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# 14. STATISTICS

## 14.1 INTRODUCTION

Everyday we come across a wide variety of informations in the form of facts, numerical figures, tables, graphs, etc. These are provided by newspapers, televisions, magazines and other means of communication. These may relate to cricket batting or bowling averages, profits of a company, temperatures of cities, expenditures in various sectors of a five year plan, polling results, and so on. These facts or figures, which are numerical or otherwise, collected with a definite purpose are called *data*. Data is the plural form of the Latin word *datum*. Of course, the word 'data' is not new for you. You have studied about data and data handling in earlier classes.

Our world is becoming more and more information oriented. Every part of our lives utilises data in one form or the other. So, it becomes essential for us to know how to extract meaningful information from such data. This extraction of meaningful information is studied in a branch of mathematics called *Statistics*.

The word 'statistics' appears to have been derived from the Latin word 'status' meaning 'a (political) state'. In its origin, statistics was simply the collection of data on different aspects of the life of people, useful to the State. Over the period of time, h owever, its scope broadened and statistics began to concern itself not only with the collection and presentation of data but also with the interpretation and drawing of inferences from the data. Statistics deals with collection, organisation, analysis and interpretation of data. The word 'statistics' has different meanings in different contexts. Let us observe the following sentences:

- May I have the latest copy of 'Educational Statistics of India'.
- I like to study 'Statistics' because it is used in day-to-day life.

In the first sentence, statistics is used in a plural sense, meaning numerical data. These may include a number of educational institutions of India, literacy rates of various states, etc. In the second sentence, the word 'statistics' is used as a singular noun, meaning the subject which deals with the collection, presentation, analysis of data as well as drawing of meaningful conclusions from the data.

In this chapter, we shall briefly discuss all these aspects regarding data.

## 14.2 COLLECTION OF DATA

Let us begin with an exercise on gathering data by performing the following activity.

## 14.3 PRESENTATION OF DATA

As soon as the work related to collection of data is over, the investigator has to find out ways to present them in a form which is meaningful, easily understood and gives its main features at a glance. Let us now recall the various ways of presenting the data through some examples.

Consider the marks obtained by 10 students in a mathematics test as given below:

55 36 95 73 60 42 25 78 75 62

The data in this form is called *raw data*.

By looking at it in this form, can you find the highest and the lowest marks?

Did it take you some time to search for the maximum and minimum scores? Wouldn't it be less time consuming if these scores were arranged in ascending or descending order? So let us arrange the marks in ascending order as

25 36 42 55 60 62 73 75 78 95

Now, we can clearly see that the lowest marks are 25 and the highest marks are 95.

The difference of the highest and the lowest values in the data is called the *range *of the data. So, the range in this case is 95 – 25 = 70.

Presentation of data in ascending or descending order can be quite time consuming, particularly when the number of observations in an experiment is large, as in the case of the next example.

Consider the marks obtained (out of 100 marks) by 30 students of Class IX of a school:

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

Recall that the number of students who have obtained a certain number of marks is called the *frequency *of those marks. For instance, 4 students got 70 marks. So the frequency of 70 marks is 4. To make the data more easily understandable, we write it in a table, as given below:

**Table 14.1**

**Table 14.1**

Marks |
Number of students (i.e., the frequency) |
---|---|

10 | 1 |

20 | 1 |

36 | 3 |

40 | 4 |

50 | 3 |

56 | 2 |

60 | 4 |

70 | 4 |

72 | 1 |

80 | 1 |

88 | 2 |

92 | 3 |

95 | 1 |

Total = 30 |

Table 14.1 is called *an ungrouped frequency distribution table, *or simply a *frequency distribution table*. Note that you can use also *tally marks *in preparing these tables, as in the next example.

100 plants each were planted in 100 schools during Van Mahotsava. After one month, the number of plants that survived were recorded as :

95 67 28 32 65 65 69 33 98 96

76 42 32 38 42 40 40 69 95 92

75 83 76 83 85 62 37 65 63 42

89 65 73 81 49 52 64 76 83 92

93 68 52 79 81 83 59 82 75 82

86 90 44 62 31 36 38 42 39 83

87 56 58 23 35 76 83 85 30 68

69 83 86 43 45 39 83 75 66 83

92 75 89 66 91 27 88 89 93 42

53 69 90 55 66 49 52 83 34 36

To present such a large amount of data so that a reader can make sense of it easily, we condense it into groups like 20-29, 30-39, . . ., 90-99 (since our data is from 23 to 98). These groupings are called 'classes' or 'class-intervals', and their size is called the * class-size *or *class width*, which is 10 in this case. In each of these classes, the least number is called the *lower class limit *and the greatest number is called the *upper class limit, *e.g., in 20-29, 20 is the 'lower class limit' and 29 is the 'upper class limit'.

