Session – 6

Measures of Central Tendency

ModeIt is the value which occurs with the maximum frequency. It is the most

typical or common value that receives the height frequency. It represents fashion and often it is used in business. Thus, it corresponds to the values of variable which occurs most frequently. The model class of a frequency distribution is the class with highest frequency. It is denoted by ‘z’.

Mode is the value of variable which is repeated the greatest number of times in the series. It is the usual, and not casual, size of item in the series. It lies at the position of greatest density.

Ex: If we say modal marks obtained by students in class test is 42, it means that the largest number of student have secured 42 marks.

If each observations occurs the same number of times, we can say that there is ‘no mode’. If two observations occur the same number of times, we can say that it is a ‘Bi-modal’. If there are 3 or more observations occurs the same number of times we say that ‘multi-modal’ case. When there is a single observation occurs mot number of times, we can say it is ‘uni-modal’ case.

For a grouped data mode can be computed by following equations with usual notations.

Mode =

where,

fm = max frequency (modal class frequency)

f1 = frequency preceding to modal class.

f2 = frequency succeeding to modal class

h = class width.

or

Mode =

1

Ex:

1. Find the modal for following data.

Marks

(CI)

No. of students

(f)

1 – 10 3

11 – 20 16

21 – 30 26

31 – 40 31 Max. frequency

41 – 50 16

51 – 60 8

f = N = 100

We shall identify the modal class being the class of maximum frequency. i.e. 31-40.

where,

fm = 31

f1 = 26

f2 = 16

h = 10

30.5

Mode (z) =

Mode =

Mode = 33.

Or

Mode = =

2

Mode = 34.30

It can be noted that there exists slightly different mode value in the second method.

Partition values Median divides in to two equal parts. There are other values also which

divides the series partitioned value (PV).

Just as one point divides as series in to two equal parts (halves), 3 points divides in to four points (Quartiles) 9 points divides in to 10 points (deciles) and 99 divide in to 100 parts (percentage). The partitioned values are useful to know the exact composition of series.

Quartiles

A measure, which divides an array, in to four equal parts is known as quartile. Each portion contain equal number of items. The first second and third point are termed as first quartile (Q1). Second quartile (Q2) and third quartile (Qs). The first quartile is also known as lower quartiles as 25% of observation of distribution below it, 75% of observations of the distribution below it and 25% of observation above it.

Calculation of quartiles

Q1 = size of

Q2 = size of

Q2 = (median) =

Measures of quartiles The quartile values are located on the principle similar to locating the median

value.

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Following table shows procedure of locating quartiles.

Measure Individual and Discrete senses Continuous series

Q1

Q2

Q3

Ex - 1: From the following marks find Q1, Median and Q3 marks

23, 48, 34, 68, 15, 36, 24, 54, 65, 75, 92, 10, 70, 61, 20, 47, 83, 19, 77

Let us arrange the data in array form.

Sl. No.

x

1. 10

2. 15

3. 19

4. 20

5. 23 Q1

6. 24

7. 34

8. 36

9. 47

10. 48 Q2

11. 54

12. 61

13. 65

14. 68

15. 70 Q3

16. 75

17. 77

18. 83

19. 92

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a. Q1 =

Q1 = Here, n = 19 items

Q1 =

Q1 = 5th item

Q1 = 23

b. Q2 =

Q2 =

10th item

Q2 = 48

c. Q3 =

Q3 = = 15th item

Q3 = 70

Ex - 2: Locate the median and quartile from the following data.

Size of shoes 4 4.5 5 5.5 6 6.5 7 7.5 8

Frequencies 20 36 44 50 80 30 30 16 14

X f cf

4 20 20

4.5 36 56

5 44 100 Q1

5.5 50 150

6 80 230 Q2

6.5 30 260 Q3

7 30 290

7.5 16 306

8 14 320

N = f = 320

5

Q1 =

Q1 =

Q1 = 80.25th item

Just above 80.25, the cf is 100. Against 100 cf, value is 5.

Q1 = 5

Q2 =

Q2 =

160.5th item

Just above 160.5, the cf is 230. Against 230 cf value is 6.

Q2 = 6

Q3 =

Q3 = 321 = 240.75th item

Just above 240.75, the cf is 260. Against 260 cf value is 6.5.

Q3 = 6.5

Ex - 3: Compute the quartiles from the following data.

