resource/jemh114-min (1).md

1062CH14

14

14.1 Introduction

In Class IX, you have studied the classification of given data into ungrouped as well as grouped frequency distributions. You have also learnt to represent the data pictorially in the form of various graphs such as bar graphs, histograms (including those of varying widths) and frequency polygons. In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median and mode. In this chapter, we shall extend the study of these three measures, i.e., mean, median and mode from ungrouped data to that of grouped data. We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution and how to draw cumulative frequency curves, called ogives.

14.2 Mean of Grouped Data

The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations. From Class IX, recall that if $x_1, x_2, \ldots, x_n$ are observations with respective frequencies $f_1, f_2, \ldots, f_n$ , then this means observation $x_1$ occurs $f_1$ times, $x_2$ occurs $f_2$ times, and so on.

Now, the sum of the values of all the observations $= f_1 x_1 + f_2 x_2 + \ldots + f_n x_n$ , and the number of observations $= f_1 + f_2 + \ldots + f_n$ .

So, the mean $\bar{x}$ of the data is given by

$$\bar{x}=\frac{f_1{x}_1+f_2{x}_2+\cdots +f_n{x}_n}{f_1+f_2+\cdots +f_n}\backslash bar\{x\}\; =\; \backslash frac\{f\_1\; x\_1\; +\; f\_2\; x\_2\; +\; \backslash dots\; +\; f\_n\; x\_n\}\{f\_1\; +\; f\_2\; +\; \backslash dots\; +\; f\_n\}$$

Recall that we can write this in short form by using the Greek letter $\Sigma$ (capital sigma) which means summation. That is,

$$\bar{x}=\frac{\sum _{i=1}^n{f}_i{x}_i}{\sum _{i=1}^n{f}_i}\backslash overline\{x\}\; =\; \backslash frac\{\backslash sum\_\{i=1\}^\{n\}\; f\_i\; x\_i\}\{\backslash sum\_\{i=1\}^\{n\}\; f\_i\}$$

which, more briefly, is written as $\overline{x} = \frac{\sum f_i x_i}{\sum f_i}$ , if it is understood that i varies from 1 to n.

Let us apply this formula to find the mean in the following example.

Example 1: The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students.

Marks obtained $(x_i)$	10	20	36	40	50	56	60	70	72	80	88	92	95
Number of students $(f_i)$	1	1	3	4	3	2	4	4	1	1	2	3	1

Solution: Recall that to find the mean marks, we require the product of each $x_i$ with the corresponding frequency $f_i$ . So, let us put them in a column as shown in Table 14.1.

Table 14.1

Marks obtained $(x_i)$	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
Total	$\Sigma f_i = 30$	$\Sigma f_i x_i = 1779$

Now, $$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{1779}{30}=59.3\backslash overline\{x\}\; =\; \backslash frac\{\backslash sum\; f\_i\; x\_i\}\{\backslash sum\; f\_i\}\; =\; \backslash frac\{1779\}\{30\}\; =\; 59.3$$

Therefore, the mean marks obtained is 59.3.

In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean.

Let us convert the ungrouped data of Example 1 into grouped data by forming class-intervals of width, say 15. Remember that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, e.g., 4 students who have obtained 40 marks would be considered in the class-interval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped frequency distribution table (see Table 14.2).

Table 14.2

Class interval	10 - 25	25 - 40	40 - 55	55 - 70	70 - 85	85 - 100
Number of students	2	3	7	6	6	6

Now, for each class-interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each class-interval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. Recall that we find the mid-point of a class (or its class mark) by finding the average of its upper and lower limits. That is,

Class mark = $$\frac{\text{Upper class limit}+\text{Lower class limit}}{2}\backslash frac\{\backslash text\{Upper\; class\; limit\}\; +\; \backslash text\{Lower\; class\; limit\}\}\{2\}$$

With reference to Table 14.2, for the class 10-25, the class mark is $\frac{10+25}{2}$ , i.e.,

17.5. Similarly, we can find the class marks of the remaining class intervals. We put them in Table 14.3. These class marks serve as our $x_i$ 's. Now, in general, for the ith class interval, we have the frequency $f_i$ corresponding to the class mark $x_i$ . We can now proceed to compute the mean in the same manner as in Example 1.

Class interval	Number of students $(f_i)$	Class mark $(x_i)$	$f_i x_i$
10 - 25	2	17.5	35.0
25 - 40	3	32.5	97.5
40 - 55	7	47.5	332.5
55 - 70	6	62.5	375.0
70 - 85	6	77.5	465.0
85 - 100	6	92.5	555.0
Total	$\Sigma f_i = 30$		$\sum f_i x_i = 1860.0$

Table 14.3

The sum of the values in the last column gives us $\sum f_i x_i$ . So, the mean $\overline{x}$ of the given data is given by

$\bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{1860.0}{30} = 62$

This new method of finding the mean is known as the Direct Method.

We observe that Tables 14.1 and 14.3 are using the same data and employing the same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the two values is because of the mid-point assumption in Table 14.3, 59.3 being the exact mean, while 62 an approximate mean.

Sometimes when the numerical values of $x_i$ and $f_i$ are large, finding the product of $x_i$ and $f_i$ becomes tedious and time consuming. So, for such situations, let us think of a method of reducing these calculations.

We can do nothing with the $f_i$ 's, but we can change each $x_i$ to a smaller number so that our calculations become easy. How do we do this? What about subtracting a fixed number from each of these $x_i$ 's? Let us try this method.

The first step is to choose one among the $x_i$ 's as the assumed mean, and denote it by 'a'. Also, to further reduce our calculation work, we may take 'a' to be that $x_i$ which lies in the centre of $x_1, x_2, \ldots, x_n$ . So, we can choose a = 47.5 or a = 62.5. Let us choose a = 47.5.

