36, 45, 3, 58, 65?
(a) 45
(b) 36
(c) 58
(d) 42
Answer: (b) 36
Question 5
Find the 73rd percentile of 4, 12, 36, 45, 3, 58, 65?
(a) 54.92
(b) 55.92
(c) 56.92
(d) 57.92
Answer: (b) 55.92
Question 6 Find the missing frequency if the median mark is 23?
X : 0-10 10-20 20-30 30-40 40-50 f : 5 8 ? 6 3
(a) 9
(b) 11
(c) 10
(d) 7
Answer: (c) 10
Question 7
The absolute sum of deviations is minimum when it is measured about
(a) Mean
(b) Median
(c) Mode
(d) None of these
Answer: (b) Median
Question 8
The variables X and Y are given by Y= 3X+6. If the median of X is 10, what is the median of Y?
(a) 3
(b) 6
(c) 18
(d) 36
Answer: (d) 36
Question 9
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
Answer: (c) 15.75
Question 10
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
Answer: (b) 10.70
Question 11 Find the 27th percentile for the following frequency distribution. X: 0-9 10-19 20-29 30-39 39-49 f: 5 7 12 10 6
(a) 13
(b) 17.79
(c) 11
(d) 11.50
Answer: (b) 17.79
Mode
Mode (Mo):
This is an industrial average.
The most frequent value of the variable is called mode.
Determination of Mode:
Case (i) When data is unclassified.
• e.g. Mo of 5, 3, 8, 9, 5, 6 is 5 which is the most frequent value.
0 1
0 1 1
0
1
1
2
limit of modalclassf modal classf modal classf modal
(note that class intervals must be continuous)
f fMo l hf f f
wherel lower
frequency of thefrequency of pre
frequency of post classh class width
−
−
−
−= + × − −
=== −= −=
Case (ii) For grouped data i.e. if frequencies corresponding to class intervals are given, then Mo is given as
Example:
Find the Mo for the following frequency distribution.
Class Intervals: 350-369, 370-389, 390-409, 410-429, 430-449, 450-469, 470-489 Frequency : 23, 38, 58, 82, 65, 31, 11
Solution Class Intervals Continuous Class Intervals Frequency
350-369 349.5-369.5 23
370-389 369.5-389.5 38
390-409 389.5-409.5 58
410-429 409.5-429.5 82
430-449 429.5-449.5 65
450-469 449.5-469.5 31
470-489 469.5-489.5 11
Solution
0 1
0 1 12
82 58409.5 202 82 58 65
. . 409.5 11.71 421.21
f fMo l h
f f f
Mo
i e Mo
−
−
−= + × − −
− = + × × − − = + =
Relation between Mean, Median And Mode:
Mode =3 Median-2 Mean • (OR)
Mean-Mode=3(Mean-Median)
MCQ’ on Mode
Question Time
Question 1 What is the Modal value for the numbers 5, 8, 6, 4, 10,15, 18, 10, 5, 10?
(a) 5
(b) 10
(c) 14
(d) None of these
Answer: (b) 10
Question 2 The Mode for the following frequency distribution:
(a) 390
(b) 394.86
(c) 395.86
(d) 396
Answer: (c) 395.86
Class Intervals: 350-369, 370-389, 390-409, 410-429, 430-449, 450-469 Frequency : 15 27 31 19 13 6
Question 6 Following is an incomplete distribution having modal mark as 44.
(a) 45
(b) 46
(c) 47
(d) 48
Answer : (d) 48
Marks: 0-20, 20-40, 40-60, 60-80, 80-100 No. of students: 5 18 X 12 5 What would be the mean marks?
Question 7 Measures of Central Tendency for given set of observations measures
(a) The scatterness of observations
(b) the central location of observations
(c) Both (a) and (b)
(d) None of these
Answer: (b) the central location of observations
Question 8
(a) The data are not classified
(b) The data are put in the form of grouped frequency distribution
(c) All the observations are not of equal importance
(d) Both (i) and (iii)
Answer: (c) All the observations are not of equal importance
Weighted averages are considered when
Geometric Mean
Geometric Mean (G)
GM is defined if no observation is zero or negative.
Geometric Mean (G)
GM of n positive observations is defined as the nth root of
their product.
Geometric Mean (G)
GM is used to calculate average growth rate or average of ratios.
Geometric Mean (G)
If Z=X.Y then GM(Z)=GM(X).GM(Y)
Geometric Mean (G)
If Z=X/Y then GM(Z)=GM(X)/GM(Y)
1
1 2 n(X .X ...X )1log log
nG
or G Xn
=
= ∑
GM of X1, X2,…Xn is given by
GM for Individual series
GM for Discrete Series
( )1 2
1
1 2. ...
1log log
nff f NnG X X X
or G f XN
=
= ∑
GM for X1-f1, X2-f2,…Xn-fn type data
Example 1
( ) ( )1 13 32 3 4 64 4G = × × = =
GM of 2, 3, and 4 is given as
Example 2.
