Please copy and paste this embed script to where you want to embed

Session – 5

Measures of Central Tendency

Combined Mean Combined arithmetic mean can be computed if we know the mean a nd number of items in each groups of the data. The following equation is used to compute combined mean. Let x 1 & x 2 are the mean of first and second group of data containing N 1 & N2 items respectively. Then Then,, comb combin ined ed mean mean = x 12 If there are 3 groups then x 123

N1 x 1 N 2 x 2 N1 N 2 N1 x 1 N 2 x 2 N 3 x 3 N1 N 2 N 3

Ex - 1:

a) Find the the means for for the entire entire group group of worker workerss for the following following data.

Group – 1

Group – 2

75

60

1000

1500

Mean wages No. of workers Given data:

N1 = 1000

N2 = 1500

x 1 75 & x 2 60

Group Mean = x 12 =

N1 x 1 N 2 x 2 N1 N 2

1000 x 75 1500 x 60 1000 1500

= x 12 Rs. 66

Ex - 2:

Compute mean for entire group.

Medical examination

No. examined

Mean weight (pounds)

A

50

113

B

60

120

C

90

115

1

Combined mean (grouped mean weight) x 123

N1 x 1 N 2 x 2 N 3 x 3 N1 N 2 N 3

(50 x 113 60 x 120 90 x 115) (50 60 90)

x 123 Mean weight 116 pounds

Merits of Arithmetic Mean

1. It is is simple simple and easy to comp compute. ute. 2. It is rigi rigidl dly y defi define ned. d. 3. It can can be used used for for furth further er calcu calculati lation. on. 4. It is based based on on all observ observati ations ons in in the serie series. s. 5. It help helpss for for direc directt compa comparis rison. on. 6. It is more more stable measur measuree of central central tendency tendency (ideal (ideal average) average)..

Limitations Limitations / Demerits Demerits of Mean

1. It is undu unduly ly affec affected ted by by extrem extremee items. items. 2. It is someti sometimes mes un-rea un-realis listic tic.. 3. It may may lead leadss to con confu fusi sion on.. 4. Suitable Suitable only only for for quantita quantitative tive data (for (for variabl variables). es). 5. It can not not be located located by graphical graphical method method or by observat observations. ions.

Geometric Mean (GM) th

The GM is n root of product of quantities of the series. It is observed by multiplying the values of items together and extracting the root of the product corresponding to the number of of items. Thus, square root of the products products of two items and cube cube root of of the products products of of the three items are are the Geometric Geometric Mean. Mean. Usually, geometric mean is never larger larger than arithmetic mean. mean. If there are are zero and negative number number in the series. If there are are zeros and negative numbers numbers in the series, the geometric means cannot be used logarithms can be used to find geometric mean to reduce large number and to save time. In the field of business business management various various problems often often arise relating to average percentage rate of change over a period of time. In such cases, the arithmetic mean is not an appropriat appropriatee average average to employ, employ, so, that we can use use geometric geometric mean in such case. GM are highly useful in the construction construction of index numbers. numbers. Geometric Mean (GM) = n x 1 x x 2 x ...........x x n When the the number number of items in the the series is larger larger than than 3, the process process of computing GM is difficult. To over over come this, a logarithm of each size is obtained.

2

The log of all the value added up and divided by number of items. The antilog of quotient obtained is the required GM.

log1 log 2 ................ log n (GM) (GM) = Antilog Antilog Anti log n Merits of GM

a. It is based based on on all the the observ observatio ations ns in the the series. series. b. It is is rigi rigidl dly y defi define ned. d. c. It is best best suite suited d for for averag averages es and ratios ratios.. d. It is less affect affected ed by extreme extreme valu values. es. e. It is useful for studying studying social and economics economics data.

Demerits of GM

a. It is is not not simpl simplee to unders understan tand. d. b. It requ require iress compu computat tation ional al skil skill. l. c. GM cannot cannot be computed computed if if any of item is zero or negative. negative. d. It has has restr restrict icted ed appl applica icatio tion. n.