The next step is to find the difference $d_i$ between a and each of the $x_i$ 's, that is, the deviation of 'a' from each of the $x_i$ 's.

i.e., $$d_i=x_i-a=x_i-47.5d\_i\; =\; x\_i\; -\; a\; =\; x\_i\; -\; 47.5$$

The third step is to find the product of $d_i$ with the corresponding $f_i$ , and take the sum of all the $f_i d_i$ 's. The calculations are shown in Table 14.4.

Class interval	Number of students $(f_i)$	Class mark $(x_i)$	$d_i = x_i - 47.5$	$f_i d_i$
10 - 25	2	17.5	-30	-60
25 - 40	3	32.5	-15	-45
40 - 55	7	47.5	0	0
55 - 70	6	62.5	15	90
70 - 85	6	77.5	30	180
85 - 100	6	92.5	45	270
Total	$\Sigma f = 30$			$\Sigma f.d. = 435$

Table 14.4

So, from Table 14.4, the mean of the deviations, $\overline{d} = \frac{\sum f_i d_i}{\sum f_i}$ .

Now, let us find the relation between $\overline{d}$ and $\overline{x}$ .

Since in obtaining $d_i$ , we subtracted 'a' from each $x_i$ , so, in order to get the mean $\overline{x}$ , we need to add 'a' to $\overline{d}$ . This can be explained mathematically as:

Mean of deviations, $$\bar{d}=\frac{\sum f_i{d}_i}{\sum f_i}\backslash overline\{d\}\; =\; \backslash frac\{\backslash sum\; f\_i\; d\_i\}\{\backslash sum\; f\_i\}$$ So, $$\bar{d}=\frac{\sum f_i(x_i-a)}{\sum f_i}\backslash overline\{d\}\; =\; \backslash frac\{\backslash sum\; f\_i\; (x\_i\; -\; a)\}\{\backslash sum\; f\_i\}$$

$$=\frac{\sum f_i{x}_i}{\sum f_i}-\frac{\sum f_{i}a}{\sum f_i}=\; \backslash frac\{\backslash sum\; f\_i\; x\_i\}\{\backslash sum\; f\_i\}\; -\; \backslash frac\{\backslash sum\; f\_i\; a\}\{\backslash sum\; f\_i\}$$

$$=\bar{x}-a\frac{\sum f_i}{\sum f_i}=\; \backslash overline\{x\}\; -\; a\; \backslash frac\{\backslash sum\; f\_i\}\{\backslash sum\; f\_i\}$$

$$=\bar{x}-a=\; \backslash overline\{x\}\; -\; a$$ So, $$\bar{x}=a+\bar{d}\backslash overline\{x\}\; =\; a\; +\; \backslash overline\{d\}$$ \ni.e., $$\bar{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}\backslash overline\{x\}\; =\; a\; +\; \backslash frac\{\backslash sum\; f\_i\; d\_i\}\{\backslash sum\; f\_i\}$$

Substituting the values of a, $\Sigma f_i d_i$ and $\Sigma f_i$ from Table 14.4, we get

$$\bar{x}=47.5+\frac{435}{30}=47.5+14.5=62\backslash overline\{x\}\; =\; 47.5\; +\; \backslash frac\{435\}\{30\}\; =\; 47.5\; +\; 14.5\; =\; 62$$ .

Therefore, the mean of the marks obtained by the students is 62.

The method discussed above is called the Assumed Mean Method.

Activity 1: From the Table 14.3 find the mean by taking each of $x_i$ (i.e., 17.5, 32.5, and so on) as 'a'. What do you observe? You will find that the mean determined in each case is the same, i.e., 62. (Why?)

So, we can say that the value of the mean obtained does not depend on the choice of a.

Observe that in Table 14.4, the values in Column 4 are all multiples of 15. So, if we divide the values in the entire Column 4 by 15, we would get smaller numbers to multiply with f_.. (Here, 15 is the class size of each class interval.)

So, let $u_i = \frac{x_i - a}{h}$ , where a is the assumed mean and h is the class size.

Now, we calculate $u_i$ in this way and continue as before (i.e., find $f_i u_i$ and then $\Sigma f_i u_i$ ). Taking h = 15, let us form Table 14.5.

Table 14.5

Class interval	$f_i$	$\boldsymbol{x}_{i}$	$d_i = x_i - a$	$u_i = \frac{x_i - a}{h}$	$f_i u_i$
			0
10 - 25	2	17.5	-30	-2	-4
25 - 40	3	32.5	-15	-1	-3
40 - 55	7	47.5	0	0	0
55 - 70	6	62.5	15	1	6
70 - 85	6	77.5	30	2	12
85 - 100	6	92.5	45	3	18
Total	$\Sigma f_i = 30$				$\Sigma f_i u_i = 29$

Let $$\bar{u}=\frac{\sum f_i{u}_i}{\sum f_i}\backslash overline\{u\}\; =\; \backslash frac\{\backslash sum\; f\_i\; u\_i\}\{\backslash sum\; f\_i\}$$

Here, again let us find the relation between $\overline{u}$ and $\overline{x}$ .

