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

Exercise 21B

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Question 1:

Find the median of:

(i) 3, 11, 7, 2, 5, 9, 9, 2, 10
(ii) 9,25, 18, 15, 6, 16, 8, 22, 21
(iii) 21, 15, 6, 25, 18, 13, 20, 9, 16, 8, 22

Answer 1:

We have to find the median of the following data.

(i)  3, 11, 7, 2, 5, 9, 9, 2 and 10

Arranging them in ascending order:

 2, 2, 3, 5, 7, 9, 9, 10, 11

Number of terms, N= 9
It is an odd number.

Median =$\left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation}$

Median=$\left(\frac{9+1}{2}\right)\mathrm{th} \mathrm{observation}$

Median = 5th observation

Median=7

(ii) 9, 25, 18, 15, 6, 16, 8, 22, 21

Arranging them in ascending order,

6, 8, 9, 15, 16, 18, 21, 22, 25

Number of terms, N=9
It is an odd number.

Median =$\left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation}$

Median =$\left(\frac{9+1}{2}\right)\mathrm{th} \mathrm{observation}$
Median = 5th observation

Median=16

(iii) 21, 15, 6, 25, 18, 13, 20, 9, 16, 8, 22

Arranging them in ascending order:

6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25

Number of terms, N = 11
It is an odd number.

Median =$\left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation}$

Median =$\left(\frac{11+1}{2}\right)$th observation

Median = 6th observation

Median=16

Question 2:

Find the median of:

(i) 10, 32, 17, 19, 21, 22, 9, 35
(ii) 55, 60, 35, 51, 29, 63, 72, 91, 85, 82

Answer 2:

We have to find the median of the following data.

(i) 10, 32, 17, 19, 21, 22, 9, 35

Arranging them in ascending  order:

9, 10, 17, 19, 21, 22, 32, 35

Number of terms, N = 8

 Median =$\frac{1}{2}\left\{\left(\frac{N}{2}\right)\mathrm{th} \mathrm{observation}+\left(\frac{N}{2}+1\right)\mathrm{th} \mathrm{observation}\right\}$

Median =$\frac{1}{2}\left(4\mathrm{th} \mathrm{observation}+5\mathrm{th} \mathrm{observation}\right)$

Median =$\frac{1}{2}\left(19+21\right)=20$

∴ Median= 20

(ii) 55, 60, 35, 51, 29, 63, 72, 91, 85, 82

Arranging them in ascending order:

29, 35, 51, 55, 60, 63, 72, 82, 85, 91

Number of terms, N =10

Median =$\frac{1}{2}\left\{\left(\frac{N}{2}\right)\mathrm{th} \mathrm{observation}+\left(\frac{N}{2}+1\right)\mathrm{th} \mathrm{observation}\right\}$

$$\begin{gathered} \mathrm{Median} = \frac{1}{2}\left(5\mathrm{th} \mathrm{observation} +6\mathrm{th} \mathrm{observation}\right) \\ \mathrm{Median} =\frac{1}{2}\left(60+63\right) \\ \therefore \mathrm{Median}=61.5 \end{gathered}$$

Question 3:

Find the median of first 15 odd numbers.

Answer 3:

First 15 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27 and 29.

Number of terms, N = 15
It is an odd number.

Median =$\left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation}$

$$\begin{gathered} \mathrm{Median} =\left(\frac{15+1}{2}\right)\mathrm{th} \mathrm{observation} \\ \mathrm{Median} =15 \end{gathered}$$

Question 4:

Find the median of first 10 even numbers.

Answer 4:

First 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18 and 20.

Number of terms, N=10

Median =$\frac{1}{2}\left\{\left(\frac{N}{2}\right)\mathrm{th} \mathrm{observation}+\left(\frac{N}{2}+1\right)\mathrm{th} \mathrm{observation}\right\}$

$$\begin{gathered} \mathrm{Median} =\frac{1}{2}\left(5\mathrm{th} \mathrm{observation}+6\mathrm{th} \mathrm{observation}\right) \\ \mathrm{Median} =\frac{1}{2}\left(10+12\right)=11 \end{gathered}$$

Question 5:

Find the median of first 50 whole numbers.

Answer 5:

First 50 whole numbers are 0, 1, 2, 3, 4 ... and 49.

