Question

Size of agriculture holding in a survey of 200 families is given in the following table:

Size of agriculture	0 – 5	5 – 10	10 – 15	15 – 20	20 – 25	25 – 30	30 – 35
Number of family	10	15	30	80	40	20	5

Compute median and mode size of the holdings.

Answer 1

Hint: Median is the value of the variable which divides the whole set of data into two equal parts. Mode is the value for which the frequency is maximum. Draw the frequency distribution table. Find the median and modal class. And then use the formula of median and mode for continuous class distribution to solve the question.

Complete step-by-step answer:
Let us form the cumulative frequency table first

Size of agriculture	Number of family	Cumulative frequency
0 – 5	10	10
5 – 10	15	25
10 – 15	30	55
15 – 20	80	135
20 – 25	40	175
25 – 30	20	195
30 – 35	5	200
Total	$N = 200$

Median is the value of the variable which divides the whole set of data into two equal parts.
For a continuous grouped data, median is given by
$M = l + \dfrac{{\dfrac{N}{2} - m}}{f} \times c$ . . . (1)
Where,
$M$ is median
$l$ is lower limit of the median class
$N$ is total number of frequencies
$f$ is frequency of the median class
$m$ is cumulative frequency of the class before the median class
$c$ then class interval of the median class
From the above formula, it is clear that we need to find the median class first to get all the rest of the values required in the formula. Median class, is a class in which cumulative frequency is just greater than $\dfrac{N}{2}$.
By observing the above table, we can write
$\dfrac{N}{2} = \dfrac{{200}}{2} = 100$
Which is close to the cumulative frequency 135. Hence, the median class is $15 - 20$.
Thus we get,
$l = 15$
$m = 55$
$f = 80$
$c = 5$
Substitute the given values in the equation (1). We get
$M = 15 + \dfrac{{100 - 55}}{{80}} \times 5$
$ = 15 + \dfrac{{45}}{{16}}$
$ \Rightarrow M = 17.81$
Therefore, median is $17.81$
Now, mode of the continuous distribution is the value, for which, frequency is maximum. It is the value around which the observations tend to be most heavily concentrated. Thus, modal class is the class which has the highest frequency.
Formula for mode is given as,
$Mode = l + \dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}} \times c$ . . . (2)
Where,
$l$ is lower limit of the mode class
${f_1}$ is frequency of modal class
${f_0}$ is frequency of the class before the modal class
${f_2}$ is frequency of the class after the modal class
$c$ is class length
By observing the above table, we can write
Maximum frequency is $80$. Therefore, modal class is $15 - 20$.
$l = 15$
${f_1} = 80$
${f_0} = 30$
 ${f_2} = 40$
$c = 5$
Substitute the given values in the equation (2). We get
$Mode = 15 + \dfrac{{80 - 30}}{{2 \times 80 - 30 - 40}} \times 5$
$ = 15 + \dfrac{{50 \times 5}}{{160 - 70}}$
$ = 15 + \dfrac{{250}}{{90}}$
$ \Rightarrow Mode = 17.77$
Therefore, mode of the given data is $17.77$

Note: This question is completely formula based. For such questions, you need to know the formula of the asked term. Otherwise, it would not be possible to solve the question. Cumulative frequency and frequency are different. Do not get confused with that. You can also solve this question using graphs (Ogives).