Combined Mean 4s2r40

Session – 5 Measures of Central Tendency

Combined Mean Combined arithmetic mean can be computed if we know the mean and 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, combined mean = x 12 

N1 x 1  N 2 x 2 N1  N 2

If there are 3 groups then x 123 

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

Ex - 1: a) Find the means for the entire group of workers for the 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

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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 simple and easy to compute. 2. It is rigidly defined. 3. It can be used for further calculation. 4. It is based on all observations in the series. 5. It helps for direct comparison. 6. It is more stable measure of central tendency (ideal average). Limitations / Demerits of Mean 1. It is unduly affected by extreme items. 2. It is sometimes un-realistic. 3. It may leads to confusion. 4. Suitable only for quantitative data (for variables). 5. It can not be located by graphical method or by observations.

Geometric Mean (GM) The GM is nth 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 items. Thus, square root of the products of two items and cube root of the products of the three items are the Geometric Mean. Usually, geometric mean is never larger than arithmetic mean. If there are zero and negative number in the series. If there are zeros and negative 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 management various problems often arise relating to average percentage rate of change over a period of time. In such cases, the arithmetic mean is not an appropriate average to employ, so, that we can use geometric mean in such case. GM are highly useful in the construction of index numbers. Geometric Mean (GM) = n x 1 x x 2 x ...........x x n When the number of items in the series is larger than 3, the process of computing GM is difficult. To over come this, a logarithm of each size is obtained.

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

 log  log 2  ................  log n  (GM) = Antilog  1  Anti log n   Merits of GM a. It is based on all the observations in the series. b. It is rigidly defined. c. It is best suited for averages and ratios. d. It is less affected by extreme values. e. It is useful for studying social and economics data. Demerits of GM a. It is not simple to understand. b. It requires computational skill. c. GM cannot be computed if any of item is zero or negative. d. It has restricted application. Ex - 1: a. Find the GM of 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 GM of data 2, 4, 8 using logarithms. Data: x1 = 2 x2 = 4 x3 = 8 N=3

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  log x i  i    1 N 

x

log x

2

0.301

4

0.602

8

0.903 logx = 1.806

  log x  GM = Antilog    N  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 Year

OH Expenses (x)

log x

2002

Base year

–

2003

132

2.126

2004

140

2.146

2005

150

2.176  log x = 6.448

  log x  GM = Antilog    N   6.448  GM = Antilog    3 

GM = 141.03 GM for discrete series GM for discrete series is given with usual notations as month:

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  log x i  i    1 N 

GM = Antilog Ex - 3:

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

Sales

I

10000

II

8000

III

12000

IV

15000

Let Base year = 100% sales. Solution:

(Rs)

Increase / decrease %ge

Conversion (x)

log (x)

100%

10000

–

–

–

II

– 20%

8000

80

80

1.903

III

+ 50%

12000

130

130

2.113

IV

+ 25%

15000

155

155

2.190

Month

Base year

I

Sales

logx = 6.206

 6.206  GM = Antilog  = 117.13  3  Average sales = 117.13 – 100 = 14.46% Ex - 4: Find GM for following 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

f = N = 22

 f log x =47.23

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  f log x  GM = Antilog    N   47.23  GM = Antilog   22  GM = 140.212 Geometric Mean for continuous series Steps: 1. Find mid value m and take log of m for each mid 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 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 – 10

3

5.5

0.740

2.220

11 – 20

16

15.5

1.190

19.040

21 – 30

26

25.5

1.406

36.556

31 – 40

31

35.5

1.550

48.050

41 – 50

16

45.5

1.658

26.528

51 – 60

8

55.5

1.744

13.954

f = N = 100

 f log m = 146.348

  f log m  GM = Antilog   N  

146.348  GM = Antilog   100  GM = 29.07

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

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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 based on all observations. 2. It is rigidly defined. 3. It is suitable in case of series having wide dispersion. 4. It is suitable for further mathematical treatment. Demerits of Harmonic Mean 1. It is not easy to compute. 2. Cannot used when one of the item is zero. 3. It cannot represent distribution. Ex: 1. The daily income of 05 families in a very rural village are given below. 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 = 67.72 0.0738

HM = 67.72

7

x

= 0.0738

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

1

48

0.0208

40

0.025

32

0.0312

1

x

x

= 0.0770 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  =

wx = 1440  wx w

1440 37

x  38.91

Both the same HM and WM are same.