Ex - 1:

a. Find Find the the GM GM of of data data 2, 4, 8 x1 = 2, x2 = 4, x3 = 8 n=3 GM = n x 1 x x 2 x x 3 GM = 3 2 x 4 x 8 GM = 3 64 4 GM = 4

b. Find Find GM of data data 2, 4, 8 usin using g logari logarithm thms. s. Data: x1 = 2 x2 = 4 x3 = 8 N=3

3

log x i i 1 N

x

log x

2

0.301

4

0.602

8

0.903 Σlogx

= 1.806

log x N

GM = Antilo Antilog g

1.806 GM = Antilog 3

GM = Antilog (0.6020) = 3.9997 GM 4

Ex - 2:

Compare the previous year the Over Head (OH) expenses which went up to 32% in year 2003, then increased by 40% in next year and 50% increase in the following year. Calculate average increase in over head expenses. Let 100% OH Expenses at base year

Y e ar

OH Expenses (x)

log x

2002

Base year

–

2003

132

2.126

2004

140

2.146

2005

150

2.176 Σ

log log x = 6.44 6.448 8

log x N

GM = Antilog

6.448 3

GM = Antilog GM = 141.03

GM for discrete series

GM for discrete series is given given with usual notations as month:

4

log x i i 1 N

GM = Antilog Ex - 3:

Consider Consider following following time series for monthly monthly sales sales of ABC company company for 4 months. Find average rate of change per monthly sales.

M on th

Sales

I

10000

II

8000

III

12000

IV

15000

Let Base year = 100% sales.

Solution:

(Rs)

Increase / decrease %ge

Conversion (x)

log (x)

Sales

Month

Base year

I

100%

10000

–

–

–

II

– 2 – 20%

8000

80

80

1.903

III

+ 50%

12000

130

130

2.113

IV

+ 25%

15000

155

155

2.190 logx Σlogx

= 6.206 6.206

6.206 = 117.13 3

GM = Antilog

Average sales = 117.13 – 117.13 – 100 100 = 14.46%

Ex - 4:

Find GM for following following data. data.

Marks

No. of students

(x)

(f)

130

log x

f log x

3

2.113

6.339

135

4

2.130

8.52

140

6

2.146

12.876

145

6

2.161

12.996

150

3

2.176

6.528

Σf

= N = 22

Σ

5

f log x =47.23

f log x N

GM = Antilog

47.23 GM = Anti Antilog log 22 GM = 140.212

Geometric Mean for continuous series Steps:

1. Find mid mid value value m and take log log of m for each mid mid value. value. 2. Multiply log m with frequency ‘f’ of each class to get f log m and sum up to obtain f log m. 3. Divide f log m by N and take take antilog antilog to get GM.

Ex:

Find out GM for given data below

Yield of wheat in

No. of farms frequency

Mid value ‘m’

log m

f log m

MT

(f)

1 – 1 – 10

3

5.5

0.740

2.220

11 – 11 – 2 20

16

15.5

1.190

19.040

21 – 21 – 3 30

26

25.5

1.406

36.556

31 – 31 – 4 40

31

35.5

1.550

48.050

41 – 41 – 5 50

16

45.5

1.658

26.528

51 – 51 – 6 60

8

55.5

1.744

13.954

Σ

f log m = 146.348

Σf

= N = 100

f log m N

GM = Antilog

146.348 GM = Antilog 100 GM = 29.07 29.07

Harmonic Mean It is the total number of items of a value divided by the sum of reciprocal of values of variable. variable. It is a specified average which solves problems problems involving variables expressed in within ‘Time rates’ that vary according to time.

6

Ex:

Speed in km/hr, min/day, price/unit.

Harmonic Mean (HM) is suitable only when time factor is variable and the act being performed remains constant. HM =

N 1

x

Merits of Harmonic Mean

1. It is is based based on all all obse observa rvatio tions. ns. 2. It is rigi rigidl dly y defi define ned. d. 3. It is suitab suitable le in case case of serie seriess having having wide wide dispers dispersion ion.. 4. It is suitable suitable for for further further mathematical mathematical treatment. treatment.

Demerits of Harmonic Mean

1. It is is not not easy easy to to comp comput ute. e. 2. Cannot Cannot used used when when one one of of the item item is is zero. zero. 3. It canno cannott repres represent ent distr distribu ibutio tion. n.