We have, $$u_i=\frac{x_i-a}{h}u\_i\; =\; \backslash frac\{x\_i\; -\; a\}\{h\}$$ Therefore, $$\bar{u}=\frac{\sum f_i}{\sum f_i}\frac{(x_i-a)}{h}=\frac{1}{h}\left[\frac{\sum f_i{x}_i-a\sum f_i}{\sum f_i}\right]\backslash overline\{u\}\; =\; \backslash frac\{\backslash sum\; f\_i\}\{\backslash sum\; f\_i\}\; \backslash frac\{(x\_i\; -\; a)\}\{h\}\; =\; \backslash frac\{1\}\{h\}\; \backslash left[\; \backslash frac\{\backslash sum\; f\_i\; x\_i\; -\; a\; \backslash sum\; f\_i\}\{\backslash sum\; f\_i\}\; \backslash right]$$

$$=\frac{1}{h}[\frac{\sum f_i{x}_i}{\sum f_i}-a\frac{\sum f_i}{\sum f_i}]=\; \backslash frac\{1\}\{h\}\; \backslash left[\; \backslash frac\{\backslash sum\; f\_i\; x\_i\}\{\backslash sum\; f\_i\}\; -\; a\; \backslash frac\{\backslash sum\; f\_i\}\{\backslash sum\; f\_i\}\; \backslash right]$$

$$=\frac{1}{h}[\bar{x}-a]=\; \backslash frac\{1\}\{h\}\; \backslash left[\; \backslash overline\{x\}\; -\; a\; \backslash right]$$ So, $$h\bar{u}=\bar{x}-ah\backslash overline\{u\}\; =\; \backslash overline\{x\}\; -\; a$$ i.e., $$\bar{x}=a+h\bar{u}\backslash overline\{x\}\; =\; a\; +\; h\backslash overline\{u\}$$ So, $$\bar{x}=a+h\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\backslash overline\{x\}\; =\; a\; +\; h\; \backslash left(\; \backslash frac\{\backslash sum\; f\_i\; u\_i\}\{\backslash sum\; f\_i\}\; \backslash right)$$

Now, substituting the values of a, h, $\Sigma f_i u_i$ and $\Sigma f_i$ from Table 14.5, we get

$$\bar{x}=47.5+15\times \left(\frac{29}{30}\right)\backslash overline\{x\}\; =\; 47.5\; +\; 15\; \backslash times\; \backslash left(\backslash frac\{29\}\{30\}\backslash right)$$

$$=47.5+14.5=62=\; 47.5\; +\; 14.5\; =\; 62$$

So, the mean marks obtained by a student is 62.

The method discussed above is called the Step-deviation method.

We note that:

• the step-deviation method will be convenient to apply if all the $d_i$ 's have a common factor.
• The mean obtained by all the three methods is the same.
• The assumed mean method and step-deviation method are just simplified forms of the direct method.
• The formula $\overline{x} = a + h\overline{u}$ still holds if a and h are not as given above, but are any non-zero numbers such that $u_i = \frac{x_i a}{h}$ .

Let us apply these methods in another example.

Example 2: The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section.

Percentage of female teachers	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65	65 - 75	75 - 85
Number of States/U.T.	6	11	7	4	4	2	1

Source: Seventh All India School Education Survey conducted by NCERT

Solution: Let us find the class marks, $x_i$ , of each class, and put them in a column (see Table 14.6):

Table 14.6

Percentage of female teachers	Number of States /U.T. $(f_i)$	$x_{i}$
15 - 25	6	20
25 - 35	11	30
35 - 45	7	40
45 - 55	4	50
55 - 65	4	60
65 - 75	2	70
75 - 85	1	80

Here we take a = 50, h = 10, then $d_i = x_i - 50$ and $u_i = \frac{x_i - 50}{10}$ .

We now find $d_i$ and $u_i$ and put them in Table 14.7.

Table 14.7

Percentage of female teachers	Number of states/U.T. $(f_i)$	$x_{i}$	$d_i = x_i - 50$	$u_i = \frac{x_i - 50}{10}$	$f_i x_i$	$f_i d_i$	$f_i u_i$
15 - 25	6	20	-30	-3	120	-180	-18
25 - 35	11	30	-20	-2	330	-220	-22
35 - 45	7	40	-10	-1	280	-70	– 7
45 - 55	4	50	0	0	200	0	0
55 - 65	4	60	10	1	240	40	4
65 - 75	2	70	20	2	140	40	4
75 - 85	1	80	30	3	80	30	3
Total	35				1390	-360	-36

From the table above, we obtain $\Sigma f_i = 35$ , $\Sigma f_i x_i = 1390$ ,

$$\mathrm{\Sigma }{f}_i{d}_i=-360,\mathrm{\Sigma }{f}_i{u}_i=-36.\backslash Sigma\; f\_i\; d\_i\; =\; -360,\; \backslash quad\; \backslash Sigma\; f\_i\; u\_i\; =\; -36.$$

Using the direct method, $$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{1390}{35}=39.71\backslash overline\{x\}\; =\; \backslash frac\{\backslash sum\; f\_i\; x\_i\}\{\backslash sum\; f\_i\}\; =\; \backslash frac\{1390\}\{35\}\; =\; 39.71$$

Using the assumed mean method,

$$\bar{x}=a+\frac{\sum f_i{d}_i}{\sum f_i}=50+\frac{(-360)}{35}=39.71\backslash overline\{x\}\; =\; a\; +\; \backslash frac\{\backslash sum\; f\_i\; d\_i\}\{\backslash sum\; f\_i\}\; =\; 50\; +\; \backslash frac\{(-360)\}\{35\}\; =\; 39.71$$

Using the step-deviation method,

$$\bar{x}=a+\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)\times h=50+\left(\frac{-36}{35}\right)\times 10=39.71\backslash overline\{x\}\; =\; a\; +\; \backslash left(\backslash frac\{\backslash sum\; f\_i\; u\_i\}\{\backslash sum\; f\_i\}\backslash right)\; \backslash times\; h\; =\; 50\; +\; \backslash left(\backslash frac\{-36\}\{35\}\backslash right)\; \backslash times\; 10\; =\; 39.71$$

Therefore, the mean percentage of female teachers in the primary schools of rural areas is 39.71.

Remark: The result obtained by all the three methods is the same. So the choice of method to be used depends on the numerical values of $x_i$ and $f_i$ . If $x_i$ and $f_i$ are sufficiently small, then the direct method is an appropriate choice. If $x_i$ and $f_i$ are numerically large numbers, then we can go for the assumed mean method or step-deviation method. If the class sizes are unequal, and $x_i$ are large numerically, we can still apply the step-deviation method by taking h to be a suitable divisor of all the $d_i$ 's.

