Measures of Central Tendency

Measures of Central Tendency

 

Earlier in this chapter, 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:

Question Numbers
1 2
3 4 5
Mary’s score 10 8 9 8 7
Hari’s score 4 7 10 10 10

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

«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨left¨ rowspacing=¨0¨»«mtr»«mtd»«mi»M«/mi»«mi»a«/mi»«mi»r«/mi»«mi»y«/mi»«mo»§apos;«/mo»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»a«/mi»«mi»v«/mi»«mi»e«/mi»«mi»r«/mi»«mi»a«/mi»«mi»g«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»s«/mi»«mi»c«/mi»«mi»o«/mi»«mi»r«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mo»§nbsp;«/mo»«mfrac»«mn»42«/mn»«mn»5«/mn»«/mfrac»«mo»=«/mo»«mn»8«/mn»«mo».«/mo»«mn»4«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»H«/mi»«mi»a«/mi»«mi»r«/mi»«mi»i«/mi»«mo»§apos;«/mo»«mi»s«/mi»«mo»§nbsp;«/mo»«mi»a«/mi»«mi»v«/mi»«mi»e«/mi»«mi»r«/mi»«mi»a«/mi»«mi»g«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»s«/mi»«mi»c«/mi»«mi»o«/mi»«mi»r«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mn»41«/mn»«mn»5«/mn»«/mfrac»«mo»=«/mo»«mn»8«/mn»«mo».«/mo»«mn»2«/mn»«/mtd»«/mtr»«/mtable»«/math»

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:

Mary’s Score 7 8 8
9 10
Hari’s Score 4 7 10 10 10

     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 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»x«/mi»«mo»§#175;«/mo»«/mover»«/math», read as ‘x bar’.
    Let us consider an example.

 

Example  :

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

number of observations is equal to=«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»s«/mi»«mi»u«/mi»«mi»m«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»a«/mi»«mi»l«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»b«/mi»«mi»s«/mi»«mi»e«/mi»«mi»r«/mi»«mi»v«/mi»«mi»a«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«/mrow»«mrow»«mi»T«/mi»«mi»o«/mi»«mi»t«/mi»«mi»a«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»b«/mi»«mi»s«/mi»«mi»e«/mi»«mi»r«/mi»«mi»v«/mi»«mi»a«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«/mrow»«/mfrac»«/math» 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 x1, second observation is x2, and so on till x5.

Also x1 = 10 means that the value of the first observation, denoted by x1, is 10.
       Similarly, x2 = 7, x3 = 13, x4 = 20 and x5 = 15.

Therefore, the mean «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»x«/mi»«mo»§#175;«/mo»«/mover»«/math»=«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mi»s«/mi»«mi»u«/mi»«mi»m«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»a«/mi»«mi»l«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»t«/mi»«mi»h«/mi»«mi»e«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»b«/mi»«mi»s«/mi»«mi»e«/mi»«mi»r«/mi»«mi»v«/mi»«mi»a«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«mo»§nbsp;«/mo»«/mrow»«mrow»«mi»T«/mi»«mi»o«/mi»«mi»t«/mi»«mi»a«/mi»«mi»l«/mi»«mo»§nbsp;«/mo»«mi»n«/mi»«mi»u«/mi»«mi»m«/mi»«mi»b«/mi»«mi»e«/mi»«mi»r«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»f«/mi»«mo»§nbsp;«/mo»«mi»o«/mi»«mi»b«/mi»«mi»s«/mi»«mi»e«/mi»«mi»r«/mi»«mi»v«/mi»«mi»a«/mi»«mi»t«/mi»«mi»i«/mi»«mi»o«/mi»«mi»n«/mi»«/mrow»«/mfrac»«/math»

                                        =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«msub»«mi»x«/mi»«mn»1«/mn»«/msub»«mo»+«/mo»«msub»«mi»x«/mi»«mn»2«/mn»«/msub»«mo»+«/mo»«msub»«mi»x«/mi»«mn»3«/mn»«/msub»«mo»+«/mo»«msub»«mi»x«/mi»«mn»4«/mn»«/msub»«mo»+«/mo»«msub»«mi»x«/mi»«mn»5«/mn»«/msub»«/mrow»«mn»2«/mn»«/mfrac»«/math»