Ex:

1. The daily daily income income of of 05 families families in in a very very rural rural village village are given given below. below. Compute Compute HM. Family

Income (x)

Reciprocal (1/x)

1

85

0.0117

2

90

0.01111

3

70

0.0142

4

50

0.02

5

60

0.016

1

HM =

=

N 1

x

5 0.0738

= 67.72

HM = 67.7 67.72 2

7

x

= 0.0738

2. A man travel travel by a car for for 3 days days he covered covered 480 480 km each each day. day. On the first first day day he drives for 10 hrs at the rate of 48 KMPH, on the second day for 12 hrs at the rate rd of 40 KMPH, and on the 3 day for 15 hrs @ 32 KMPH. KMPH. Compute HM and weighted mean and compare them. Harmon Harmonic ic Mean Mean x

1

48

0.0208

40

0.025

32

0.0312

x

1 = 0.0770 x

Data: 10 hrs @ 48 KMPH 12 hrs @ 40 KMPH 15 hrs @ 32 KMPH HM =

=

N 1

x

3 0.0770

HM = 38.91

Weighted Mean w

x

wx

10

48

480

12

40

480

15

32

480

w = 37

Weighted Mean = x =

Σwx

wx w

1440 37

x 38.91 Both the same HM and WM are same.

8

= 1440

3. Find Find HM for the follow following ing data. data.

1 m

Reciprocal

1 m

f

Class (CI)

Frequency (f)

Mid point (m)

0 – 1 – 10

5

5

0.2

1

10 – 10 – 2 20

15

15

0.0666

0.999

20 – 20 – 3 30

25

25

0.04

1

30 – 30 – 4 40

8

35

0.0285

0.228

40 – 40 – 5 50

7

45

0.0222

0.1554

Σf

HM =

=

1 f = 3.3824 m

= 60

N

1 f m 60 3.3824

HM = 17.73

Relationship between Mean, Geometric Mean and Harmonic Mean. 1. If all the the items in in a variable variable are are the same, same, the the arithmetic arithmetic mean, mean, harmon harmonic ic mean mean and Geometric mean are equal. i.e., x GM HM . 2. If the size size vary, vary, mean will will be greater greater than than GM and GM will be greater greater than than HM. This is because of the property that geometric mean to give larger weight to smaller item and of the HM to give largest weight to smallest item. Hence, x GM HM .

Median Median is the value of that item in a series which divides the array into two equal parts, one consisting of all the values less than it and other consisting of all the values more than it. Median is a positional average. average. The number of of items below it is equal to the number. The number of items below it is equal to the number of items above it. It occupies occupies central position. Thus, Median is defined defined as the mid value value of of the variants. variants. If the the values values are arranged in ascending or descending order of their magnitude, median is the middle value of the number of variant is odd and average of two middle values if the number of variants is even. th

Ex:

If 9 students are stand in the order of of their heights; the 5 student from either side shall be the one whose height will be Median height of of the students group. group. Thus, median of group is given by an equation.

9

N 1 2

Median = Ex

1. Find Find the medi median an for for follow following ing data. data. 22

20

25

31

26

24

23

Arrange the given data in array a rray form (either in ascending or descending order). 20

22

23

25

24

26

31

N 1 th 7 1 8 th item = = Median = 4 item. 2 2 4

Median is given by

2. Find Find median median for follow following ing data. data.

20

21

22

24

N 1 th item = 2

Median is given by

28

32

6 1 th Median = 3.5 item. 2 Median

rd

The item lies between 3 and 4. So, there are two values 22 and 24. The median value will be the mean values of these two values. values.

22 24 = 23 2

Median =

Discrete Series – Median

In discrete series, the values are (already) in the form of array and the frequencies are recorded against each value. However, to determine the size of N 1 th median item, a separate column is to be prepared for cumulative 2 frequencies. The median median size is first located with reference to the cumulative frequency which covers the size first. Then, against that that cumulative frequency, frequency, the value will be located as median.