Example 3: The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?

Number of wickets	20 - 60	60 - 100	100 - 150	150 - 250	250 - 350	350 - 450
Number of bowlers	7	5	16	12	2	3

Solution : Here, the class size varies, and the $x_i$ 's are large. Let us still apply the step-deviation method with a = 200 and h = 20. Then, we obtain the data as in Table 14.8.

$u_i = \frac{d_i}{20}$ Number of Number of $d_i = x_i - 200$ $u_i f_i$ $\boldsymbol{x}_{\cdot}$ wickets bowlers taken $(f_i)$ 20 - 60 7 40 -160-8-5660 - 100 5 80 -120-6 -3016 100 - 150-3.75-60125 -75150 - 25012 200 0 0 0 2 5 250 - 350 300 100 10 3 400 350 - 450 200 10 30 45 Total -106

Table 14.8

So, $$\bar{u}=\frac{-106}{45}\backslash overline\{u\}\; =\; \backslash frac\{-106\}\{45\}$$ . Therefore, $\overline{x} = 200 + 20\left(\frac{-106}{45}\right) = 200 - 47.11 = 152.89$ .

This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89.

Now, let us see how well you can apply the concepts discussed in this section!

Activity 2 :

Divide the students of your class into three groups and ask each group to do one of the following activities.

• 1. Collect the marks obtained by all the students of your class in Mathematics in the latest examination conducted by your school. Form a grouped frequency distribution of the data obtained.
• 2. Collect the daily maximum temperatures recorded for a period of 30 days in your city. Present this data as a grouped frequency table.
• 3. Measure the heights of all the students of your class (in cm) and form a grouped frequency distribution table of this data.

After all the groups have collected the data and formed grouped frequency distribution tables, the groups should find the mean in each case by the method which they find appropriate.

EXERCISE 14.1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants	0 - 2	2 - 4	4 - 6	6 - 8	8 - 10	10 - 12	12 - 14
Number of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean, and why?

2. Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in `)	500 - 520	520 -540	540 - 560	560 - 580	580 -600
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Daily pocket allowance (in `)	11 - 13	13 - 15	15 - 17	17 - 19	19 - 21	21 - 23	23 - 25
Number of children	7	6	9	13	f	5	4

4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Number of heartbeats per minute	65 - 68	68 - 71	71 - 74	74 - 77	77 - 80	80 - 83	83 - 86
Number of women	2	4	3	8	7	4	2

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes	50 - 52	53 - 55	56 - 58	59 - 61	62 - 64
Number of boxes	15	110	135	115	25

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in ₹)	100 - 150	150 - 200	200 - 250	250 - 300	300 - 350
Number of households	4	5	12	2	2

Find the mean daily expenditure on food by a suitable method.

7. To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Concentration of SO 2 (in ppm)	Frequency
0.00 - 0.04	4
0.04 - 0.08	9
0.08 - 0.12	9
0.12 - 0.16	2
0.16 - 0.20	4
0.20 - 0.24	2

Find the mean concentration of SO₂ in the air.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Number of days	0 - 6	6 - 10	10 - 14	14 - 20	20 - 28	28 - 38	38 - 40
Number of students	11	10	7	4	4	3	1

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %)	45 - 55	55 - 65	65 - 75	75 - 85	85 - 95
Number of cities	3	10	11	8	3

14.3 Mode of Grouped Data

Recall from Class IX, a mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency. Further, we discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining a mode of grouped data. It is possible that more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to problems having a single mode only.

Let us first recall how we found the mode for ungrouped data through the following example.

Example 4 : The wickets taken by a bowler in 10 cricket matches are as follows:

2 6 4 5 0 2 1 3 2 3

Find the mode of the data.

Solution : Let us form the frequency distribution table of the given data as follows:

Number of wickets	0	1	2	3	4	5	6
Number of matches	1	1	3	2	1	1	1

Clearly, 2 is the number of wickets taken by the bowler in the maximum number (i.e., 3) of matches. So, the mode of this data is 2.

In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula:

Mode = $$l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times hl\; +\; \backslash left(\backslash frac\{f\_1\; -\; f\_0\}\{2f\_1\; -\; f\_0\; -\; f\_2\}\backslash right)\; \backslash times\; h$$

where l = lower limit of the modal class,

h = size of the class interval (assuming all class sizes to be equal)

$f_1$ = frequency of the modal class,

$f_0$ = frequency of the class preceding the modal class,

$f_2$ = frequency of the class succeeding the modal class.

Let us consider the following examples to illustrate the use of this formula.

Example 5: A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:

Family size	1-3	3 - 5	5 - 7	7 - 9	9 - 11
Number of families	7	8	2	2	1

Find the mode of this data.

Solution : Here the maximum class frequency is 8, and the class corresponding to this frequency is 3 - 5. So, the modal class is 3 - 5.

Now

modal class = 3 - 5, lower limit (l) of modal class = 3, class size (h) = 2

frequency $(f_1)$ of the modal class = 8,

frequency $(f_0)$ of class preceding the modal class = 7,

frequency $(f_2)$ of class succeeding the modal class = 2.

Now, let us substitute these values in the formula:

Mode = $$l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times hl\; +\; \backslash left(\backslash frac\{f\_1\; -\; f\_0\}\{2f\_1\; -\; f\_0\; -\; f\_2\}\backslash right)\; \backslash times\; h$$

= $3 + \left(\frac{8 - 7}{2 \times 8 - 7 - 2}\right) \times 2 = 3 + \frac{2}{7} = 3.286$

Therefore, the mode of the data above is 3.286.

Example 6: The marks distribution of 30 students in a mathematics examination are given in Table 14.3 of Example 1. Find the mode of this data. Also compare and interpret the mode and the mean.

