Class 9 Measures of Central Tendency
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Class 9 Chapter - 14 STATISTICS

Measures of Central Tendency

Topic covered

`color{red} ♦` Measures of Central Tendency

Measures of Central Tendency

Previously, we represented the data in various forms through frequency distribution tables, bar graphs, histograms and frequency polygons.

Now, the question arises if we always need to study all the data to ‘make sense’ of it, or if we can make out some important features of it by considering only certain representatives of the data.

This is possible, by using measures of central tendency or averages.

Consider a situation when two students Mary and Hari received their test copies. The test had five questions, each carrying ten marks. Their scores were as follows:



Upon getting the test copies, both of them found their average scores as follows:

Mary’s average score ` = (42)/5 = 8.4`

Hari’s average score ` = (41)/5 = 8.2`

Since Mary’s average score was more than Hari’s, Mary claimed to have performed better than Hari, but Hari did not agree. He arranged both their scores in ascending order and found out the middle score as given below:




Hari said that since his middle-most score was 10, which was higher than Mary’s middle-most score, that is 8, his performance should be rated better.

But Mary was not convinced. To convince Mary, Hari tried out another strategy. He said he had scored 10 marks more often (3 times) as compared to Mary who scored `10` marks only once. So, his performance was better.

Now, to settle the dispute between Hari and Mary, let us see the three measures they adopted to make their point.

The average score that Mary found in the first case is the mean. The ‘middle’ score that Hari was using for his argument is the median. The most often scored mark that Hari used in his second strategy is the mode.

Now, let us first look at the mean in detail.

The mean (or average) of a number of observations is the sum of the values of all the observations divided by the total number of observations.

It is denoted by the symbol `bar x` , read as ‘x bar’.
Q 3220078811

`5` people were asked about the time in a week they spend in doing social work in their community. They said `10, 7, 13, 20` and `15` hours, respectively. Find the mean (or average) time in a week devoted by them for social work.
Class 9 Chapter 14 Example 10
Solution:

We have already studied in our earlier classes that the mean of a certain

number of observations is equal to ` text( Sum of all the observations)/text(Total number of observations)` . To simplify our

working of finding the mean, let us use a variable xi to denote the ith observation. In
this case, i can take the values from `1` to `5`. So our first observation is `x_1`, second
observation is `x_2`, and so on till `x_5`.
Also `x_1 = 10` means that the value of the first observation, denoted by `x_1,` is `10`.

Similarly, `x_2 = 7, x_3 = 13, x_4 = 20` and `x_5 = 15`.

Therefore, the mean ` bar x = text(Sum of all the observations)/text(Total number of observations)`

` = ( x_1 + x_2 + x_3 + x_4 + x_5)/5`

` = ( 10 + 7 + 13 + 20 + 15)/5 = (65)/5 = 13`

So, the mean time spent by these `5` people in doing social work is ` 13` hours in a week.
Now, in case we are finding the mean time spent by `30` people in doing social
work, writing `x_1 + x_2 + x_3 + . . . + x_(30)` would be a tedious job.We use the Greek symbol
`Σ` (for the letter Sigma) for summation. Instead of writing `x_1 + x_2 + x_3 + . . . + x_(30)`, we
write ` Σ_(i - 1)^(30) x_i` which is read as ‘the sum of `x_i` as `i ` varies from `1` to `30`’.