Also, recall that using tally marks, the data above can be condensed in tabular form as follows:

**Table 14.2**

**Table 14.2**

Number of plants survived |
Tally Marks |
Number of schools (frequency) |
---|---|---|

20-29 | ||| | 3 |

30-39 | |||| | 14 |

40-49 | || | 12 |

50-59 | ||| | 8 |

60-69 | ||| | 18 |

70-79 | 10 | |

80-89 | ||| | 23 |

90-99 | || | 12 |

Total=100 |

Presenting data in this form simplifies and condenses data and enables us to observe certain important features at a glance. This is called *a grouped frequency distribution table*. Here we can easily observe that 50% or more plants survived in 8 + 18 + 10 + 23 + 12 = 71 schools.

We observe that the classes in the table above are non-overlapping. Note that we could have made more classes of shorter size, or fewer classes of larger size also. For instance, the intervals could have been 22-26, 27-31, and so on. So, there is no hard and fast rule about this except that the classes should not overlap.

Let us now consider the following frequency distribution table which gives the weights of 38 students of a class:

**Table 14.3**

**Table 14.3**

Weight (in kg) |
Number of Students |
---|---|

31-35 | 9 |

36-40 | 5 |

41-45 | 14 |

46-50 | 3 |

51-55 | 1 |

56-60 | 2 |

61-65 | 2 |

66-70 | 1 |

71-75 | 1 |

Total = 38 |

Now, if two new students of weights 35.5 kg and 40.5 kg are admitted in this class, then in which interval will we include them? We cannot add them in the ones ending with 35 or 40, nor to the following ones. This is because there are gaps in between the upper and lower limits of two consecutive classes. So, we need to divide the intervals so that the upper and lower limits of consecutive intervals are the same. For this, we find the difference between the upper limit of a class and the lower limit of its succeeding class. We then add half of this difference to each of the upper limits and subtract the same from each of the lower limits.

For example, consider the classes 31 - 35 and 36 - 40.

The lower limit of 36 - 40 = 36

The upper limit of 31 - 35 = 35

The difference = 36 – 35 = 1

So, half the difference = ½ =0.5

So the new class interval formed from 31 - 35 is (31 – 0.5) - (35 + 0.5), i.e., 30.5 - 35.5. Similarly, the new class formed from the class 36 - 40 is (36 – 0.5) - (40 + 0.5), i.e., 35.5 - 40.5.

Continuing in the same manner, the continuous classes formed are: 30.5-35.5, 35.5-40.5, 40.5-45.5, 45.5-50.5, 50.5-55.5, 55.5-60.5, 60.5 - 65.5, 65.5 - 70.5, 70.5 - 75.5.

Now it is possible for us to include the weights of the new students in these classes. But, another problem crops up because 35.5 appears in both the classes 30.5 - 35.5 and 35.5 - 40.5. In which class do you think this weight should be considered?

If it is considered in both classes, it will be counted twice.

*By convention*, we consider 35.5 in the class 35.5 - 40.5 and not in 30.5 - 35.5. Similarly, 40.5 is considered in 40.5 - 45.5 and not in 35.5 - 40.5.

So, the new weights 35.5 kg and 40.5 kg would be included in 35.5 - 40.5 and 40.5 - 45.5, respectively. Now, with these assumptions, the new frequency distribution table will be as shown below:

**Table 14.4**

**Table 14.4**

Weight (in kg) |
Number of Students |
---|---|

30.5-35.5 | 9 |

35.5-40.5 | 6 |

40.5-45.5 | 15 |

45.5-50.5 | 3 |

50.5-55.5 | 1 |

55.5-60.5 | 2 |

60.5-65.5 | 2 |

65.5-70.5 | 1 |

70.5-75.5 | 1 |

Total = 40 |

Now, let us move to the data collected by you in Activity 1. This time we ask you to present these as frequency distribution tables.

## 14.4 GRAPHICAL REPRESENTATION OF DATA

The representation of data by tables has already been discussed. Now let us turn our attention to another representation of data, i.e., the graphical representation. It is well said that one picture is better than a thousand words. Usually comparisons among the individual items are best shown by means of graphs. The representation then becomes easier to understand than the actual data. We shall study the following graphical representations in this section.