Solution : Refer to Table 14.3 of Example 1. Since the maximum number of students (i.e., 7) have got marks in the interval 40 - 55, the modal class is 40 - 55. Therefore,

the lower limit (l) of the modal class = 40,

the class size (h) = 15,

the frequency $(f_1)$ of modal class = 7,

the frequency $(f_0)$ of the class preceding the modal class = 3,

the frequency $(f_2)$ of the class succeeding the modal class = 6.

Now, using the formula:

Mode = $$l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times hl\; +\; \backslash left(\backslash frac\{f\_1\; -\; f\_0\}\{2f\_1\; -\; f\_0\; -\; f\_2\}\backslash right)\; \backslash times\; h$$ ,

we get

Mode = $$40+\left(\frac{7-3}{14-6-3}\right)\times 15=5240\; +\; \backslash left(\backslash frac\{7-3\}\{14-6-3\}\backslash right)\; \backslash times\; 15\; =\; 52$$

So, the mode marks is 52.

Now, from Example 1, you know that the mean marks is 62.

So, the maximum number of students obtained 52 marks, while on an average a student obtained 62 marks.

Remarks:

• 1. In Example 6, the mode is less than the mean. But for some other problems it may be equal or more than the mean also.
• 2. It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the average of the marks obtained by most

of the students. In the first situation, the mean is required and in the second situation, the mode is required.

Activity 3 : Continuing with the same groups as formed in Activity 2 and the situations assigned to the groups. Ask each group to find the mode of the data. They should also compare this with the mean, and interpret the meaning of both.

Remark : The mode can also be calculated for grouped data with unequal class sizes. However, we shall not be discussing it.

EXERCISE 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years)	5 - 15	15 - 25	25 - 35	35 - 45	45 - 55	55 - 65
Number of patients	6	11	21	23	14	5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

Lifetimes (in hours)	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency	10	35	52	61	38	29

Determine the modal lifetimes of the components.

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :

Expenditure (in `)	Number of families
1000 - 1500	24
1500 - 2000	40
2000 - 2500	33
2500 - 3000	28
3000 - 3500	30
3500 - 4000	22
4000 - 4500	16
4500 - 5000	7

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher	Number of states / U.T.
15 - 20	3
20 - 25	8
25 - 30	9
30 - 35	10
35 - 40	3
40 - 45	0
45 - 50	0
50 - 55	2

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored	Number of batsmen
3000 - 4000	4
4000 - 5000	18
5000 - 6000	9
6000 - 7000	7
7000 - 8000	6
8000 - 9000	3
9000 - 10000	1
10000 - 11000	1

Find the mode of the data.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data :

Number of cars	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60	60 - 70	70 - 80
Frequency	7	14	13	12	20	11	15	8

14.4 Median of Grouped Data

As you have studied in Class IX, the median is a measure of central tendency which gives the value of the middle-most observation in the data. Recall that for finding the median of ungrouped data, we first arrange the data values of the observations in

ascending order. Then, if n is odd, the median is the $\left(\frac{n+1}{2}\right)$ th observation. And, if n

is even, then the median will be the average of the $\frac{n}{2}$ th and the $\left(\frac{n}{2}+1\right)$ th observations.

Suppose, we have to find the median of the following data, which gives the marks, out of 50, obtained by 100 students in a test:

Marks obtained	20	29	28	33	42	38	43	25
Number of students	6	28	24	15	2	4	1	20

First, we arrange the marks in ascending order and prepare a frequency table as follows:

Table 14.9

Marks obtained	Number of students (Frequency)
20	6
25	20
28	24
29	28
33	15
38	4
42	2
43	1
Total	100

Here n = 100, which is even. The median will be the average of the $\frac{n}{2}$ th and the $\left(\frac{n}{2} + 1\right)$ th observations, i.e., the 50th and 51st observations. To find these observations, we proceed as follows:

Table 14.10

Marks obtained	Number of students
20	6
upto 25	6 + 20 = 26
upto 28	26 + 24 = 50
upto 29	50 + 28 = 78
upto 33	78 + 15 = 93
upto 38	93 + 4 = 97
upto 42	97 + 2 = 99
upto 43	99 + 1 = 100

Now we add another column depicting this information to the frequency table above and name it as cumulative frequency column.

Table 14.11

Marks obtained	Number of students	Cumulative frequency
20	6	6
25	20	26
28	24	50
29	28	78
33	15	93
38	4	97
42	2	99
43	1	100

From the table above, we see that:

50th observation is 28 (Why?)

51st observation is 29

So, Median = $$\frac{28+29}{2}=28.5\backslash frac\{28\; +\; 29\}\{2\}\; =\; 28.5$$

Remark : The part of Table 14.11 consisting Column 1 and Column 3 is known as Cumulative Frequency Table. The median marks 28.5 conveys the information that about 50% students obtained marks less than 28.5 and another 50% students obtained marks more than 28.5.

Now, let us see how to obtain the median of grouped data, through the following situation.

Consider a grouped frequency distribution of marks obtained, out of 100, by 53 students, in a certain examination, as follows:

Table 14.12

Marks	Number of students
0 - 10	5
10 - 20	3
20 - 30	4
30 - 40	3
40 - 50	3
50 - 60	4
60 - 70	7
70 - 80	9
80 - 90	7
90 - 100	8

From the table above, try to answer the following questions:

How many students have scored marks less than 10? The answer is clearly 5.

How many students have scored less than 20 marks? Observe that the number of students who have scored less than 20 include the number of students who have scored marks from 0 - 10 as well as the number of students who have scored marks from 10 - 20. So, the total number of students with marks less than 20 is 5 + 3, i.e., 8. We say that the cumulative frequency of the class 10 -20 is 8.