10

Ex:

Find the median for the students’ marks. Obtained in statistics Marks (x)

No. of students (f)

Cumulative frequency

10

5

5

20

5

10

30

3

13

40

15

28

50

30

58

60

10

68

Just above 34 is 58. 58. Agains Againstt 58 c.f. the value is 50 which is median value

N = 68

Ex:

In a class 15 15 students students,, 5 students students were failed in a test. The marks of 10 students students who have passed were 9, 6, 7, 7, 8, 9, 6, 5, 4, 7, 8. Find the Median marks of 15 students. Marks

No. of students (f)

cf

0 1 2 5

3 4

1

6

5

1

7

6

2

9

7

2

11

8

2

13

9

2

15

Σf

Medi Median an = Me =

N 1th

15 1 2

2

= 15

item

th

=8

th

Me 8 item covers in cf of 9. the marks against cf 9 is 6 and hence hence Median = 6

11

Continuous Series

The procedure is different to get median in continuous continuous series. The class intervals are already in the form of array and the frequency are recorded against each class interval. For determining the size, we should take

n

th

item and median class 2 located accordingly with reference to the cumulative frequency, which covers the size first. When the median class is located, the median median value is to be interpolated interpolated using formula given below. Median =

Where

h N C f 2

1

0

2 point of previous class.

where,

0

is left end point of N/2 class and l 1is right end

h = Class width, f = frequency of median clas C = Cumulative frequency frequency of class preceding preceding the median class.

Ex:

Find the median for following data. data. The class marks obtained by 50 students are are as follows. Frequency (f)

10 – 10 – 1 15

6

6

15 – 15 – 2 20

18

24

20 – 20 – 2 25

9

33 N/2 class

25 – 25 – 3 30

10

43

30 – 30 – 3 35

4

47

35 – 35 – 4 40

3

50

Σf

N 2

50 2

Cum. frequency (cf)

CI

= N = 50

25

Cum. Cum. frequ frequenc ency y just just abov abovee 25 is 33 33 and hence, hence, 20 – 20 – 25 25 is median class.

0

1 2

20 20 2

20

20

h = 20 – 20 – 15 15 = 5

12

f=9 c = 24 Median =

h N C f 2

5

Median = 20 = 20

9

25 24

5 9

Median = 20.555

Ex:

Find the median median for followi following ng data.

Mid values (m)

115

125

135

145

155

165

175

185

195

Frequencies (f)

6

25

48

72

116

60

38

22

3

The interval of mid-values of CI and magnitudes of class intervals are same i.e. 10. So, half of 10 is deducted from and added to mid-values will give us the lower and upper limits. Thus, classes are. 115 – 115 – 5 5 = 110 (lower limit) 115 – 115 – 5 5 = 120 (upper limit) similarly for all mid values we can get CI. Frequency (f)

110 – 110 – 1 120

6

6

120 – 120 – 1 130

25

31

130 – 130 – 1 140

48

79

140 – 140 – 1 150

72

151

150 – 150 – 1 160

116

267

160 – 160 – 1 170

60

327

170 – 170 – 1 180

38

365

180 – 180 – 1 190

22

387

190 – 190 – 2 200

3

390

Σf

N 2

Cum. frequency (cf)

CI

= N = 390 390

390 2

195 Cum. frequency just above 195 is 267.

13

N/2 class

Median class = 150 – 150 – 160 160

=

150 150 2

= 150

h = 116 116 N/2 = 195 C = 151 h = 10 Median =

h N C f 2

Median = 150

10 116

195 151

Median Median = 153.8 153.8

Merits of Median

a. It is simple simple,, easy to to comput computee and under understan stand. d. value is not affected by extreme variables. b. It’s value c. It is capabl capablee for furthe furtherr algebr algebraic aic treatm treatment ent.. d. It can be determi determined ned by by inspectio inspection n for arrayed arrayed data. data. e. It can can be be found found graph graphica ically lly also. also. f. It indi indicate catess the the valu valuee of middle middle item. item.

Demerits of Median

a. It may not be be representati representative ve value value as it ignores ignores extreme extreme values. values. b. It can’t be determined precisely when its size falls between the two values. c. It is not useful useful in cases where where large large weights weights are to be be given given to extreme extreme values. values.