• X : 2 4 8 16 • f : 2 3 3 2
Find the GM of the following frequency distribution.
Solution ( )
( )
( )
( )
1 2
1
1 2
12 3 3 2 (2 3 3 2)
12 6 9 8 10
125 2.510
. ...
2 .4 .8 .16
2 .2 .2 .2
2 2 4 2
5.66
nff f NnG X X X
or G
or G
or G
G
+ + +
=
=
=
= = =
=
Harmonic Mean
Harmonic Mean (H)
HM is defined if no observation is zero.
Harmonic Mean (H)
HM of n observations is defined as the reciprocal of Arithmetic Mean of the
reciprocals of the observations.
Harmonic Mean (H)
This is used to calculate the average speeds.
Harmonic Mean (H) Combined HM of two groups having n1 and n2
observations and H1 and H2 as their respective Harmonic Means is given by
( )1 2
1 2
1 2
n nH
n nH H
+=
+
Case (i)
1nH
X
=
∑
HM of X1, X2, … Xn is
Example 1
3 3 36 3.27111 1 1 1 11122 4 6
nH
X
= = = = = + +
∑ ∑
HM of 2, 4, and 6 is
Find the HM of the following data X : 2 4 8 16 f : 2 3 3 2
Example 2
( )2 3 3 24.44
2 3 3 22 4 16
:
8
NHf
Solut
X
ion+ + +
= = = + + +
∑ ∑
Case(ii) If the data is of X1-f1, X2-f2,…Xn-fn type
data, then HM is given by
NH where N ffX
= = ∑∑
Relation Between AM, GM and HM: 2( ) ( ) ( )
.
G A H
MoreoverAM GM HMNote that equality holds if all the observations are equal
= ×
≥ ≥
Weighted Average
Weighted Average:
If all the observations are not equally important i.e. weights are assigned to them, then Weighted AM, Weighted GM and Weighted
HM are calculated.
Weighted AM is
WX
AMW
= ∑∑
Weighted GM is
log
logW X
GM AnteW
=
∑∑
Weighted HM is
WHM
WX
= ∑∑
Example 1
Find the weighted AM and weighted HM for the following data
X : 1 2 3 4 5 6 7 8 9 10 W : 12 22 32 42 52 62 72 82 92 102
Solution 2 2 2
2 2 2
2
3 3 3
2 2 2
1 1 2 2 ...1 2 ...
(n 1)1 2 ... 2
(n 1)(2n 1)1 2 ...6
3n(n 1)2(2n 1)
WX n nWeighted AMW n
nn
nn
or AM
× + × + + ×= =
+ + +
+ + + + = =+ ++ + +
+=
+
∑∑
Solution 2 2 2
2 2 2
2 2 2
1 2 ...1 2 ...1 2
(n 1)(2n 1)1 2 ... 6
(n 1)1 2 ...2
(2n 1)3
W nWeighted HMW nX n
nn
nn
or HM
+ + += =
+ + +
+ + + + + = =
++ + +
+=
∑∑
Question Time MCQ’ on GM & HM
Question 1
(a) AM
(b) GM
(c) HM
(d) Both (b) & (c)
Answer: (b) GM
Which measure of central tendency is used to find average rates.
Question 2
(a) AM ≥ GM ≥ HM
(b) HM ≥ GM ≥ AM
(c) AM > GM > HM
(d) GM > AM > HM
Answer: (a) AM ≥ GM ≥ HM
Which of the following results hold for a set of distinct positive observations?
Question 3
(a) 2.00
(b) 3.33
(c) 2.90
(d) -3.33
Answer: (c) 2.90
The Harmonic Mean for the numbers 2,3,5 is
Question 4
(a) 24
(b) 12
(c) 36
(d) 18
Answer: (b) 12
What is the Geometric Mean of the numbers 4,12, 36?
Question 5
(a) 6 and 7
(b) 9 and 4
(c) 10 and 3
(d) 8 and 5
Answer (b) 9 and 4
If the AM and GM for the two numbers are 6.50 and 6 respectively then the two numbers are
Question 6
(a) 16
(b) 4.10
(c) 4.05
(d) 4.00
Answer: (d) 4
If the AM and HM for two numbers are 5 and 3.2 respectively then the GM will be
Question 7
(a) 150
(b) log 10 x log15
(c) log 150
(d) none of these
Answer: (a) 150
If GM of X is 10 and GM of Y is 15, then the value of GM of XY is
Question 8
(a) 40
(b) 50
(c) 20
(d) 30
Answer: (a) 40
If AM of a variable X is 10 and GM of X is 20, then the HM of X is
Question 9
(a) n
(b) 2n
(c) 2/(n+1)
(d) n(n+1)/2
Answer: (c) 2/(n+1)
The HM of 1, 1/2, 1/3 … 1/n is
Thank You