Similarly, we can compute the cumulative frequencies of the other classes, i.e., the number of students with marks less than 30, less than 40, . . ., less than 100. We give them in Table 14.13 given below:

Table 14.13

Marks obtained	Number of students (Cumulative frequency)
Less than 10	5
Less than 20	5 + 3 = 8
Less than 30	8 + 4 = 12
Less than 40	12 + 3 = 15
Less than 50	15 + 3 = 18
Less than 60	18 + 4 = 22
Less than 70	22 + 7 = 29
Less than 80	29 + 9 = 38
Less than 90	38 + 7 = 45
Less than 100	45 + 8 = 53

The distribution given above is called the cumulative frequency distribution of the less than type. Here 10, 20, 30, . . . 100, are the upper limits of the respective class intervals.

We can similarly make the table for the number of students with scores, more than or equal to 0, more than or equal to 10, more than or equal to 20, and so on. From Table 14.12, we observe that all 53 students have scored marks more than or equal to 0. Since there are 5 students scoring marks in the interval 0 - 10, this means that there are 53 – 5 = 48 students getting more than or equal to 10 marks. Continuing in the same manner, we get the number of students scoring 20 or above as 48 – 3 = 45, 30 or above as 45 – 4 = 41, and so on, as shown in Table 14.14.

Table 14.14

Marks obtained	Number of students (Cumulative frequency)
More than or equal to 0	53
More than or equal to 10	53 – 5 = 48
More than or equal to 20	48 – 3 = 45
More than or equal to 30	45 – 4 = 41
More than or equal to 40	41 – 3 = 38
More than or equal to 50	38 – 3 = 35
More than or equal to 60	35 – 4 = 31
More than or equal to 70	31 – 7 = 24
More than or equal to 80	24 – 9 = 15
More than or equal to 90	15 – 7 = 8

The table above is called a cumulative frequency distribution of the more than type. Here 0, 10, 20, . . ., 90 give the lower limits of the respective class intervals.

Now, to find the median of grouped data, we can make use of any of these cumulative frequency distributions.

Let us combine Tables 14.12 and 14.13 to get Table 14.15 given below:

Table 14.15

Marks	Number of students ( f )	Cumulative frequency (cf)
0 - 10	5	5
10 - 20	3	8
20 - 30	4	12
30 - 40	3	15
40 - 50	3	18
50 - 60	4	22
60 - 70	7	29
70 - 80	9	38
80 - 90	7	45
90 - 100	8	53

Now in a grouped data, we may not be able to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in

a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves. But which class should this be?

To find this class, we find the cumulative frequencies of all the classes and $\frac{n}{2}$ . We now locate the class whose cumulative frequency is greater than (and nearest to) $\frac{n}{2}$ . This is called the median class. In the distribution above, n = 53. So, $\frac{n}{2} = 26.5$ . Now 60 - 70 is the class whose cumulative frequency 29 is greater than (and nearest to) $\frac{n}{2}$ , i.e., 26.5.

Therefore, 60 - 70 is the median class.

After finding the median class, we use the following formula for calculating the median.

Median = $$l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times hl\; +\; \backslash left(\backslash frac\{\backslash frac\{n\}\{2\}\; -\; cf\}\{f\}\backslash right)\; \backslash times\; h$$ ,

where

l = lower limit of median class,

n = number of observations,

cf = cumulative frequency of class preceding the median class,

f = frequency of median class,

h = class size (assuming class size to be equal).

Substituting the values $\frac{n}{2} = 26.5$ , l = 60, cf = 22, f = 7, h = 10

in the formula above, we get

Median = $$60+\left(\frac{26.5-22}{7}\right)\times 1060\; +\; \backslash left(\backslash frac\{26.5\; -\; 22\}\{7\}\backslash right)\; \backslash times\; 10$$

= $60 + \frac{45}{7}$
= $66.4$

So, about half the students have scored marks less than 66.4, and the other half have scored marks more than 66.4.

Example 7 : A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained:

Height (in cm)	Number of girls
Less than 140	4
Less than 145	11
Less than 150	29
Less than 155	40
Less than 160	46
Less than 165	51

Find the median height.

Solution : To calculate the median height, we need to find the class intervals and their corresponding frequencies.

The given distribution being of the less than type, 140, 145, 150, . . ., 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140 - 145, 145 - 150, . . ., 160 - 165. Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e., the frequency of class interval below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140 - 145 is 11 – 4 = 7. Similarly, the frequency of 145 - 150 is 29 – 11 = 18, for 150 - 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with the given cumulative frequencies becomes:

Table 14.16

Class intervals	Frequency	Cumulative frequency
Below 140	4	4
140 - 145	7	11
145 - 150	18	29
150 - 155	11	40
155 - 160	6	46
160 - 165	5	51

Now $$n=51n\; =\; 51$$ . So, $\frac{n}{2} = \frac{51}{2} = 25.5$ . This observation lies in the class 145 - 150. Then, $l$ (the lower limit) = 145,

cf (the cumulative frequency of the class preceding 145 - 150) = 11,

f (the frequency of the median class 145 - 150) = 18,

h (the class size) = 5.

Using the formula, Median = $$l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times hl\; +\; \backslash left(\backslash frac\{\backslash frac\{n\}\{2\}\; -\; cf\}\{f\}\backslash right)\; \backslash times\; h$$ , we have

Median = $$145+\left(\frac{25.5-11}{18}\right)\times 5145\; +\; \backslash left(\backslash frac\{25.5\; -\; 11\}\{18\}\backslash right)\; \backslash times\; 5$$

= $145 + \frac{72.5}{18} = 149.03$ .

So, the median height of the girls is 149.03 cm.

This means that the height of about 50% of the girls is less than this height, and 50% are taller than this height.

Example 8: The median of the following data is 525. Find the values of x and y, if the total frequency is 100.