14

View more...
Measures of Central Tendency

Combined Mean Combined arithmetic mean can be computed if we know the mean a nd number of items in each groups of the data. The following equation is used to compute combined mean. Let x 1 & x 2 are the mean of first and second group of data containing N 1 & N2 items respectively. Then Then,, comb combin ined ed mean mean = x 12 If there are 3 groups then x 123

N1 x 1 N 2 x 2 N1 N 2 N1 x 1 N 2 x 2 N 3 x 3 N1 N 2 N 3

Ex - 1:

a) Find the the means for for the entire entire group group of worker workerss for the following following data.

Group – 1

Group – 2

75

60

1000

1500

Mean wages No. of workers Given data:

N1 = 1000

N2 = 1500

x 1 75 & x 2 60

Group Mean = x 12 =

N1 x 1 N 2 x 2 N1 N 2

1000 x 75 1500 x 60 1000 1500

= x 12 Rs. 66

Ex - 2:

Compute mean for entire group.

Medical examination

No. examined

Mean weight (pounds)

A

50

113

B

60

120

C

90

115

1

Combined mean (grouped mean weight) x 123

N1 x 1 N 2 x 2 N 3 x 3 N1 N 2 N 3

(50 x 113 60 x 120 90 x 115) (50 60 90)

x 123 Mean weight 116 pounds

Merits of Arithmetic Mean

1. It is is simple simple and easy to comp compute. ute. 2. It is rigi rigidl dly y defi define ned. d. 3. It can can be used used for for furth further er calcu calculati lation. on. 4. It is based based on on all observ observati ations ons in in the serie series. s. 5. It help helpss for for direc directt compa comparis rison. on. 6. It is more more stable measur measuree of central central tendency tendency (ideal (ideal average) average)..

Limitations Limitations / Demerits Demerits of Mean

1. It is undu unduly ly affec affected ted by by extrem extremee items. items. 2. It is someti sometimes mes un-rea un-realis listic tic.. 3. It may may lead leadss to con confu fusi sion on.. 4. Suitable Suitable only only for for quantita quantitative tive data (for (for variabl variables). es). 5. It can not not be located located by graphical graphical method method or by observat observations. ions.

Geometric Mean (GM) th

The GM is n root of product of quantities of the series. It is observed by multiplying the values of items together and extracting the root of the product corresponding to the number of of items. Thus, square root of the products products of two items and cube cube root of of the products products of of the three items are are the Geometric Geometric Mean. Mean. Usually, geometric mean is never larger larger than arithmetic mean. mean. If there are are zero and negative number number in the series. If there are are zeros and negative numbers numbers in the series, the geometric means cannot be used logarithms can be used to find geometric mean to reduce large number and to save time. In the field of business business management various various problems often often arise relating to average percentage rate of change over a period of time. In such cases, the arithmetic mean is not an appropriat appropriatee average average to employ, employ, so, that we can use use geometric geometric mean in such case. GM are highly useful in the construction construction of index numbers. numbers. Geometric Mean (GM) = n x 1 x x 2 x ...........x x n When the the number number of items in the the series is larger larger than than 3, the process process of computing GM is difficult. To over over come this, a logarithm of each size is obtained.

2

The log of all the value added up and divided by number of items. The antilog of quotient obtained is the required GM.

log1 log 2 ................ log n (GM) (GM) = Antilog Antilog Anti log n Merits of GM

a. It is based based on on all the the observ observatio ations ns in the the series. series. b. It is is rigi rigidl dly y defi define ned. d. c. It is best best suite suited d for for averag averages es and ratios ratios.. d. It is less affect affected ed by extreme extreme valu values. es. e. It is useful for studying studying social and economics economics data.

Demerits of GM

a. It is is not not simpl simplee to unders understan tand. d. b. It requ require iress compu computat tation ional al skil skill. l. c. GM cannot cannot be computed computed if if any of item is zero or negative. negative. d. It has has restr restrict icted ed appl applica icatio tion. n.