(A) Bar graphs

(B) Histograms of uniform width, and of varying widths

(C) Frequency polygons

**(A) Bar graphs**

In earlier classes, you have already studied and constructed bar graphs. Here we shall discuss them through a more formal approach. Recall that a bar graph is a pictorial representation of data in which usually bars of uniform width are drawn with equal spacing between them on one axis (say, the *x*-axis), depicting the variable. The values of the variable are shown on the other axis (say, the *y*-axis) and the heights of the bars depend on the values of the variable.

In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared for the data so obtained:

**Fig. 14.1**

Observe the bar graph given above and answer the following questions:

- How many students were born in the month of November?
- In which month were the maximum number of students born?

Note that the variable here is the 'month of birth', and the value of the variable is the 'Number of students born'.

- 4 students were born in the month of November.
- The Maximum number of students were born in the month of August.

Let us now recall how a bar graph is constructed by considering the following example.

**Table 14.5**

**Table 14.5**

Heads |
Expenditure (in thousand rupees) |
---|---|

Grocery | 4 |

Rent | 5 |

Education of children | 5 |

Medicine | 2 |

Fuel | 2 |

Entertainment | 1 |

Miscellaneous | 1 |

Draw a bar graph for the data above.

We draw the bar graph of this data in the following steps. Note that the unit in the second column is thousand rupees. So, '4' against 'grocery' means Rs 4000.

- We represent the Heads (variable) on the horizontal axis choosing any scale, since the width of the bar is not important. But for clarity, we take equal widths for all bars and maintain equal gaps in between. Let one Head be represented by one unit.
- We represent the expenditure (value) on the vertical axis. Since the maximum expenditure is Rs 5000, we can choose the scale as 1 unit = Rs 1000.
- To represent our first Head, i.e., grocery, we draw a rectangular bar with width 1 unit and height 4 units.
- Similarly, other Heads are represented leaving a gap of 1 unit in between two consecutive bars.

The bar graph is drawn in Fig. 14.2.

**Fig. 14.2**

Here, you can easily visualise the relative characteristics of the data at a glance, e.g., the expenditure on education is more than double that of medical expenses. Therefore, in some ways it serves as a better representation of data than the tabular form.

Let us now see how a frequency distribution table for *continuous *class intervals can be represented graphically.

**(B) Histogram**

This is a form of representation like the bar graph, but it is used for continuous class intervals. For instance, consider the frequency distribution Table 14.6, representing the weights of 36 students of a class:

**Table 14.6**

**Table 14.6**

Weights in kg |
Number of Students |
---|---|

30.5-35.5 | 9 |

35.5-40.5 | 6 |

40.5-45.5 | 15 |

45.5-50.5 | 3 |

50.5-55.5 | 1 |

55.5-60.5 | 2 |

Total=36 |

Let us represent the data given above graphically as follows:

- We represent the weights on the horizontal axis on a suitable scale. We can choose the scale as 1 cm = 5 kg. Also, since the first class interval is starting from 30.5 and not zero, we show it on the graph by marking a
*kink*or a break on the axis. - We represent the number of students (frequency) on the vertical axis on a suitable scale. Since the maximum frequency is 15, we need to choose the scale to accomodate this maximum frequency.
- We now draw rectangles (or rectangular bars) of width equal to the class-size and lengths according to the frequencies of the corresponding class intervals. For example, the rectangle for the class interval 30.5 - 35.5 will be of width 1 cm and length 4.5 cm.
- In this way, we obtain the graph as shown in Fig. 14.3:

**Fig. 14.3**

Observe that since there are no gaps in between consecutive rectangles, the resultant graph appears like a solid figure. This is called a *histogram*, which is a graphical representation of a grouped frequency distribution with continuous classes. Also, unlike a bar graph, the width of the bar plays a significant role in its construction.

Here, in fact, areas of the rectangles erected are proportional to the corresponding frequencies. However, since the widths of the rectangles are all equal, the lengths of the rectangles are proportional to the frequencies. That is why, we draw the lengths according to (iii) above.

Now, consider a situation different from the one above.

A teacher wanted to analyse the performance of two sections of students in a mathematics test of 100 marks. Looking at their performances, she found that a few students got under 20 marks and a few got 70 marks or above. So she decided to group them into intervals of varying sizes as follows: 0 - 20, 20 - 30, . . ., 60 - 70, 70 - 100. Then she formed the following table:

**Table 14.7**

**Table 14.7**

Marks |
Number of Students |
---|---|

0-20 | 7 |

20-30 | 10 |

30-40 | 10 |

40-50 | 20 |

50-60 | 20 |

60-70 | 15 |

70-above | 8 |

Total=90 |

A histogram for this table was prepared by a student as shown in Fig. 14.4.