Class interval	Frequency
0 - 100	2
100 - 200	5
200 - 300	x
300 - 400	12
400 - 500	17
500 - 600	20
600 - 700	у
700 - 800	9
800 - 900	7
900 - 1000	4

Solution:

Class intervals	Frequency	Cumulative frequency
0 - 100	2	2
100 - 200	5	7
200 - 300	x	7 + x
300 - 400	12	19 + x
400 - 500	17	36 + x
500 - 600	20	56 + x
600 - 700	y	56 + x + y
700 - 800	9	65 + x + y
800 - 900	7	72 + x + y
900 - 1000	4	76 + x + y

It is given that n = 100

So, $$76+x+y=10076\; +\; x\; +\; y\; =\; 100$$ , i.e., $x + y = 24$ (1)

The median is 525, which lies in the class 500 - 600

So, $$l=500l\; =\; 500$$ , $f = 20$ , $cf = 36 + x$ , $h = 100$

Using the formula:

Median = $$l+\left(\frac{\frac{n}{2}-cf}{f}\right)hl\; +\; \backslash left(\backslash frac\{\backslash frac\{n\}\{2\}\; -\; cf\}\{f\}\backslash right)h$$ , we get

$$525=500+\left(\frac{50-36-x}{20}\right)\times 100525\; =\; 500\; +\; \backslash left(\backslash frac\{50\; -\; 36\; -\; x\}\{20\}\backslash right)\; \backslash times\; 100$$

i.e., $$525-500=(14-x)\times 5525\; -\; 500\; =\; (14\; -\; x)\; \backslash times\; 5$$

i.e., $$25=70-5x25\; =\; 70\; -\; 5x$$

i.e., $$5x=70-25=455x\; =\; 70\; -\; 25\; =\; 45$$

So, $$x=9x\; =\; 9$$

Therefore, from (1), we get 9 + y = 24

i.e., $$y=15y\; =\; 15$$

Now, that you have studied about all the three measures of central tendency, let us discuss which measure would be best suited for a particular requirement.

The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes, i.e., the largest and the smallest observations of the entire data. It also enables us to compare two or more distributions. For example, by comparing the average (mean) results of students of different schools of a particular examination, we can conclude which school has a better performance.

However, extreme values in the data affect the mean. For example, the mean of classes having frequencies more or less the same is a good representative of the data. But, if one class has frequency, say 2, and the five others have frequency 20, 25, 20, 21, 18, then the mean will certainly not reflect the way the data behaves. So, in such cases, the mean is not a good representative of the data.

In problems where individual observations are not important, and we wish to find out a 'typical' observation, the median is more appropriate, e.g., finding the typical productivity rate of workers, average wage in a country, etc. These are situations where extreme values may be there. So, rather than the mean, we take the median as a better measure of central tendency.

In situations which require establishing the most frequent value or most popular item, the mode is the best choice, e.g., to find the most popular T.V. programme being watched, the consumer item in greatest demand, the colour of the vehicle used by most of the people, etc.

Remarks :

1. There is a empirical relationship between the three measures of central tendency :

3 Median = Mode + 2 Mean

2. The median of grouped data with unequal class sizes can also be calculated. However, we shall not discuss it here.

EXERCISE 14.3

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)	Number of consumers
65 - 85	4
85 - 105	5
105 - 125	13
125 - 145	20
145 - 165	14
165 - 185	8
185 - 205	4

2. If the median of the distribution given below is 28.5, find the values of x and y.

Class interval	Frequency
0 - 10	5
10 - 20	x
20 - 30	20
30 - 40	15
40 - 50	y
50 - 60	5
Total	60

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Age (in years)	Number of policy holders
Below 20	2
Below 25	6
Below 30	24
Below 35	45
Below 40	78
Below 45	89
Below 50	92
Below 55	98
Below 60	100

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

Length (in mm)	Number of leaves
118 - 126	3
127 - 135	5
136 - 144	9
145 - 153	12
154 - 162	5
163 - 171	4
172 - 180	2

Find the median length of the leaves.

(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

5. The following table gives the distribution of the life time of 400 neon lamps :

Life time (in hours)	Number of lamps
1500 - 2000	14
2000 - 2500	56
2500 - 3000	60
3000 - 3500	86
3500 - 4000	74
4000 - 4500	62
4500 - 5000	48

Find the median life time of a lamp.

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	1 - 4	4 - 7	7 - 10	10 - 13	13 - 16	16 - 19
Number of surnames	6	30	40	16	4	4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Weight (in kg)	40 - 45	45 - 50	50 - 55	55 - 60	60 - 65	65 - 70	70 - 75
Number of students	2	3	8	6	6	3	2

14.5 Graphical Representation of Cumulative Frequency Distribution

As we all know, pictures speak better than words. A graphical representation helps us in understanding given data at a glance. In Class IX, we have represented the data through bar graphs, histograms and frequency polygons. Let us now represent a cumulative frequency distribution graphically.

For example, let us consider the cumulative frequency distribution given in Table 14.13.

Recall that the values 10, 20, 30, . . ., 100 are the upper limits of the respective class intervals. To represent the data in the table graphically, we mark the upper limits of the class intervals on the horizontal axis (x-axis) and their corresponding cumulative frequencies on the vertical axis ( y-axis), choosing a convenient scale. The scale may not be the same on both the axis. Let us now plot the points corresponding to the ordered pairs given by (upper limit, corresponding cumulative frequency), i.e., (10, 5), (20, 8), (30, 12), (40, 15),

Fig. 14.1

(50, 18), (60, 22), (70, 29), (80, 38), (90, 45), (100, 53) on a graph paper and join them by a free hand smooth curve. The curve we get is called a cumulative frequency curve, or an ogive (of the less than type). (See Fig. 14.1)

The term 'ogive' is pronounced as 'ojeev' and is derived from the word ogee. An ogee is a shape consisting of a concave arc flowing into a convex arc, so forming an S-shaped curve with vertical ends. In architecture, the ogee shape is one of the characteristics of the 14th and 15th century Gothic styles.