Ex - 1:

a. Find Find the the GM GM of of data data 2, 4, 8 x1 = 2, x2 = 4, x3 = 8 n=3 GM = n x 1 x x 2 x x 3 GM = 3 2 x 4 x 8 GM = 3 64 4 GM = 4

b. Find Find GM of data data 2, 4, 8 usin using g logari logarithm thms. s. Data: x1 = 2 x2 = 4 x3 = 8 N=3

3

log x i i 1 N

x

log x

2

0.301

4

0.602

8

0.903 Σlogx

= 1.806

log x N

GM = Antilo Antilog g

1.806 GM = Antilog 3

GM = Antilog (0.6020) = 3.9997 GM 4

Ex - 2:

Compare the previous year the Over Head (OH) expenses which went up to 32% in year 2003, then increased by 40% in next year and 50% increase in the following year. Calculate average increase in over head expenses. Let 100% OH Expenses at base year

Y e ar

OH Expenses (x)

log x

2002

Base year

–

2003

132

2.126

2004

140

2.146

2005

150

2.176 Σ

log log x = 6.44 6.448 8

log x N

GM = Antilog

6.448 3

GM = Antilog GM = 141.03

GM for discrete series

GM for discrete series is given given with usual notations as month:

4

log x i i 1 N

GM = Antilog Ex - 3:

Consider Consider following following time series for monthly monthly sales sales of ABC company company for 4 months. Find average rate of change per monthly sales.

M on th

Sales

I

10000

II

8000

III

12000

IV

15000

Let Base year = 100% sales.

Solution:

(Rs)

Increase / decrease %ge

Conversion (x)

log (x)

Sales

Month

Base year

I

100%

10000

–

–

–

II

– 2 – 20%

8000

80

80

1.903

III

+ 50%

12000

130

130

2.113

IV

+ 25%

15000

155

155

2.190 logx Σlogx

= 6.206 6.206

6.206 = 117.13 3

GM = Antilog

Average sales = 117.13 – 117.13 – 100 100 = 14.46%

Ex - 4:

Find GM for following following data. data.

Marks

No. of students

(x)

(f)

130

log x

f log x

3

2.113

6.339

135

4

2.130

8.52

140

6

2.146

12.876

145

6

2.161

12.996

150

3

2.176

6.528

Σf

= N = 22

Σ

5

f log x =47.23

f log x N

GM = Antilog

47.23 GM = Anti Antilog log 22 GM = 140.212

Geometric Mean for continuous series Steps:

1. Find mid mid value value m and take log log of m for each mid mid value. value. 2. Multiply log m with frequency ‘f’ of each class to get f log m and sum up to obtain f log m. 3. Divide f log m by N and take take antilog antilog to get GM.

Ex:

Find out GM for given data below

Yield of wheat in

No. of farms frequency

Mid value ‘m’

log m

f log m

MT

(f)

1 – 1 – 10

3

5.5

0.740

2.220

11 – 11 – 2 20

16

15.5

1.190

19.040

21 – 21 – 3 30

26

25.5

1.406

36.556

31 – 31 – 4 40

31

35.5

1.550

48.050

41 – 41 – 5 50

16

45.5

1.658

26.528

51 – 51 – 6 60

8

55.5

1.744

13.954

Σ

f log m = 146.348

Σf

= N = 100

f log m N

GM = Antilog

146.348 GM = Antilog 100 GM = 29.07 29.07

Harmonic Mean It is the total number of items of a value divided by the sum of reciprocal of values of variable. variable. It is a specified average which solves problems problems involving variables expressed in within ‘Time rates’ that vary according to time.

6

Ex:

Speed in km/hr, min/day, price/unit.

Harmonic Mean (HM) is suitable only when time factor is variable and the act being performed remains constant. HM =

N 1

x

Merits of Harmonic Mean

1. It is is based based on all all obse observa rvatio tions. ns. 2. It is rigi rigidl dly y defi define ned. d. 3. It is suitab suitable le in case case of serie seriess having having wide wide dispers dispersion ion.. 4. It is suitable suitable for for further further mathematical mathematical treatment. treatment.

Demerits of Harmonic Mean

1. It is is not not easy easy to to comp comput ute. e. 2. Cannot Cannot used used when when one one of of the item item is is zero. zero. 3. It canno cannott repres represent ent distr distribu ibutio tion. n.

