Mean of Grouped Data

If the values of the variable (i.e., observations or variates) be x\(_{1}\), x\(_{2}\), x\(_{3}\), x\(_{4}\), ....., x\(_{n}\) and their corresponding frequencies are f\(_{1}\), f\(_{2}\), f\(_{3}\), f\(_{4}\), ....., f\(_{n}\) then the mean of the data is given by

Mean = A (or \(\overline{x}\)) = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3}  + x_{4}f_{4} + .... + x_{n}f_{n}}{ f_{1} +  f_{2} + f_{3} + f_{4} + ..... + f_{n}}\)

Symbolically, A = \(\frac{\sum{x_{i}. f_{i}}}{\sum f_{i}}\); i = 1, 2, 3, 4, ...., n.

In words,

Mean = \(\frac{\textbf{Sum of products of the Variables and their corresponding Frequencies}}{\textbf{Total Frequency}}\)

This is the formula for finding the mean of the grouped data by direct method. 

For Example:

The number of Mobile sold is given in the table below. Find the mean of the number of Mobile sold.

Number of Mobile Sold

2

5

6

10

12

Number of Shops

6

10

8

1

5

Solution:

Here, x\(_{1}\) = 2, x\(_{2}\) = 5, x\(_{3}\) = 6, x\(_{4}\) = 10, x\(_{5}\) = 12.

f\(_{1}\) = 6, f\(_{2}\) = 10, f\(_{3}\) = 8, f\(_{4}\) = 1, f\(_{5}\) = 5.

Therefore, mean = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3} + x_{4}f_{4} + x_{5}f_{5}}{f_{1} + f_{2} + f_{3} + f_{4} + f_{5}}\)

                         = \(\frac{2 × 6 + 5 × 10 + 6 × 8 + 10 × 1 + 12 × 5}{6 + 10 + 8 + 1 + 5}\)

                         = \(\frac{12 + 50 + 48  10 + 60}{30}\)

                         = \(\frac{180}{30}\)

                         = 6.

Therefore, mean number of Mobile sold is 6.


Short-cut method for finding the mean of grouped data:

We know that the direct method of finding mean for grouped data gives

mean A =  \(\frac{\sum{x_{i}. f_{i}}}{\sum f_{i}}\)

where x\(_{1}\), x\(_{2}\), x\(_{3}\), x\(_{4}\), ....., x\(_{n}\) are variates and f\(_{1}\), f\(_{2}\), f\(_{3}\), f\(_{4}\), ....., f\(_{n}\)  are their corresponding frequencies.

Let a = a number taken as assumed mean from which the diviation of the variate is di = xi - a.

Then, A =\(\frac{\sum{(a + d_{i})f_{i}}}{\sum f_{i}}\)

            = \(\frac{\sum{af_{i}} + \sum{d_{i}f_{i}}}{\sum f_{i}}\)

            =  \(\frac{a\sum{f_{i}} + \sum{d_{i}f_{i}}}{\sum f_{i}}\)

            = a + \(\frac{\sum{d_{i}f_{i}}}{\sum f_{i}}\)

Therefore, A = a + \(\frac{\sum{d_{i}f_{i}}}{\sum f_{i}}\), where di = xi - a.


For Example: 

Find the mean of the following distribution using the short-cut method.


Variate

20

40

60

80

100

Frequency

15

22

18

30

16


Solution:

Putting the calculated values in a tabular form, we have the following.

Variate

Frequency

Deviation di from assumed mean a = 60, i.e., (xi - a)

dixi

20

15

-40

-600

40

22

-20

-440

60

18

0

0

80

30

20

600

100

16

40

640


\(\sum f_{i}\) = 101


\(\sum d_{i}f_{i}\) = 200


Therefore, mean A = a + \(\frac{\sum{d_{i}f_{i}}}{\sum f_{i}}\)

                            = 60 + \(\frac{200}{101}\)

                            = 61\(\frac{99}{101}\)

                            = 61.98.


Solved Examples on Mean of Grouped Data or Mean of the Arrayed Data:

1. A class has 20 students whose ages (in years) are as follows.

14, 13, 14, 15, 12, 13, 13, 14, 15, 12, 15, 14, 12, 16, 13, 14, 14, 15, 16, 12

Find the mean ago of the students of the class.