Next, again we consider the cumulative frequency distribution given in Table 14.14 and draw its ogive (of the more than type).

Recall that, here 0, 10, 20, . . ., 90 are the lower limits of the respective class intervals 0 - 10, 10 - 20, . . ., 90 - 100. To represent 'the more than type' graphically, we plot the lower limits on the x-axis and the corresponding cumulative frequencies on the y-axis. Then we plot the points (lower limit, corresponding cumulative frequency), i.e., (0, 53), (10, 48), (20, 45), (30, 41), (40, 38), (50, 35), (60, 31), (70, 24), (80, 15), (90, 8), on a graph paper, and join them by a free hand smooth curve.

The curve we get is a cumulative frequency curve, or an ogive (of the more than type). (See Fig. 14.2)

Remark: Note that both the ogives (in Fig. 14.1 and Fig. 14.2) correspond to the same data, which is given in Table 14.12.

Now, are the ogives related to the median in any way? Is it possible to obtain the median from these two cumulative frequency curves corresponding to the data in Table 14.12? Let us see.

One obvious way is to locate

$$\frac{n}{2}=\frac{53}{2}=26.5\backslash frac\{n\}\{2\}\; =\; \backslash frac\{53\}\{2\}\; =\; 26.5$$ on the y-axis (see Fig.

14.3). From this point, draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis determines the median of the data (see Fig. 14.3).

Another way of obtaining the median is the following:

Draw both ogives (i.e., of the less than type and of the more than type) on the same axis. The two ogives will intersect each other at a point. From this point, if we draw a perpendicular on the x-axis, the point at which it cuts the x-axis gives us the median (see Fig. 14.4).

Fig. 14.3

10

10

10

20

10

20

30

40

10

10

10

10

10

10

10

10

1

Example 9: The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution:

Profit (Rs in lakhs)	Number of shops (frequency)
More than or equal to 5	30
More than or equal to 10	28
More than or equal to 15	16
More than or equal to 20	14
More than or equal to 25	10
More than or equal to 30	7
More than or equal to 35	3

Draw both ogives for the data above. Hence obtain the median profit.

Solution : We first draw the coordinate axes, with lower limits of the profit along the horizontal axis, and the cumulative frequency along the vertical axes. Then, we plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7) and (35, 3). We join these points with a smooth curve to get the 'more than' ogive, as shown in Fig. 14.5.

Now, let us obtain the classes, their frequencies and the cumulative frequency from the table above.

Fig. 14.5

Table 14.17

Classes	5 - 10	10 - 15	15 - 20	20 - 25	25 - 30	30 - 35	35 - 40
No. of shops	2	12	2	4	3	4	3
Cumulative frequency	2	14	16	20	23	27	30

Using these values, we plot the points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27), (40, 30) on the same axes as in Fig. 14.5 to get the 'less than' ogive, as shown in Fig. 14.6.

The abcissa of their point of intersection is nearly 17.5, which is the median. This can also be verified by using the formula. Hence, the median profit (in lakhs) is ` 17.5.

Remark : In the above examples, it may be noted that the class intervals were continuous. For drawing ogives, it should be ensured that the class intervals are continuous. (Also see constructions of histograms in Class IX)

EXERCISE 14.4

1. The following distribution gives the daily income of 50 workers of a factory.

Daily income (in ₹)	100 - 120	120 - 140	140 - 160	160 - 180	180 - 200
Number of workers	12	14	8	6	10

Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.

2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Weight (in kg)	Number of students
Less than 38	0
Less than 40	3
Less than 42	5
Less than 44	9
Less than 46	14
Less than 48	28
Less than 50	32
Less than 52	35

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

3. The following table gives production yield per hectare of wheat of 100 farms of a village.

Production yield (in kg/ha)	50-55	55-60	60-65	65 - 70	70-75	75 - 80
Number of farms	2	8	12	24	38	16

Change the distribution to a more than type distribution, and draw its ogive.

14.6 Summary

In this chapter, you have studied the following points:

1. The mean for grouped data can be found by:

(i) the direct method : $$\bar{x}=\frac{\sum f_i{x}_i}{\sum f_i}\backslash overline\{x\}\; =\; \backslash frac\{\backslash sum\; f\_i\; x\_i\}\{\backslash sum\; f\_i\}$$

• (ii) the assumed mean method: $\overline{x} = a + \frac{\sum f_i d_i}{\sum f_i}$
• (iii) the step deviation method : $\overline{x} = a + \left(\frac{\sum f_i u_i}{\sum f_i}\right) \times h$ ,

with the assumption that the frequency of a class is centred at its mid-point, called its class mark.

2. The mode for grouped data can be found by using the formula:

Mode = $$l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times hl\; +\; \backslash left(\backslash frac\{f\_1\; -\; f\_0\}\{2f\_1\; -\; f\_0\; -\; f\_2\}\backslash right)\; \backslash times\; h$$

where symbols have their usual meanings.

• 3. The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class.
• 4. The median for grouped data is formed by using the formula:

$$Median=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h,Median\; =\; l\; +\; \backslash left(\backslash frac\{\backslash frac\{n\}\{2\}\; -\; cf\}\{f\}\backslash right)\; \backslash times\; h,$$

where symbols have their usual meanings.

• 5. Representing a cumulative frequency distribution graphically as a cumulative frequency curve, or an ogive of the less than type and of the more than type.
• 6. The median of grouped data can be obtained graphically as the x-coordinate of the point of intersection of the two ogives for this data.

A NOTE TO THE READER

For calculating mode and median for grouped data, it should be ensured that the class intervals are continuous before applying the formulae. Same condition also apply for construction of an ogive. Further, in case of ogives, the scale may not be the same on both the axes.