CI 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80

Frequency (f)

5 8 7 12 28 20 10 10

First quartile (Q1) = and Q3 =

and (Q2) = Median =

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CI f cf

0-10 5 5

10-20 8 13

20-30 7 20

30-40 12 32 Q1

40-50 28 60 Q2

50-60 20 80 Q3

60-70 10 90

70-80 10 100

N = f = 100

a. First locate Q1 for ¼ N

¼ N = 25

= 30

h = 10

f = 12

c = 20

(Q1) =

=

Q1 =

Q1 = 34.16

b. Locate Q2 (Median)

Q2 corresponds to N/2 = 50,

Q2 =

Q2 =

Q2 = 46.42

Q3 corresponds to ¾ N = 75,

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Q3 =

Q3 =

Q3 = 57.5

Deciles The deciles divide the arrayed set of variates into ten portions of equal

frequency and they are some times used to characterize the data for some specific purpose. In this process, we get nine decile values. The fifth decile is nothing but a median value. We can calculate other deciles by following the procedure which is used in computing the quartiles.

Formula to compute deciles.

Percentiles Percentile value divides the distribution into 100 parts of equal frequency. In

this process, we get ninety-nine percentile values. The 25th, 50th and 75th percentiles are nothing but quartile first, median and third quartile values respectively.

Formula to compute percentiles is given below:

P25 = P26 = and so, on

Ex:

Find the decile 7 and 60th percentile for the given data of patients visited to hospital on a particular day.

CI f Cf

10-20 1 1

20-30 3 4

30-40 11 15

40-50 21 36

50-60 43 79 P60

60-70 32 111 D70

70-80 9 120

f = N = 120

a. D7 =

8

h = 10, f = 32

c = 79

D7 =

7th Decile = D7 = 61.562

b. 60th percentile

P60 =

h = 10

f = 43

c = 36

P60 =

P60 =

P60 = 58.37

SOME NUMERICAL EXAMPLES1. Show that following distribution is symmetrical about the average. Also shows

that median is the mid-way between lower and upper quartiles.

X 2 3 4 5 6 7 8 9 10

Frequency 2 9 29 57 80 57 29 9 2

To show the given distribution is symmetrical, Mean, Median and Mode must be same.

To show median is mid-way between the lower and upper quartile i.e., Q2 – Q1

= Q3 – Q2.

Mid-point Class interval f d = (x – 6) fd cf

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x CI Cum. freq.

2 1.5 – 2.5 2 -4 -8 2

3 2.5 – 3.5 9 -3 -27 11

4 3.5 – 4.5 29 -2 -58 40

5 4.5 – 5.5 57 -1 -57 97 Q1 class

6 5.5 – 6.5 80 0 0 177 Q2 class

7 6.5 – 7.5 57 1 57 234 Q3 class

8 7.5 – 8.5 29 2 58 263

9 8.5 – 9.5 9 3 27 272

10 8.5 – 10.5 2 4 8 274

N=274 fd = 0

Let A = 6

Mean =

=

Mean = 6.

Median

Q2 =

C = 97

Q2 =

Q2 = 5.5 + 0.5

Median = Q2 = 6.

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Mode

Mode = Modal class 5.5 – 6.5

Mode =

Mode = 6.

Since, Mean = Mode = Median. The given distribution is symmetrical.

Q1 calculation

Q1 =

Q1 =

Q1 = 7.

Now, Q2 – Q1 = Q3 – Q2

i.e. 6 – 5 = 7 – 5

2 = 2

2. Find the mean for the set of observations given below.

6, 7, 5, 4

=

3. Find the mean for the following data.

CI f xi fx

0-10 3 5.5 16.5

11-20 16 15.5 248

21-30 26 23.5 683

31-40 31 35.5 1180.5

41-50 16 45.5 728

51-60 8 55.5 444

N = f = 100 3300

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4. Find the mean profit of the organisation for the given data below:

Profit CI f xi fx

100-200 10 150 1500

200-300 18 250 4500

300-400 20 350 7000

400-500 26 450 11700

500-600 30 550 16500

600-700 28 650 18200

700-800 18 750 13500

N = f = 150 72900

x1 =

x1 =

x1 = 150

=

486

Step Deviation Method

x = a + hd d =

a = Arbitrary constant

h = class width

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Profit CI f xi d fd

100-200 10 150 -3 -30

200-300 18 250 -2 -36

300-400 20 350 -1 -20

400-500 26 450 0 0

500-600 30 550 +1 30

600-700 28 650 +2 56

700-800 18 750 +3 34

N = f = 150 fm = 54

5. In an office there are 84 employees and there salaries are given below.

Salary 2430 2590 2870 3390 4720 5160

Employees 4 28 31 16 3 2

1. Find the mean salary of the employees

2. What is the total salary of the employees?

=

Rs. 2975.36

1. 2975.36

2. Total salary = 2,49,930 (Rs.)

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6. The average marks secured by 36 students was 52 but it was discovered that on item 64 was misread as 46. Find the correct me of the marks.

fx = 52 x 36 = 1872

fx = fx - incorrect + correct

correct = 1872 – 46 64 = 1890

52.5

7. The mean of 100 items is 46, later it was discovered that an item 16 was misread as 61 and another item 43 was misread as 34 and also found that the total number of items are 90 not 100 find the correct mean value.

fx = 4600

fx = fx - incorrect + correct

= 4600– 61 - 34 + 16 + 43

= 4564

= 50.71

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8. Calculate the mean for the following data.