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3. Find HM for the following data. Class (CI)

Frequency (f)

Mid point (m)

1 Reciprocal   m

1 f  m

0 – 10

5

5

0.2

1

10 – 20

15

15

0.0666

0.999

20 – 30

25

25

0.04

1

30 – 40

8

35

0.0285

0.228

40 – 50

7

45

0.0222

0.1554

f = 60

HM =

=

1  f   = 3.3824 m

N 1 f  m 60 3.3824

HM = 17.73

Relationship between Mean, Geometric Mean and Harmonic Mean. 1. If all the items in a variable are the same, the arithmetic mean, harmonic mean and Geometric mean are equal. i.e., x  GM  HM . 2. If the size vary, mean will be greater than GM and GM will be greater 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. The number 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 central position. Thus, Median is defined as the mid value of the variants. If the 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. Ex: If 9 students are stand in the order of their heights; the 5th student from either side shall be the one whose height will be Median height of the students group. Thus, median of group is given by an equation.

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 N  1 Median =    2  Ex 1. Find the median for following data. 22

20

25

31

26

24

23

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

22

23

25

24

26

31

 N  1 th  7  1 8 th Median is given by  item =    = 4 Median = 4 item. 2 2     2. Find median for following data. 20

21

22

24

28

32

 N  1 th  6  1 Median is given by  item =  Median = 3.5th item.    2   2 

The item lies between 3rd and 4. So, there are two values 22 and 24. The median value will be the mean values of these two values.  22  24  Median =   = 23  2 

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 size is first located with reference to the cumulative frequency which covers the size first. Then, against that cumulative frequency, the value will be located as median.

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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. Against 58 c.f. the value is 50 which is median value

N = 68

Ex: In a class 15 students, 5 students were failed in a test. The marks of 10 students who have ed were 9, 6, 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

f = 15 Median = Me =

N  1th item 2

15  1 = 8th 2

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

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Continuous Series The procedure is different to get median in continuous series. The class intervals are already in the form of array and the frequency are recorded against each th n class interval. For determining the size, we should take 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 value is to be interpolated using formula given below. h N   C  f 2 

Median =  

 0  1 where,  0 is left end point of N/2 class and l1is right end 2 point of previous class.

Where  

h = Class width, f = frequency of median clas C = Cumulative frequency of class preceding the median class. Ex: Find the median for following data. The class marks obtained by 50 students are as follows. Cum. frequency (cf)

CI

Frequency (f)

10 – 15

6

6

15 – 20

18

24

20 – 25

9

33  N/2 class

25 – 30

10

43

30 – 35

4

47

35 – 40

3

50

f = N = 50 N 50   25 2 2

Cum. frequency just above 25 is 33 and hence, 20 – 25 is median class. 

 0  1 2

20  20  20 2   20

h = 20 – 15 = 5

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f=9 c = 24

h N   C  f 2 

Median =  

5 25  24 9

Median = 20  = 20 

5 9

Median = 20.555 Ex: Find the median for following 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 – 5 = 110 (lower limit) 115 – 5 = 120 (upper limit) similarly for all mid values we can get CI. Cum. frequency (cf)

CI

Frequency (f)

110 – 120

6

6

120 – 130

25

31

130 – 140

48

79

140 – 150

72

151

150 – 160

116

267

160 – 170

60

327

170 – 180

38

365

180 – 190

22

387

190 – 200

3

390

f = N = 390 N 390  2 2

 195 Cum. frequency just above 195 is 267.

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N/2 class

Median class = 150 – 160  =

150  150 = 150 2

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

h f

Median = 150 

N   2  C 10 195 151 116

Median = 153.8 Merits of Median a. It is simple, easy to compute and understand. b. It’s value is not affected by extreme variables. c. It is capable for further algebraic treatment. d. It can be determined by inspection for arrayed data. e. It can be found graphically also. f. It indicates the value of middle item. Demerits of Median a. It may not be representative value as it ignores extreme values. b. It can’t be determined precisely when its size falls between the two values. c. It is not useful in cases where large weights are to be given to extreme values.

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