So, ` bar x = ( Σ_(i - 1)^(30) x_i)/(30)`

Similarly, for n observations ` bar x = ( Σ_(i - 1)^(n) x_i)/(n)`
Q 3250078814

Find the mean of the marks obtained by `30` students of Class IX of a school, given in Example `2`.
Class 9 Chapter 14 Example 11
Solution:

Now, ` bar x = ( x _1 + x_2 + x_3 + ......... + x_(30))/(30)`

` sum _( i - 1)^(30) x_i = 10 + 20 + 36 + 92 + 95 + 40 + 50 + 56 + 60 + 70 + 92 + 88`
`80 + 70 + 72 + 70 + 36 + 40 + 36 + 40 + 92 + 40 + 50 + 50`
`56 + 60 + 70 + 60 + 60 + 88 = 1779`

So, ` bar x = ( 1779)/(30) = 59.3`

Is the process not time consuming? Can we simplify it? Note that we have formed
a frequency table for this data (see Table 14.1).
The table shows that `1` student obtained `10` marks, `1` student obtained `20` marks, `3`
students obtained `36` marks, `4 ` students obtained `40` marks, `3` students obtained `50`
marks, `2` students obtained `56` marks, `4` students obtained `60` marks, `4` students obtained
`70` marks, `1` student obtained `72` marks, `1` student obtained `80` marks, `2` students obtained
`88` marks, `3` students obtained `92` marks and `1` student obtained `95` marks.
So, the total marks obtained `= (1 xx 10) + (1 xx 20) + (3 xx 36) + (4 xx 40) + (3 xx 50)`
`+ (2 xx 56) + (4 xx 60) + (4 xx 70) + (1 xx 72) + (1 xx 80)`
`+ (2 xx 88) + (3 xx 92) + (1 xx 95)`
`= f_1 x_1 + . . . + f_(13) x_(13)`, where `f_i` is the frequency of the ith
entry in Table 14.1.

In brief, we write this as ` sum_( i -1)^(13) f_i x_i`

So, the total marks obtained ` = sum_( i -1)^(13) f_i x_i`

`= 10 + 20 + 108 + 160 + 150 + 112 + 240 + 280 + 72 + 80`
`+ 176 + 276 + 95`
`= 1779`
Now, the total number of observations

` = sum_( i -1)^(13) f_i x_i`

`= f_1 + f_2 + . . . + f_(13)`
`= 1 + 1 + 3 + 4 + 3 + 2 + 4 + 4 + 1 + 1 + 2 + 3 + 1`
`= 30`

So , the mean `bar x = text( Sum of all the observations) /text( Total number of observations) = (( sum_( i -1)^(13) f_i x_i)/( sum_( i -1)^(13) f_i))`

` = ( 1779)/(30) = 59.3`

This process can be displayed in the following table, which is a modified form of
Table 14.1.

Thus, in the case of an ungrouped frequency distribution, you can use the formula

` bar x = ( sum_( i -1)^(13) f_i x_i)/( sum_( i -1)^(n) f_i )`

for calculating the mean.
Let us now move back to the situation of the argument between Hari and Mary,
and consider the second case where Hari found his performance better by finding the
middle-most score. As already stated, this measure of central tendency is called the
median.
The median is that value of the given number of observations, which divides it into
exactly two parts. So, when the data is arranged in ascending (or descending) order
the median of ungrouped data is calculated as follows:

(i) When the number of observations (n) is odd, the median is the value of the
` ((n +1)/2)^(th)` observation. For example, if `n = 13`, the value of the ` ((13 + 1)/2)^(th)` .i.e
the 7th observation will be the median


(ii) When the number of observations (n) is even, the median is the mean of the
`(n/2)^(th) ` and the `(n/2 + 1)^(th)` observations. For example, if `n = 16`, the mean of the
values of the ` ((16)/2)^(th)` and the ` ((16)/2 + 1 )^(th)` observations, i.e., the mean of the
values of the 8th and 9th observations will be the median
Q 3260078815

The heights (in cm) of `9` students of a class are as follows:
155 160 145 149 150 147 152 144 148
Find the median of this data.
Class 9 Chapter 14 Example 12
Solution:

First of all we arrange the data in ascending order, as follows:

144 145 147 148 149 150 152 155 160

Since the number of students is `9`, an odd number, we find out the median by finding

the height of the ` ( (n +1)/2) th = ( (9 +1)/2) th = ` the `5`th student, which is `149` cm.
So, the median, i.e., the medial height is `149` cm.
Q 3270078816

The points scored by a Kabaddi team in a series of matches are as follows:
`"17, 2, 7, 27, 15, 5, 14, 8, 10, 24, 48, 10, 8, 7, 18, 28"`
Find the median of the points scored by the team.
Class 9 Chapter 14 Example 13
(A)

11

(B)

12

(C)

14

(D)

10

Solution:

Arranging the points scored by the team in ascending order, we get

`"2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48"`.

