Additional Questions - Chapter 8 - Statistics - 9th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

You can Download the Additional Questions - Chapter 8 - Statistics - 9th Maths Guide Samacheer Kalvi Solutions - Tamil Medium with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

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