Solution:

In the data, only five different numbers appear respectively. So, we write the frequencies of the variates as below.


Age (in years)

(x\(_{i}\))

12

13

14

15

16

Total

Number of Students

(f\(_{i}\))

4

4

6

4

2

20


Therefore, mean A = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3}  + x_{4}f_{4} + x_{5}f_{5}}{ f_{1} +  f_{2} + f_{3} + f_{4} + f_{5}}\)

= \(\frac{12 × 4 + 13 × 4 + 14 × 6 + 15 × 4 + 16 × 2}{4 + 4 + 6 + 4 + 2}\)

= \(\frac{48 + 52 + 84 + 60 + 32}{20}\)

= \(\frac{276}{20}\)

= 13.8

Therefore, the mean age of the students of the class = 13.8 years.


2. The weights (in kg) of 30 boxes are as given below.

40, 41, 41, 42, 44, 47, 49, 50, 48, 41, 43, 45, 46, 47, 49, 41, 40, 43, 46, 47, 48, 48, 50, 50, 40, 44, 44, 47, 48, 50.

Find the mean weight of the boxes by preparing a frequency table of the arrayed data.

Solution:

The frequency table for the given data is 

Weight (in Kg)

(xi)

Tally Mark

Frequency

(fi)

xifi

40

///

3

120

41

////

4

164

42

/

1

42

43

//

2

86

44

///

3

132

45

/

1

45

46

//

2

92

47

////

4

188

48

////

4

192

49

//

2

98

50

////

4

200

\(\sum f_{i}\) = 30

\(\sum x_{i}f_{i}\) = 1359

By formula, mean = \(\frac{\sum{x_{i}f_{i}}}{\sum f_{i}}\)

                           = \(\frac{1359}{30}\)

                           = 45.3.

Therefore, the mean weight of the boxes = 45.3 kg.


3. Four variates are 2, 4, 6 and 8. The frequencies of the first three variates are 3, 2 and 1 respectively. If the mean of the variates is 4 then find the frequency of the fourth variate. 

Solution:

Let the frequency of the fourth variate (8) be f. Then, 

mean A = \(\frac{x_{1}f_{1} + x_{2}f_{2} + x_{3}f_{3}  + x_{4}f_{4}}{ f_{1} +  f_{2} + f_{3} + f_{4}}\)

⟹ 4 = \(\frac{2 × 3 + 4 × 2 + 6 × 1 + 8 × f}{3 + 2 + 1 + f}\)

⟹ 4 = \(\frac{6 + 8 + 6 + 8f}{6 + f}\)

⟹ 24 + 4f = 20 + 8f

⟹ 4f = 4

⟹ f = 1

Therefore, the frequency of 8 is 1.

Formula for Finding the Mean of the Grouped Data

4. Find the mean of the following data.


Variate (x) 

1

2

3

4

5

Cumulative Frequency

3

5

9

12

15


Solution:

The frequency table and calculations involved in finding the mean are given below.

Variate

(xi)

Cumulative Frequency

Frequency

(fi)

xifi

1

3

3

3

2

5

2

4

3

9

4

12

4

12

3

12

5

15

3

15

\(\sum f_{i}\) = 15

\(\sum x_{i}f_{i}\) = 46

Therefore, mean = \(\frac{\sum{x_{i}f_{i}}}{\sum f_{i}}\)

                         = \(\frac{46}{15}\)

                         = 3.07.


5. Find the mean mark from the following frequency table by using the short-cut method.


Marks Obtained

30

35

40

45

50

Number of Students

45

26

12

10

7


Solution:

Taking the assumed mean a = 40, the calculations will be as follows.

Marks Obtained

(xi)

Number of Students

(fi)

Deviation di = xi - a = xi - 40

difi

30

45

-10

-450

35

26

-5

-130

40

12

0

0

45

10

5

50

50

7

10

70

\(\sum f_{i}\) = 100

\(\sum d_{i}f_{i}\) = -460

Therefore, mean = a + \(\frac{\sum{d_{i}f_{i}}}{\sum f_{i}}\)

                         = 40 + \(\frac{-460}{100}\)

                         = 40 - 4.6

                         = 35.4.

Therefore, the mean mark is 35.4.






9th Grade Math

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