ML Aggarwal Solution Class 10 Chapter 21 Measures of Central Tendency Exercise 21.3

 Exercise 21.3

Question 1

Find the mode of the following sets of numbers ;

(i) 3, 2, 0, 1, 2, 3, 5, 3

(ii) 5, 7, 6, 8, 9, 0, 6, 8, 1, 8

(iii) 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7

Sol :

(i) ∵ The number 3 occurs maximum times

Mode = 3

(ii) ∵ The number 8 occurs maximum times

Mode = 8

(iii) ∵ The number 5, occurs maximum times

Mode = 5


Question 2

Calculate the mean, the median and the mode of the numbers : 3, 2, 6, 3, 3, 1, 1, 2

Sol :

Arranging in ascending order 1, 1, 2, 2, 3, 3, 3, 6

(i) Mean $=\frac{\sum x_{i}}{n}$

$=\frac{1+1+2+2+3+3+3+6}{8}$

$=\frac{21}{8}$

= 2.625


(ii) Here n = 8 which is even

$\therefore$ Median $=\frac{1}{2}\left[\frac{n}{2} \mathrm{th}+\left(\frac{n}{2}+1\right) \mathrm{th}\right]$

$=\frac{1}{2}\left[\frac{8}{2}+\left(\frac{8}{2}+1\right)\right]$

$=\frac{1}{2}$(4th term+5th term)

$=\frac{1}{2}[2+3]=\frac{5}{2}=2 \cdot 5$


(iii) Here 3 occurs maximum times

∴Mode=3


Question 3

Find the mean, median and mode of the following distribution : 8, 10, 7, 6, 10, 11, 6, 13, 10

Sol :
$\operatorname{Mean}=\frac{8+10+7+6+10+11+6+13+10}{2}$

$=\frac{81}{9}=9$

Given nos. in ascending order are as follows:

6, 6, 7, 8, 10, 10, 10, 11, 13

Median $=\frac{n+1}{2}$ th term

$=\frac{9+1}{2}=5$ th term=10

Mode = 10 (having highest frequency 3 times)


Question 4

Calculate the mean, the median and the mode of the following numbers : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2

Sol :

Arranging in ascending order 1, 2, 3, 3, 3, 4, 5, 5, 6, 7

(i) Mean $=\frac{\sum x_{i}}{n}$

$=\frac{1+2+3+3+3+4+5+5+6+7}{8}$

$=\frac{39}{10}$

=3.9


(ii)

(Here n=10 which is even)

$\therefore$ Median $=\frac{1}{2}\left[\frac{10}{2}\right.$ th term $+\left(\frac{10}{2}+1\right)$ th term $]$

$=\frac{1}{2}$ (5th +6 th terms)

$=\frac{1}{2}(3+4)=\frac{7}{2}$

=3.5


(iii) Here 3 occurs maximum times

∴Mode=3


Question 5

The marks of 10 students of a class in an examination arranged in ascending order are as follows: 13, 35, 43, 46, x, x +4, 55, 61,71, 80

If the median marks is 48, find the value of x. Hence, find the mode of the given data. (2017)

Sol :

Given marks are 13, 35, 43, 46, x, x + 4, 55, 61, 71, 80

n = 10 (even), median = 48

Median $=\left(\frac{n}{2}\right)^{\text {th }}$ observation $+\left(\frac{n}{2}+1\right)^{\operatorname{th}}$ observation
$=\frac{\text { 5th observation }+6 \text { th observation }}{2}$
$=\frac{x+x+4}{2}$
$\therefore 48=\frac{2 x+4}{2}$

$\Rightarrow 48=\frac{2(x+2)}{2}$

$\Rightarrow x=48-2=46$

Now , put the value of x in 6th observation i.e. x+4

∴x+4=46+4=50

∴The numbers are : 13,35,43,46,50,55,61,71,8

Since 46 has highest frequency

∴Mode=46


Question 6

A boy scored the following marks in various class tests during a term each test being marked out of 20: 15, 17, 16, 7, 10, 12, 14, 16, 19, 12, 16

(i) What are his modal marks ?

(ii) What are his median marks ?

(iii) What are his mean marks ?

Sol :

Arranging in ascending order 7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19

(ii) Here n = 11 which is odd

∴Median $=\frac{n+1}{2}$ th 

$=\frac{11+1}{2}$=6th term
∴Median=15

(i) Modal class is 16 as it occurs is maximum times

(ii) Mean $=\frac{\sum x_{i}}{n}$
$=\frac{7+10+12+12+14+15+16+16+16+17+19}{11}$
$=\frac{154}{11}$
=14

Question 7

Find the mean, median and mode of the following marks obtained by 16 students in a class test marked out of 10 marks : 0, 0, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 6, 7, 8
Sol :
Here, n = 16

(i) ∴Mean$=\frac{\sum x_{i}}{n}$
$=\frac{0+0+2+2+3+3+3+4+5+5+5+5+6+6+7+8}{16}$
$=\frac{64}{16}=4$

(ii) Median 
$=\frac{1}{2}\left[\frac{16}{2} t h+\left(\frac{16}{2}+1\right) t h\right.$ term $]$
$\therefore$ Median $=\frac{1}{2}$ (8th +9 th term)
$=\frac{1}{2}(4+5)=\frac{9}{2}$
=4.5