**Fig. 14.4**

Carefully examine this graphical representation. Do you think that it correctly represents the data? No, the graph is giving us a misleading picture. As we have mentioned earlier, the areas of the rectangles are proportional to the frequencies in a histogram. Earlier this problem did not arise, because the widths of all the rectangles were equal. But here, since the widths of the rectangles are varying, the histogram above does not give a correct picture. For example, it shows a greater frequency in the interval 70 - 100, than in 60 - 70, which is not the case.

So, we need to make certain modifications in the lengths of the rectangles so that the areas are again proportional to the frequencies.

The steps to be followed are as given below:

- Select a class interval with the minimum class size. In the example above, the minimum class-size is 10.
- The lengths of the rectangles are then modified to be proportionate to the class size 10.

For, instance, when the class size is 20, the length of the rectangle is 7. So when the class size is 10, the length of the rectangle will be ^{7}/_{10} = 3.5. Similarly, proceeding in this manner, we get the following table:

**Table 14.8**

**Table 14.8**

Marks |
frequency |
Width of class |
Length of the rectangle |
---|---|---|---|

0-20 | 7 | 20 | ^{7}/_{20} x 10 = 3.5 |

20-30 | 10 | 10 | ^{10}/_{10} x 10 = 10 |

30-40 | 10 | 10 | ^{10}/_{10} x 10 = 10 |

40-50 | 20 | 10 | ^{20}/_{10} x 10 = 20 |

50-60 | 20 | 10 | ^{20}/_{10} x 10 = 20 |

60-70 | 15 | 10 | ^{15}/_{10} x 10 = 15 |

70-100 | 8 | 30 | ^{8}/_{30} x 10 = 2.67 |

Total=90 |

Since we have calculated these lengths for an interval of 10 marks in each case, we may call these lengths as "proportion of students per 10 marks interval".

So, the correct histogram with varying width is given in Fig. 14.5.

**Fig. 14.5**

**(C) Frequency Polygon**

There is yet another visual way of representing quantitative data and its frequencies. This is a polygon. To see what we mean, consider the histogram represented by Fig. 14.3. Let us join the mid-points of the upper sides of the adjacent rectangles of this histogram by means of line segments. Let us call these mid-points B, C, D, E, F and G. When joined by line segments, we obtain the figure BCDEFG (see Fig. 14.6). To complete the polygon, we assume that there is a class interval with frequency zero before 30.5 - 35.5, and one after 55.5 - 60.5, and their mid-points are A and H, respectively. ABCDEFGH is the frequency polygon corresponding to the data shown in Fig. 14.3. We have shown this in Fig. 14.6.

**Fig. 14.6**

Although, there exists no class preceding the lowest class and no class succeeding the highest class, addition of the two class intervals with zero frequency enables us to make the area of the frequency polygon the same as the area of the histogram. Why is this so? (**Hint : **Use the properties of congruent triangles.)

Now, the question arises: how do we complete the polygon when there is no class preceding the first class? Let us consider such a situation.

Consider the marks, out of 100, obtained by 51 students of a class in a test, given in Table 14.9.

**Table 14.9**

**Table 14.9**

Marks |
Number of Students |
---|---|

0-10 | 5 |

10-20 | 10 |

20-30 | 4 |

30-40 | 6 |

40-50 | 7 |

50-60 | 3 |

60-70 | 2 |

70-80 | 2 |

80-90 | 3 |

90-100 | 9 |

Total=51 |

Draw a frequency polygon corresponding to this frequency distribution table.

Let us first draw a histogram for this data and mark the mid-points of the tops of the rectangles as B, C, D, E, F, G, H, I, J, K, respectively. Here, the first class is 0-10. So, to find the class preceeding 0-10, we extend the horizontal axis in the negative direction and find the mid-point of the imaginary class-interval (–10) - 0. The first end point, i.e., B is joined to this mid-point with zero frequency on the negative direction of the horizontal axis. The point where this line segment meets the vertical axis is marked as A. Let L be the mid-point of the class succeeding the last class of the given data. Then OABCDEFGHIJKL is the frequency polygon, which is shown in Fig. 14.7.

**Fig. 14.7**

* Frequency polygons can also be drawn independently without drawing histograms. *For this, we require the mid-points of the class-intervals used in the data. These mid-points of the class-intervals are called **class-marks**.

To find the class-mark of a class interval, we find the sum of the upper limit and lower limit of a class and divide it by 2. Thus,

**Class-mark =**

^{Upper Limit + Lower Limit }/

_{2}

Let us consider an example.

