RS Aggarwal 2019,2020 solution class 7 chapter 21 Collection and Organisation of Data(Mean, Median and Mode) Exercise 21C

Exercise 21C

Page-269

Question 1:

Find the mode of the data:

(i) 10, 8, 4, 7, 8, 11, 15, 8, 6, 8
(ii) 27, 23, 39, 18, 27, 21, 27, 27, 40, 36, 27

Answer 1:

We have to find the mode of the given data.

Mode - It is that value of the variables that occurs most frequently.

(i) 10, 8, 4, 7, 8, 11, 15, 8, 6, 8

Here, 8 occurs most frequently. Hence, the mode of the data is 8.

(ii) 27, 23, 39, 18, 27, 21, 27, 27, 40, 36, 27

Here, 27 occurs most frequently. Hence, the mode of the data is 27.

Question 2:

The ages (in years) of 11 cricket players are given below:
28, 34, 32, 41, 36, 32, 32, 38, 32, 40, 31.

Find the mode of the ages.

Answer 2:

Following are the ages (in years) of 11 cricket players:

28, 34, 32, 41, 36, 32, 32, 38, 32, 40, 31

Mode is the value of the variable that occurs most frequently.

Here, 32 occurs maximum number of times.

Hence, 32 is the mode of the ages.

Question 3:

Daily wages of 45 workers in a factory are given below:
 

Daily wages (in Rs)	100	125	150	175	200
Number of workers	6	8	9	12	10

Find the median and the mean.
Using empirical formula, calculate its mode.

Answer 3:

$$\begin{gathered} \mathrm{Here}, N \mathrm{is} 45, \mathrm{which} \mathrm{is} \mathrm{odd}. \\ \mathrm{Median} =\left\{\left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation}\right\} \\ =\left\{\frac{45+1}{2}\right\}\mathrm{observation} \\ =23 \mathrm{th} \mathrm{observation} \\ \mathrm{Median} =150 \\ \mathrm{Mean} =\frac{\sum (f_{\mathrm{i}}\times {\mathit{x}}_{\mathrm{i}})}{\sum f_{\mathrm{i}}}=\frac{7050}{45}=156.67 \\ \mathrm{Mode}=3\left(\mathrm{Median}\right)-2\left(\mathrm{Mean}\right) \\ =3\left(150\right)-2(156.67) \\ =450-313.34 \\ =136.6 \end{gathered}$$

Hence, the median is 150, the mean is 156.67 and the mode is 136.6.

Question 4:

The following table shows the marks obtained by 41 students of a class.

Marks obtained	15	17	20	22	25	30
Number of students	2	5	10	12	8	4

Find the median and mean marks.
Using empirical formula, calculate its mode.

Answer 4:

$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{terms} \left(N\right) is 41, \mathrm{which} \mathrm{is}\text{ odd}. \\ \mathrm{Median} =\left\{\left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation}\right\} \\ =\left\{21\mathrm{th} \mathrm{observation}\right\} \\ =22 \\ \mathrm{Median} =22 \\ \mathrm{Mean} =\frac{\sum \mathit{(}{f}_i\mathit{\times }{x}_i\mathit{)}}{\sum f_{\mathrm{i}}} \\ =\frac{899}{41} \\ \mathrm{Mean} =21.92 \\ \mathrm{Using} \mathrm{empirical} \mathrm{formula}: \\ \mathrm{Mode} = 3\left(\mathrm{Median}\right)-2\left(\mathrm{Mean}\right) \\ = 66-43.84 \\ \mathrm{Mode}=22.16 \end{gathered}$$

Hence, the median is 22, the mean is 21.92 and the mode is 22.16.

Question 5:

The following table shows the weight of 12 players:
 

Weight (in kg)	48	50	52	54	58
Number of players	4	3	2	2	1

Find the median and mean weights.
Using empirical formula, calculate its mode.

Answer 5:

We will prepare the table given below:



$$\begin{gathered} \mathrm{Number} \mathrm{of} \mathrm{terms} \left(N\right) \mathrm{is} 12, \mathrm{which} \mathrm{is} \mathrm{an} \mathrm{even} \mathrm{number}. \\ \mathrm{Median} =\frac{1}{2}\left\{\left(\frac{N}{2}\right)\mathrm{th} \mathrm{observation}+\left(\frac{N}{2}+1\right)\mathrm{th} \mathrm{observation}\right\} \\ =\left\{6\mathrm{th} \mathrm{observation}+7\mathrm{th} \mathrm{observation}\right\} \\ =\frac{1}{2}\left\{50+50\right\} \\ \mathrm{Median} =50 \\ \mathrm{Mean} =\frac{\sum \mathit{(}{f}_i\mathit{\times }{x}_i\mathit{)}}{\sum f_{\mathrm{i}}} \\ =\frac{612}{12} \\ \mathrm{Mean} = 51 \\ \mathrm{Using} \mathrm{empirical} \mathrm{formula}: \\ \mathrm{Mode} = 3\left(\mathrm{Median}\right)-2\left(\mathrm{Mean}\right) \\ = 150-102 \\ \mathrm{Mode}=48 \end{gathered}$$

Hence, the median is 50, the mean is 51 and the mode is 48.