Additional Questions - Chapter 8 Statistics 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions
Exercise 8.1
Question 1.
The following data gives the number of residents in an area based on their age. Find the average age of the residents.

$\text { Mean }=\overline{\mathrm{X}}=\frac{\sum f x}{\Sigma f}=\frac{688}{30}=22.93$
Hence the average age $=22.93$
 

Question $2 .$
Find the mean for the following frequency table :

$\begin{aligned}
\operatorname{Mean} &=\overline{\mathrm{X}}=\mathrm{A}+\frac{\sum f d}{\sum f}=170+\left(\frac{-1200}{58}\right) \\
&=170-20.689 \Rightarrow \overline{\mathrm{X}}=149.311
\end{aligned}$

Question $3 .$
Find the mean for the following distribution using step Deviation Method.

Solution:

$\text { Mean }=\overline{\mathrm{X}}=\mathrm{A}+\frac{\sum f d}{\sum f} \times c=28+\left(\frac{-22}{100}\right) \times 8=28-1.76=-20.24$
 

Exercise 8.2
Question 1 .
For the following up grouped data $8,15,14,19,11,16,10,8,17,20$. Find the median.
Solution:
Arrange the values in ascending order $8,8,10,11,14,15,16,17,19,20$
The number of values $=10$
Median $=$ Average of $\left(\frac{10}{2}\right)^{\text {th }}$ and $\left(\frac{10}{2}+1\right)^{\text {th }}$ values
$=$ Average of $5^{\text {th }}$ and $6^{\text {th }}$ values $=\frac{14+15}{2}=\frac{29}{2}=14.5$
 

Question $2 .$
The following table gives the weekly expenditure of 200 families. Find the median of the weekly expenditure.

Solution:

Median class $=\left(\frac{\mathrm{N}}{2}\right)^{\text {th }}$ Value $=\left(\frac{200}{2}\right)^{\text {th }}$ Value $=100^{\text {th }}$ Value
Median Class $=2000-3000$
$\begin{aligned}
\frac{\mathrm{N}}{2} &=100, l=2000 \\
m &=70, c=1000, f=56 \\
\text { Median } &=l+\frac{\left(\frac{\mathrm{N}}{2}-m\right)}{f} \times c=2000+\left(\frac{100-70}{56}\right) \times 1000 \\
&=2000+\left(\frac{30}{56}\right) \times 1000=2000+535.7=2535.7
\end{aligned}$

Question $3 .$
The median of the following data is 24 . Find the value of $x$.

Solution:

Since the median is 24 and median class is $20-30$
$l=20, \mathrm{~N}=55+x, m=30, c=10, f=x$
$\operatorname{Median}=l+\left(\frac{\frac{\mathrm{N}}{2}-m}{f}\right) \times c$
$24=20+\left(\frac{\frac{55+x}{2}-30}{x}\right) \times 10$
$4=\frac{5 x-25}{x}$
$4 x=5 x-25$
$5 x-4 x=25$
$x=25$
 

Question $4 .$
The following are the scores obtained by 11 players in a cricket match $7,21,45,12,56,35,25,0,58,66$, 29. Find the median score.
Solution:
Let us arrange the values in ascending order $0,7,12,21,25,29,35,45,56,58,66$
The number of values $=11$ which is odd.
Median $=\left(\frac{11+1}{2}\right)^{\text {th }}$ value $=\left(\frac{12}{2}\right)^{\text {th }}$ value $=6$ th value $=29$
 

Exercise 8.3
Question $1 .$
Find the mode of the given data : $65,65,71,71,72,75,82,72,47,72$.

Solution:
In the given data 72 occurs thrice. Hence the mode is $72 .$
 

Question $2 .$
Find the mode:

Solution:
7 has the maximum frequency 21 . Therefore 7 is the mode.
 

Question $3 .$
Find the mode for the following data.

Solution:

$\begin{aligned}
l &=40, c=10, f=50, f_{1}=36, f_{2}=20 \\
\text { Mode } &=l+\left(\frac{f-f_{1}}{2 f-f_{1}-f_{2}}\right) \times c=40+\left(\frac{36-50}{2(50)-36-20}\right) \times 10 \\
&=40+\left(\frac{14}{100-36-20}\right) \times 10=40+\frac{14}{44} \times 10=40+\frac{35}{11}=40+3.18 \\
\therefore \text { Mode } &=43.18
\end{aligned}$

Question $4 .$
In a distribution, the mean and mode are 46 and 40 respectively. Calculate the median.
Solution:
Given, Mean $=46$ and mode $=40$
Using mode $\approx 3$ median $-2$ mean ,

$40 \approx 3$ Median $-2(46)$
3 Median $\approx 40+92$
Therefore, Median $\approx \frac{132}{3}=44$
 

Exercise $8.4$
Multiple Choice Questions :
Question 1.
The mean of first 10 natural numbers.
(1) 25
(2) 55
(3) $5.5$
(4) $2.5$
Hint :Mean $=\frac{\sum x}{n}=\frac{1+2+3+4+5+6+7+8+9+10}{10}=\frac{55}{10}=5.5$
Solution:
(3) $5.5$
 

Question 2.
The mean of a distribution is 23 , the median is 24 and the mode is $25.5$. It is most likely that this distribution is :
(1) Positively skewed
(2) Symmetrical
(3) Asymptotic
(4) Negatively skewed
Hint: For Negatively skewed means is likely to be less than mode and median
Solution:
(4) Negatively skewed
 

Question $3 .$
The middle value of an ordered array of numbers is the
(1) Mode
(2) Mean
(3) Median
(4) Mid point
Solution:
(3) Median

Question $4 .$
The weights of students in a school is a :
(1) Discrete variable
(2) Continuous variable
(3) Qualitative variable
(4) None of these
Solution:
(2) Continuous variable
 

Question $5 .$
The first hand and unorganized form data is called
(1) Secondary data
(2) Organised data
(3) Primary data
(4) None of these
Solution:
(3) Primary data