Ex:

1. The daily daily income income of of 05 families families in in a very very rural rural village village are given given below. below. Compute Compute HM. Family

Income (x)

Reciprocal (1/x)

1

85

0.0117

2

90

0.01111

3

70

0.0142

4

50

0.02

5

60

0.016

1

HM =

=

N 1

x

5 0.0738

= 67.72

HM = 67.7 67.72 2

7

x

= 0.0738

2. A man travel travel by a car for for 3 days days he covered covered 480 480 km each each day. day. On the first first day day he drives for 10 hrs at the rate of 48 KMPH, on the second day for 12 hrs at the rate rd of 40 KMPH, and on the 3 day for 15 hrs @ 32 KMPH. KMPH. Compute HM and weighted mean and compare them. Harmon Harmonic ic Mean Mean x

1

48

0.0208

40

0.025

32

0.0312

x

1 = 0.0770 x

Data: 10 hrs @ 48 KMPH 12 hrs @ 40 KMPH 15 hrs @ 32 KMPH HM =

=

N 1

x

3 0.0770

HM = 38.91

Weighted Mean w

x

wx

10

48

480

12

40

480

15

32

480

w = 37

Weighted Mean = x =

Σwx

wx w

1440 37

x 38.91 Both the same HM and WM are same.

8

= 1440

3. Find Find HM for the follow following ing data. data.

1 m

Reciprocal

1 m

f

Class (CI)

Frequency (f)

Mid point (m)

0 – 1 – 10

5

5

0.2

1

10 – 10 – 2 20

15

15

0.0666

0.999

20 – 20 – 3 30

25

25

0.04

1

30 – 30 – 4 40

8

35

0.0285

0.228

40 – 40 – 5 50

7

45

0.0222

0.1554

Σf

HM =

=

1 f = 3.3824 m

= 60

N

1 f m 60 3.3824

HM = 17.73

Relationship between Mean, Geometric Mean and Harmonic Mean. 1. If all the the items in in a variable variable are are the same, same, the the arithmetic arithmetic mean, mean, harmon harmonic ic mean mean and Geometric mean are equal. i.e., x GM HM . 2. If the size size vary, vary, mean will will be greater greater than than GM and GM will be greater greater than than HM. This is because of the property that geometric mean to give larger weight to smaller item and of the HM to give largest weight to smallest item. Hence, x GM HM .

Median Median is the value of that item in a series which divides the array into two equal parts, one consisting of all the values less than it and other consisting of all the values more than it. Median is a positional average. average. The number of of items below it is equal to the number. The number of items below it is equal to the number of items above it. It occupies occupies central position. Thus, Median is defined defined as the mid value value of of the variants. variants. If the the values values are arranged in ascending or descending order of their magnitude, median is the middle value of the number of variant is odd and average of two middle values if the number of variants is even. th

Ex:

If 9 students are stand in the order of of their heights; the 5 student from either side shall be the one whose height will be Median height of of the students group. group. Thus, median of group is given by an equation.

9

N 1 2

Median = Ex

1. Find Find the medi median an for for follow following ing data. data. 22

20

25

31

26

24

23

Arrange the given data in array a rray form (either in ascending or descending order). 20

22

23

25

24

26

31

N 1 th 7 1 8 th item = = Median = 4 item. 2 2 4

Median is given by

2. Find Find median median for follow following ing data. data.

20

21

22

24

N 1 th item = 2

Median is given by

28

32

6 1 th Median = 3.5 item. 2 Median

rd

The item lies between 3 and 4. So, there are two values 22 and 24. The median value will be the mean values of these two values. values.

22 24 = 23 2

Median =

Discrete Series – Median

In discrete series, the values are (already) in the form of array and the frequencies are recorded against each value. However, to determine the size of N 1 th median item, a separate column is to be prepared for cumulative 2 frequencies. The median median size is first located with reference to the cumulative frequency which covers the size first. Then, against that that cumulative frequency, frequency, the value will be located as median.