Number of terms, N= 50
It is an even number.

$$\begin{gathered} \mathrm{Median} =\frac{1}{2}\left\{\left(\frac{N}{2}\right)\mathrm{th} \mathrm{observation}+\left(\frac{N}{2}+1\right)\mathrm{th} \mathrm{observation}\right\} \\ =\frac{1}{2}\left\{25\mathrm{th} \mathrm{observation} +26\mathrm{th} \mathrm{observation}\right\} \\ =\frac{1}{2}\left\{24+25\right\} \\ = 24.5 \end{gathered}$$

Question 6:

The marks of 15 students (out of 50) in an examination are
20, 22, 26, 31, 40, 19, 17, 19, 25, 29, 23, 17, 24, 21, 35

Find the median marks.

Answer 6:

Marks of the students (out of 50) in an examination are given below:

20, 22, 26, 31, 40, 19, 17, 19, 25, 29, 23, 17, 24, 21, 35

Arranging the marks in ascending order:

17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40

Number of terms, N=15
This is an odd number.

$$\begin{gathered} \mathrm{Median} = \left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation} \\ \mathrm{Median}=\left(\frac{15+1}{2}\right)\mathrm{th} \mathrm{observation} \\ \mathrm{Median} =8\mathrm{th} \mathrm{observation} \\ \mathrm{Median} = 23 \end{gathered}$$

Hence, the median marks are 23.

Question 7:

The ages (in years) of 10 teachers in a school are
34, 37, 53, 46, 52, 43, 31, 36, 40, 50.

Find the median age.

Answer 7:

Ages (in years) of 10 teachers in a school are given below:

34, 37, 53, 46, 52, 43, 31, 36, 40, 50

Arranging them in ascending order:

31, 34, 36, 37, 40, 43, 46, 50, 52, 53

Number of terms, N=10
It is an even number.

$$\begin{gathered} \mathrm{Median}=\frac{1}{2}\left\{\left(\frac{N}{2}\right)\mathrm{th} \mathrm{observation}+\left(\frac{N}{2}+1\right)\mathrm{th} \mathrm{observation}\right\} \\ \mathrm{Median} =\frac{1}{2}\left\{5\mathrm{th} \mathrm{observation} +6\mathrm{th} \mathrm{observation}\right\} \\ \mathrm{Median} =\frac{1}{2}\left\{40+43\right\} \\ \mathrm{Median} =41.5 \end{gathered}$$

Hence, the median age is 41.5 years.

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Question 8:

Find the median weight for the following data:
 

Weight (in kg)	45	46	48	50	52	54	55
Number of boys	8	5	6	9	7	4	2

Answer 8:

Cumulative frequency table:

 

Number of terms, N = 41
It is an odd number.

$$\begin{gathered} \mathrm{Median} =\left\{\left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation}\right\} \\ =\left\{\left(\frac{41+1}{2}\right)\mathrm{th} \mathrm{observation}\right\} \\ =\left\{21\mathrm{th} \mathrm{observation}\right\} \\ =50 \mathrm{kg} \end{gathered}$$

Hence, the median weight is 50 kg.

Question 9:

Calculate the median for the following data:
 

Marks	17	20	22	15	30	25
Number of students	5	9	4	3	10	6

Answer 9:

Arranging the terms in ascending order, we have:



Cumulative frequency table:


Number of terms, N = 37

$$\begin{gathered} \mathrm{Median} = \left\{\left(\frac{N+1}{2}\right)\mathrm{th} \mathrm{observation}\right\} \\ =\left\{\left(\frac{37+1}{2}\right)\mathrm{th} \mathrm{observation}\right\} \\ =19\mathrm{th} \mathrm{observation} \\ =22 \end{gathered}$$

Hence, the median is 22.

Question 10:

The heights (in cm) of 50 students of a class are given below:
 

Height (in cm)	156	154	155	151	157	152	153
Number of students	8	4	10	6	7	3	12

Find the median height.

Answer 10:

Arranging the terms in ascending order:



Cumulative frequency table:



Number of terms, N = 50

$$\begin{gathered} \mathrm{Median} =\frac{1}{2}\left\{\left(\frac{N}{2}\right)\mathrm{th} \mathrm{observation}+\left(\frac{N}{2}+1\right)\mathrm{th} \mathrm{observation}\right\} \\ =\frac{1}{2}\left\{25\mathrm{th} \mathrm{observation}+26\mathrm{th} \mathrm{observation}\right\} \\ =\frac{1}{2}\left\{154+155\right\} \end{gathered}$$

Median =154.5