                                       =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»10«/mn»«mo»+«/mo»«mn»7«/mn»«mo»+«/mo»«mn»13«/mn»«mo»+«/mo»«mn»20«/mn»«mo»+«/mo»«mn»15«/mn»«/mrow»«mn»5«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mn»65«/mn»«mn»5«/mn»«/mfrac»«mo»=«/mo»«mn»13«/mn»«/math»

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 x1 + x2 + x3 + . . . + x30 would be a tedious job.We use the Greek symbol
            Σ (for the letter Sigma) for summation. Instead of writing x1 + x2 + x3 + . . . + x30, we

write «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«munderover»«mrow»«mo»§#8721;«/mo»«msub»«mo largeop=¨true¨»x«/mo»«mi»i«/mi»«/msub»«mo»§nbsp;«/mo»«/mrow»«mrow»«mi»i«/mi»«mo»-«/mo»«mn»1«/mn»«/mrow»«mn»30«/mn»«/munderover»«/math» which is read as ‘the sum of xi as i varies from 1 to 30’.

                                            so     «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨left¨ rowspacing=¨0¨»«mtr»«mtd»«msub»«mi»x«/mi»«mi»i«/mi»«/msub»«mo»§nbsp;«/mo»«munderover»«mo»§#8721;«/mo»«mrow»«mi»i«/mi»«mo»-«/mo»«mn»1«/mn»«/mrow»«mn»30«/mn»«/munderover»«msub»«mo largeop=¨true¨»x«/mo»«mrow»«mi»i«/mi»«mo»§nbsp;«/mo»«/mrow»«/msub»«/mtd»«/mtr»«mtr»«mtd»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mo»§nbsp;«/mo»«mover»«mn»30«/mn»«mo»§#175;«/mo»«/mover»«/mtd»«/mtr»«/mtable»«/math»

Similarly : fro n observation =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mover»«mi»x«/mi»«mo»§#175;«/mo»«/mover»«/math» «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnalign=¨left¨ rowspacing=¨0¨»«mtr»«mtd/»«/mtr»«mtr»«mtd/»«/mtr»«mtr»«mtd»«munderover»«mrow»«mo»§#8721;«/mo»«msub»«mi»x«/mi»«mi»i«/mi»«/msub»«/mrow»«mrow»«mi»i«/mi»«mo»-«/mo»«mn»1«/mn»«/mrow»«mn»11«/mn»«/munderover»«/mtd»«/mtr»«mtr»«mtd»«mo»§nbsp;«/mo»«mover»«mi»n«/mi»«mo»§#175;«/mo»«/mover»«/mtd»«/mtr»«/mtable»«/math»

 

Example  :

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.

Solution : Mean=«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»5000«/mn»«mo»+«/mo»«mn»5000«/mn»«mo»+«/mo»«mn»5000«/mn»«mo»+«/mo»«mn»5000«/mn»«mo»+«/mo»«mn»15000«/mn»«/mrow»«mn»5«/mn»«/mfrac»«mo»=«/mo»«mfrac»«mn»35000«/mn»«mn»5«/mn»«/mfrac»«mo»=«/mo»«mn»7000«/mn»«/math»

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
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mo»(«/mo»«mfrac»«mrow»«mn»5«/mn»«mo»+«/mo»«mn»1«/mn»«/mrow»«mn»2«/mn»«/mfrac»«mo»)«/mo»«mi»t«/mi»«mi»h«/mi»«mo»§nbsp;«/mo»«mo»=«/mo»«mfrac»«mn»6«/mn»«mn»2«/mn»«/mfrac»«mi»t«/mi»«mi»h«/mi»«/math» = 3rd 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.

 

Example  :

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
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.

 

Example  :

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.
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 «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»16«/mn»«mn»2«/mn»«/mfrac»«/math» th and «math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mn»16«/mn»«mn»2«/mn»«/mfrac»«mo»+«/mo»«mn»1«/mn»«/math»th, i.e., the 8th and 9th terms.
So, the median is the mean of the values of the 8th and 9th terms.
i.e, the median =«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mfrac»«mrow»«mn»10«/mn»«mo»+«/mo»«mn»14«/mn»«/mrow»«mn»2«/mn»«/mfrac»«mo»=«/mo»«mn»12«/mn»«/math»
           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.

 

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