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Statistics

Mean of Grouped Data Using the Direct Method

We have studied three measures of central tendency - mean, median, and mode. But do you know that there is an empirical relationship between these three measures.

The relation between them is given by

Mean – Mode = 3(Mean – Median)

or

3 Median = Mode + 2 Mean

We can use this relationship to find the value of any of them, when the other two are known. It can also be used to check if the values of the three central tendencies obtained are correct or not as they will satisfy this relation. However, this relation is not always true.

First of all, let us verify it by taking an example.

Example:

The following data shows the numbers of days for which the patients were hospitalized.

No. of Patients

No. of days attending the hospital

0 − 10

10 − 20

20 − 30

30 − 40

40 − 50

50 − 60

2

6

9

7

4

2

Justify the relationship between mean, median, and mode of the given data.

Solution:

To verify the relation between mean, median, and mode, we have to find out their values from the given data and then substitute them to see the authenticity of the relationship.

Firstly, let us find the mean of the given data.

Mean

To find out the mean, we construct the following table.

No. of Patients

No. of days for which the patients were hospitalized: fi

Class Mark: x­i

fi × x­i

0 − 10

10 − 20

20 − 30

30 − 40

40 − 50

50 − 60

2

6

9

7

4

2

5

15

25

35

45

55

10

90

225

245

180

110

 

= 30

 

= 860

We know that, Mean =

=

Now, let us find the mode of the given set of data.

Mode

In the given table, 20 − 30 is the modal class as this group has the maximum frequency.

From the table,

l = 20 (Lower limit of the modal class)

h = 10 (Class size)

f1 = 9 (Frequency of the modal class)

f0 = 6 (Frequency of the class preceding the modal class)

f2 = 7 (Frequency of the class succeeding the modal class)

Now, Mode =

Mode

Mode = 26

Median

To find out the median, we construct the following table.

No. of Patients

Frequency

Cumulative Frequency

0 − 10

10 − 20

20 − 30

30 − 40

40 − 50

50 − 60

2

6

9

7

4

2

2

8

17

24

28

30

   

30

Here, n = 30

The cumulative frequency greater than 15 is 17.

Hence, the median class is 20 − 30.

From the table,

l = 20 (Lower limit of median class)

cf = 8 (Cumulative frequency of class preceding the median class)

f = 9 (Frequency of median class)

h = 10 (Class size of the groups)

Now, Median

Median =

Median =

Now, Mode + 2 Mean = 26 + 2 ×

= 26 +

=

But, 3 Median = 3 × = = Mode + 2 Mean

Hence it is verified that,

3 Median = Mode + 2 Mean

 

Let us go through few more examples to learn the application of the concept. 

Example 1:

For a certain frequency distribution, mean and median are obtained as 123.5 and 127 respectively. What is the value of mode?

Solution:

We have the relation

3 Median = Mode + 2 Mean

⇒ 3(127) = Mode + 2(123.5) 

⇒ 381 = Mode + 247

⇒ Mode = 381 – 247

⇒ Mode = 134

Example 2:

For a certain frequency distribution, median and mode are obtained as 55 and 70 respectively. What is the value of mean?

Solution:

We have the relation

3 Median = Mode + 2 Mean

⇒ 3(55) = 70 + 2 Mean

⇒ 165 = 70 + 2 Mean

⇒ 2 Mean = 165 – 70

⇒ 2 Mean  = 95

⇒ Mean  = 47.5  

Example 3:

For a certain frequency distribution, mean is 6 more than mode. Find the relation between median and mode. 

Solution:

It is given that

Mean – Mode = 6    ...(1)

We have the relation

Mean – Mode = 3(Mean – Median)

⇒ 6 = 3(Mean – Median)

⇒ Mean – Median = 2    ...(2)

On subtracting (2) from (1), we get

Mean – Mode – (Mean – Median) = 6 – 2

⇒ Mean – Mode – Mean + Median = 4

⇒ Median – Mode = 4   

Thus, median is 4 more than mode.

A cricket player played 7 matches and scored 88, 72, 90, 94, 56, 62, and 76 runs. What is the average score or mean score of the player in these 7 matches?

We know that,

“The mean is the sum of all observations divided by the number of observations”.

We can use the following formula to find the mean of the given data.

Using this formula, the mean of scores obtained by the player

 = 76.86

Thus, the average score of the player is 76.86. There are many situations where the number of observations is very large and then it is not possible to use this formula to find the mean of the given data.

Let us consider such a situation.

We have the marks distribution of 100 students of a class in a Mathematics test. The distribution is given in the form of groups like 0 − 10, 10 − 20 …. To find the mean in such situations, we follow another method. This method is explained in the given video.

Let us solve some more examples to understand the concept better.

Example 1:

The following table shows the various income brackets of the employees of a company.

Income of employees (in Rupees) 10000 − 15000 15000 − 20000 20000 − 25000

25000 −

30000

30000 −

35000

35000 −

40000

40000 −

45000

No. of employees

32

18

13

7

14

5

11

Find the average monthly income of each employee.

Solution:

Firstly, we will calculate the class marks of each class interval and then the product of the class marks with the corresponding frequencies. By doing so, we obtain the following table.

Income of employees

(in Rupees)

No. of employees: fi

Class Mark: xi

xi × fi

10000 − 15000

15000 − 20000

20000 − 25000

25000 − 30000

30000 − 35000

35000 − 40000

40000 − 45000

32

18

13

7

14

5

11

12500

17500

22500

27500

32500

37500

42500

400000

315000

292500

192500

455000

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