measures of central tendency

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

Case (i) Arithmetic Mean (AM)

1 2 ...(i) nX X X

+ + +=

Case (ii) For Discrete Series

i.e. For x1-f1, x2-f2,…xn-fn type data

If values of X are large enough

1, 2,3....

fdX A X C

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

= + + + + + + + + =

= = = =

∑∑

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

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

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

X A Xd

C− −

Solution

( )43419.50 20

308416.71

fdX A X C

−= +

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

( )( )

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

n X n XX

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

40, 520060, 6800

40 5200 60 680040 60

n Xn XThen combined mean salary is givenby

n X n XX

n nX X

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

Mid-value (X)

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

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

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.

Answer: (b) 5

AM of 5,5,5,5,5, is

Question.3.

(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

(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

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

(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

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.

(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

36 132. . 20 10 25

N cThen Md l h

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

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

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

(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

40 54. . 10 10 16.25

N cThen Q l h

− = + ×

− = + × =

3 40 304. . 40 10 40

N cThen Q l h

− = + ×

× − = + × =

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

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)

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 40 2310. . 30 10

7i. . 30 7.14 37.14

N cThen D l h

− = + ×

× − = + ×

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

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

(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 40 1310020 10

1020 7.8 27.8

N cP l h

× − = + ×

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

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

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

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

D n th value

e th value value

i e value

= + ×

= + × =

= + −= + −

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

731100

73i. . 10 1 8.03100

. . 8 0.03(9 value 8 value)i.e. 90 0.03(120 90)i.e. 90.9

th th th

P n th value

e th value value

i e value

= + ×

= + × =

= + −= + −

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.

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

(b) 11

(c) 10

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?

(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 (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.

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 12

82 58409.5 202 82 58 65

. . 409.5 11.71 421.21

f fMo l h

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?

(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 2 n(X .X ...X )1log log

or G Xn

GM of X1, X2,…Xn is given by

GM for Individual series

GM for Discrete Series

( )1 2

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 ( )

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

nff f NnG X X X

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

n nH H

Case (i)

HM of X1, X2, … Xn is

Example 1

3 3 36 3.27111 1 1 1 11122 4 6

= = = = = + +

∑ ∑

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

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( ) ( ) ( )

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

= ∑∑

Weighted GM is

logW X

GM AnteW

∑∑

Weighted HM is

= ∑∑

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

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

× + × + + ×= =

+ + + + = =+ ++ + +

∑∑

Solution 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

+ + += =

+ + + + + = =

++ + +

∑∑

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

(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

measures of central tendency - icai knowledge …. measures of central tendency. the tendency of a...

Documents

elementary statistics measures of central tendency...

chapter 3 descriptive measures measures central tendency...

measures of central tendency. what are measures of central...

basic concept measures of central tendency measures of...

chapter 3 measures of central tendency. 3.1 defining central...

measures of central tendency measures of dispersion

measures of central tendency measures of location measures

measures of central...

measures of central tendency - chromeias.com€¦ · 1....

chapter 3.4 measures of central tendency measures of central...

measures of central tendency -...

measures of central tendency