Value Frequency

< 10 4

< 20 10

< 30 15

< 40 25

< 50 30

CI f ‘m’ mid point fm

0-10 4 5 20

10-20 10 15 150

20-30 15 25 375

30-40 25 35 875

40-50 30 45 1350

f = 84 fx 2770

32.97

9. For a given frequency table, find out the missing data. The average accident are 1.46.

No. of accidents Frequency

0 46

1 ?

2 ?

3 25

4 10

5 5

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No. of accidents (x)

Frequency

(f)fx

0 46 0

1 ? f1

2 ? 2f1

3 25 75

4 10 40

5 5 25

N = 200 fx = 140 + f1 + 2f2

1.46 =

292 = 140 + f1 + 2f2

f1 + 2f2 = 152 ----(1)

w.k.t. N = f

200 = 86 + f1 + f2

f1 + f2 = 114 ----(2)

f1 + 2f2 = 152 ----(1)

f1 + f2 = 114 ----(2) (1) – (2)

---------------------------------

f2 = 38

---------------------------------

f 2 = 38

f1 + f2 = 114

f1 + 114 – 38

f1 = 76

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10. Find out the missing values of the variate for the following data with mean is 31.87.

xi F

12 8

20 16

27 48

33 90

? 30

54 8

N = 200

xi f fx

12 8 96

20 16 320

27 48 1296

33 90 2970

x 30 30x

54 8 432

N = 200 fx = 5114 + 30x

31.87

fx = 6374 ----(1)

fx = 5114 + 30x ----(2)

(1) = (2)

6374 = 5114 + 30x

6374 - 5114 = 30x

30x = 1260

x = 42.

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11. The average rainfall of a city from Monday to Saturday is 0.3 inches. Due to heavy rainfall Sunday the average rainfall for the week increased to 0.5 inches. What is the rainfall on Sunday?

Given: Mon – Sat = 0.3”

Sun = 0.5”

fx1 = 1.8

fx2 = 3.5

Rainfall on Sunday = fx2 – fx1

= 3.5 – 1.8

= 1.7”

12. The average salary of male employees in a firm was Rs. 520 and that of females Rs. 420 the mean of salary of all the employees as a whole is Rs. 500. Find the percentage of male and female employees.

Given:

n1 = Male persons. n2 = Female persons.

500n1 + 500n2 = 520n1 – 420n2

80n2 = 20n1

n1 = 4n2

Let n1 + n2 = 100

4n2 + n2 = 100

5n2 = 100

n2 = 20% Female

n1 = 80% Male

20% and 80% are male and females in the firm.

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13. The A-M of two observations is 25 and there GM is 15. Find the HM.

Given:

AM = 25

a + b = 50

GM = 15

GM =

GM =

15 =

(15)2 = ( )2

ab = 225

HM = ?

HM =

HM =

HM =

HM = 9

a + b = 50

ab = 225

a =

HM = 9

14. The GM is 60 an HM is 28.24. Find AM for two observations.

AM GM HM

= 127.475

60 =

602 = ab

ab = 3600

28.24 =

a + b =

=

a + b = 254.95

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15. Calculate the missing frequency from the data if the median is 50.

CI f cf

10-20 2 2

20-30 8 10

30-40 6 16

40-50 ? f1 16+f1

50-60 15 31+f1 median class

60-70 10 41+f1

f = 41 + f1

Q =

50 = 50 + 50 – 50 =

0 = 0 =

0 =

16 + f1 = ½ (41 + f1)

2 (16 + f1) = 41 + f1

32 + 2f1 = 41 + f1

f1 = 9

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SOURCES AND REFERENCES

1. Statistics for Management, Richard I Levin, PHI / 2000.

2. Statistics, RSN Pillai and Bagavathi, S. Chands, Delhi.

3. An Introduction to Statistical Method, C.B. Gupta, & Vijaya Gupta, Vikasa Publications, 23e/2006.

4. Business Statistics, C.M. Chikkodi and Salya Prasad, Himalaya Publications, 2000.

5. Statistics, D.C. Sancheti and Kappor, Sultan Chand and Sons, New Delhi, 2004.

6. Fundamentals of Statistics, D.N. Elhance and Veena and Aggarwal, KITAB Publications, Kolkata, 2003.

7. Business Statistics, Dr. J.S. Chandan, Prof. Jagit Singh and Kanna, Vikas Publications, 2006.

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