10

Ex:

Find the median for the students’ marks. Obtained in statistics Marks (x)

No. of students (f)

Cumulative frequency

10

5

5

20

5

10

30

3

13

40

15

28

50

30

58

60

10

68

Just above 34 is 58. 58. Agains Againstt 58 c.f. the value is 50 which is median value

N = 68

Ex:

In a class 15 15 students students,, 5 students students were failed in a test. The marks of 10 students students who have passed were 9, 6, 7, 7, 8, 9, 6, 5, 4, 7, 8. Find the Median marks of 15 students. Marks

No. of students (f)

cf

0 1 2 5

3 4

1

6

5

1

7

6

2

9

7

2

11

8

2

13

9

2

15

Σf

Medi Median an = Me =

N 1th

15 1 2

2

= 15

item

th

=8

th

Me 8 item covers in cf of 9. the marks against cf 9 is 6 and hence hence Median = 6

11

Continuous Series

The procedure is different to get median in continuous continuous series. The class intervals are already in the form of array and the frequency are recorded against each class interval. For determining the size, we should take

n

th

item and median class 2 located accordingly with reference to the cumulative frequency, which covers the size first. When the median class is located, the median median value is to be interpolated interpolated using formula given below. Median =

Where

h N C f 2

1

0

2 point of previous class.

where,

0

is left end point of N/2 class and l 1is right end

h = Class width, f = frequency of median clas C = Cumulative frequency frequency of class preceding preceding the median class.

Ex:

Find the median for following data. data. The class marks obtained by 50 students are are as follows. Frequency (f)

10 – 10 – 1 15

6

6

15 – 15 – 2 20

18

24

20 – 20 – 2 25

9

33 N/2 class

25 – 25 – 3 30

10

43

30 – 30 – 3 35

4

47

35 – 35 – 4 40

3

50

Σf

N 2

50 2

Cum. frequency (cf)

CI

= N = 50

25

Cum. Cum. frequ frequenc ency y just just abov abovee 25 is 33 33 and hence, hence, 20 – 20 – 25 25 is median class.

0

1 2

20 20 2

20

20

h = 20 – 20 – 15 15 = 5

12

f=9 c = 24 Median =

h N C f 2

5

Median = 20 = 20

9

25 24

5 9

Median = 20.555

Ex:

Find the median median for followi following ng data.

Mid values (m)

115

125

135

145

155

165

175

185

195

Frequencies (f)

6

25

48

72

116

60

38

22

3

The interval of mid-values of CI and magnitudes of class intervals are same i.e. 10. So, half of 10 is deducted from and added to mid-values will give us the lower and upper limits. Thus, classes are. 115 – 115 – 5 5 = 110 (lower limit) 115 – 115 – 5 5 = 120 (upper limit) similarly for all mid values we can get CI. Frequency (f)

110 – 110 – 1 120

6

6

120 – 120 – 1 130

25

31

130 – 130 – 1 140

48

79

140 – 140 – 1 150

72

151

150 – 150 – 1 160

116

267

160 – 160 – 1 170

60

327

170 – 170 – 1 180

38

365

180 – 180 – 1 190

22

387

190 – 190 – 2 200

3

390

Σf

N 2

Cum. frequency (cf)

CI

= N = 390 390

390 2

195 Cum. frequency just above 195 is 267.

13

N/2 class

Median class = 150 – 150 – 160 160

=

150 150 2

= 150

h = 116 116 N/2 = 195 C = 151 h = 10 Median =

h N C f 2

Median = 150

10 116

195 151

Median Median = 153.8 153.8

Merits of Median

a. It is simple simple,, easy to to comput computee and under understan stand. d. value is not affected by extreme variables. b. It’s value c. It is capabl capablee for furthe furtherr algebr algebraic aic treatm treatment ent.. d. It can be determi determined ned by by inspectio inspection n for arrayed arrayed data. data. e. It can can be be found found graph graphica ically lly also. also. f. It indi indicate catess the the valu valuee of middle middle item. item.

Demerits of Median

a. It may not be be representati representative ve value value as it ignores ignores extreme extreme values. values. b. It can’t be determined precisely when its size falls between the two values. c. It is not useful useful in cases where where large large weights weights are to be be given given to extreme extreme values. values.

14

Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website.