There are 16 terms. So there are two middle terms, i.e. the ` (16)/2 th` and ` ((16)/2 + 1) th `, i.e

the `8` th and `9` th terms.

So, the median is the mean of the values of the `8`th and `9`th terms.

i.e, the median ` = ( 10+ 14)/2 = 12`

So, the medial point scored by the Kabaddi team is `12`.

Let us again go back to the unsorted dispute of Hari and Mary.

The third measure used by Hari to find the average was the mode.

The mode is that value of the observation which occurs most frequently, i.e., an
observation with the maximum frequency is called the mode.

The readymade garment and shoe industries make great use of this measure of
central tendency. Using the knowledge of mode, these industries decide which size of
the product should be produced in large numbers.
Let us illustrate this with the help of an example
Correct Answer is `=>` (B) 12
Q 3280078817

Find the mode of the following marks (out of `10`) obtained by `20` students:
`"4, 6, 5, 9, 3, 2, 7, 7, 6, 5, 4, 9, 10, 10, 3, 4, 7, 6, 9, 9"`
Class 9 Chapter 14 Example 14
(A)

8

(B)

7

(C)

9

(D)

6

Solution:

We arrange this data in the following form :

`"2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9, 10, 10"`

Here `9` occurs most frequently, i.e., four times. So, the mode is `9`.
Correct Answer is `=>` (C) 9
Q 3200078818

Consider a small unit of a factory where there are `5` employees : a supervisor and four labourers. The labourers draw a salary of Rs. `5,000` per month each while the supervisor gets Rs. `15,000` per month. Calculate the mean, median and mode of the salaries of this unit of the factory.
Class 9 Chapter 14 Example 15
(A)

7000 , 5000 , 5000

(B)

6000 , 4000 , 3000

(C)

1000 , 3000 , 2000

(D)

8000 , 9000 , 9000

Solution:

Mean ` = ( 5000 + 5000 + 5000 + 5000 + 15000)/5 = (35000)/5 = 7000`

So, the mean salary is Rs. `7000` per month.

To obtain the median, we arrange the salaries in ascending order:

5000, 5000, 5000, 5000, 15000

Since the number of employees in the factory is `5`, the median is given by the

` ( 5-1)/2` th ` 6/2` th `= 3` th observation. Therefore, the median is Rs. `5000` per month.
To find the mode of the salaries, i.e., the modal salary, we see that `5000` occurs the
maximum number of times in the data `5000, 5000, 5000, 5000, 15000`. So, the modal
salary is Rs. `5000` per month.
Now compare the three measures of central tendency for the given data in the
example above. You can see that the mean salary of Rs `7000` does not give even an
approximate estimate of any one of their wages, while the medial and modal salaries
of Rs. `5000` represents the data more effectively.
Extreme values in the data affect the mean. This is one of the weaknesses of the
mean. So, if the data has a few points which are very far from most of the other
points, (like 1,7,8,9,9) then the mean is not a good representative of this data. Since the
median and mode are not affected by extreme values present in the data, they give a
better estimate of the average in such a situation.
Again let us go back to the situation of Hari and Mary, and compare the three
measures of central tendency.

This comparison helps us in stating that these measures of central tendency are not
sufficient for concluding which student is better. We require some more information to
conclude this, which you will study about in the higher classes
Correct Answer is `=>` (A) 7000 , 5000 , 5000
 
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