In a city, the weekly observations made in a study on the cost of living index are given in the following table:

**Table 14.10**

**Table 14.10**

Cost of living index |
Number of weeks |
---|---|

140-150 | 5 |

150-160 | 10 |

160-170 | 20 |

170-180 | 9 |

180-190 | 6 |

190-200 | 2 |

Total = 52 |

Draw a frequency polygon for the data above (without constructing a histogram).

Since we want to draw a frequency polygon without a histogram, let us find the class-marks of the classes given above, that is of 140 - 150, 150 - 160,.... For 140 - 150, the upper limit = 150, and the lower limit = 140

So, the class-mark = ^{150 + 140}/_{2} =^{290}/_{2} = 145.

Continuing in the same manner, we find the class-marks of the other classes as well. So, the new table obtained is as shown in the following table:

**Table 14.10**

**Table 14.10**

Classes |
Class-marks |
frequency |
---|---|---|

140-150 | 145 | 5 |

150-160 | 155 | 10 |

160-170 | 165 | 20 |

170-180 | 175 | 9 |

180-190 | 185 | 6 |

190-200 | 195 | 2 |

Total = 52 |

We can now draw a frequency polygon by plotting the class-marks along the horizontal axis, the frequencies along the vertical-axis, and then plotting and joining the points B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6) and G(195, 2) by line segments. We should not forget to plot the point corresponding to the class-mark of the class 130 - 140 (just before the lowest class 140 - 150) with zero frequency, that is, A(135, 0), and the point H (205, 0) occurs immediately after G(195, 2). So, the resultant frequency polygon will be ABCDEFGH (see Fig. 14.8).

**Fig. 14.8**

Frequency polygons are used when the data is continuous and very large. It is very useful for comparing two different sets of data of the same nature, for example, comparing the performance of two different sections of the same class.

## 14.5 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:

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:

Mary's score | 7 | 8 | 9 | 10 | |
---|---|---|---|---|---|

Hari's score | 4 | 7 | 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 , read as '

**bar'.**

*x*Let us consider an 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.

We have already studied in our earlier classes that the mean of a certain number of observations is equal to ^{Sum of all observations}/_{Total number of observations}. To simplify our working of finding the mean, let us use a variable *x _{i } *to denote the

*i*th 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 *= ^{Sum of all observations}/_{Total number of observations}*

= ^{ 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 , which is read as 'the sum of *x _{i} as i *varies from 1 to 30.

So, =^{}/_{30}

Similarly, for *n *observations =^{}/_{n}

Find the mean of the marks obtained by 30 students of Class IX of a school, given in Example 2.

Now, = ^{ x1 + x2 + . . . + x30 } /_{30}

= 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, = ^{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 × 10) + (1 × 20) + (3 × 36) + (4 × 40) + (3 × 50) + (2 × 56) + (4 × 60) + (4 × 70) + (1 × 72) + (1 × 80) + (2 × 88) + (3 × 92) + (1 × 95)

= *f*_{1}*x*_{1} + . . . + *f*_{13}*x*_{13 }, where *fi *is the frequency of the *i*th entry in Table 14.1.

In brief, we write this as = *f*_{1} + *f*_{2} + . . . + *f* _{13}

= 10 + 20 + 108 + 160 + 150 + 112 + 240 + 280 + 72 + 80 + 176 + 276 + 95 = 1779

Now, the total number of observations = = *f*_{1 }+ *f*_{2} + . . . + *f*_{13} = 1 + 1 + 3 + 4 + 3 + 2 + 4 + 4 + 1 + 1 + 2 + 3 + 1 = 30

So, the mean = ^{Sum of all observations}/_{Total number of observations} = = ^{1779}/_{30}

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

**Table 14.12**

**Table 14.12**

Marks (x_{i}) |
Number of students (f_{i}) |
f_{i} x_{i} |
---|---|---|

10 | 1 | 10 |

20 | 1 | 20 |

36 | 3 | 108 |

40 | 4 | 160 |

50 | 3 | 150 |

56 | 2 | 112 |

60 | 4 | 240 |

70 | 4 | 280 |

72 | 1 | 72 |

80 | 1 | 80 |

88 | 2 | 176 |

92 | 3 | 276 |

95 | 1 | 95 |

= 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:

- 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 [see Fig. 14.9 (i)]. - 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 [see Fig. 14.9 (ii)].

**Fig. 14.9**The Number Line

Let us illustrate this with the help of some examples.

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.

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 5th student, which is 149 cm.

So, the median, i.e. the medial height is 149 cm.

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.

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 8th and 9th terms.

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

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

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.

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.

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 = 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.

Measures of central tendency |
Day |
Mary |
---|---|---|

mean | 8.2 | 8.4 |

Median | 10 | 8 |

Mode | 10 | 8 |

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.