(iii) ∵Here 5 occurs in maximum times 
∴Mode=5

Question 8

Find the mode and median of the following frequency distribution :
x 10 11 12 13 14 15
f 1 4 7 5 9 3
Sol :
x f Cumulative frequency
10 1 1
11 4 5
12 7 12
13 5 17
14 9 26
15 3 29
Here n=29 (odd)

Median $=$ Value of $\left(\frac{29+1}{2}\right)$ th observation
=value of 15th observation
=13

Since , the frequency corresponding to 14 is maximum, so 14 is the mode

Question 9

The marks obtained by 30 students in a class assessment of 5 marks is given below:
Marks 0 1 2 3 4 5
No. of students 1 3 6 10 5 5
Calculate the mean, median and mode of the above distribution.
Sol :
Marks 0 1 2 3 4 5
No. of students 1 3 6 10 5 5
Comulative frequency 1 4 10 20 25 30
Mean $=\frac{\Sigma f x}{\Sigma f}$
$=\frac{0 \times 1+1 \times 3+2 \times 6+3 \times 10+4 \times 5+5 \times 5}{1+3+6+10+5+5}$
$=\frac{90}{30}$
=3 

There are a total of 30 observations in the data
The median is the arithmetic mean of $\left(\frac{n}{2}\right)^{\text {th }}$ and $\left(\frac{n}{2}+1\right)^{\text {th }}$ observation in case of even number of observations=Arithmetic mean of $\left(\frac{30}{2}\right)^{\text {th }}$ and $\left(\frac{30}{2}+1\right)^{\text {th }}$

=Arithmetic mean of 15th and 16th observation will be the median

Mean$=\frac{3+3}{2}=3$

Frequency is the highest for the observation $x_{i}=3$
Mode =3

Question 10

The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.
Marks obtained 5 6 7 8 9 10
No. of students 3 9 6 4 2 1
Sol :
Marks obtained 
(xi)		 No. of students
(fi)		 fixi
5 3 15
6 9 54
7 6 42
8 4 32
9 2 18
10 1 10
Total ∑fi=25 ∑fixi=171

$\operatorname{Mean} \bar{x}=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{171}{25}$
$\bar{x}=6.84$
Marks obtained 
 Frequency Cumulative frequency
5 3 3
6 9 12
7 6 18
8 4 22
9 2 24
10 1 25
Total Total=25

Number of terms=25 (odd)

Median $=\left(\frac{25+1}{2}\right)$ th term
$=\left(\frac{26}{2}\right)$ th term
=13th term

∴Median=7
Mode=Marks with maximum frequency is 6
∴Mode=6

Question 11

At a shooting competition, the scores of a competitor were as given below :
Marks obtained 0 1 2 3 4 5
No. of shots 0 3 6 4 7 5
(i) What was his modal score ?
(ii) What was his median score ?
(iii) What was his total score ?
(iv) What was his mean ?
Solution:
Writing the given distribution in cumulative frequency distribution:
Score
(x)		 No. of shots
(f)		 c.f f.x
0 0 0 0
1 3 3 3
2 6 9 12
3 4 13 12
4 7 20 28
5 5 25 25
Total 25 80

(i) Model score is 4 as it occurs in maximum times i.e. 7
∴Mode=4

(ii) Here n=25 which is an odd number

$\therefore$ Median $=\frac{25+1}{2}=13$ th term
Hence median=3

(iii) Total scores=80

(iv) Mean $=\frac{\sum f x}{\sum f}=\frac{80}{25}=3 \cdot 2$


Question 12

(i) Using step-deviation method, calculate the mean marks of the following distribution.
(ii) State the modal class.
Class interval 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90
Frequency 5 20 10 10 9 6 12 8
Sol :
C.I.xifiAssumed mean
A=67.5
$u_{i}=\frac{x_{i}-\mathrm{A}}{h}$, h=5

fiui
50-5552.55-3-15
55-6057.520-2-40
60-6562.510-1-10
65-7067.51000
70-7572.5919
75-8077.56212
80-8582.512336
85-9087.58432
Total8024

(i) Mean $A+\frac{\Sigma f_{i} u_{i}}{\sum f_{i}} \times h$
$=67.5+\frac{24}{80} \times 5$
=67.5+1.5=69

(ii) Modal class=55-60

Question 13

The following table gives the weekly wages (in Rs.) of workers in a factory :
Weekly wages
(in Rs)		50-5555-6060-6565-7070-7575-8080-8585-90
No. of workers520101096128
Calculate:
(i) The mean.
(ii) the modal class
(iii) the number of workers getting weekly wages below Rs. 80.
(iv) the number of workers getting Rs. 65 or more but less than Rs. 85 as weekly wages.
Sol :
Representing the given distribution in cumulative frequency distribution
Weekly wagesNo. of workers
(f)		Class marks
(x)		c.f

f x
50-55552.55262.5
55-602057.5251150.0
60-651062.535625.0
65-701067.545675.0
70-75972.554652.5
75-80677.560465.0
80-851282.572990.0
85-90887.580700.0
Total80
5520.0

(i) Mean $=\frac{\sum f x}{\sum f}=\frac{5520}{80}$
=69

(ii) Modal class :
∵frequency of class 55-60 is maximum i.e. 20
∴Class 55-60 is the modal class

(iii) No. of workers getting weekly wages below Rs 80=60

(iv) No. of workers getting Rs 65 or more but less than 85 as weekly wages
=72-35=37

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