Exercise 8.3 - Chapter 8 Statistics 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex 8.3
Question 1.
The monthly salary of 10 employees in a factory are given below : $₹ 5000$, ₹ 7000 , $₹ 5000$, ₹ 7000 , ₹ 8000 , $₹ 7000,17000$, ₹ 8000 , ₹ 7000, ₹ 5000 Find the mean, median and mode.
Solution:
The monthly salary of 10 employees are ₹ 5000 , ₹ 7000 , ₹ 5000 , ₹ 7000 , ₹ 8000 , ₹ 7000 , ₹ 7000 , ₹ 8000, ₹ $7000, ₹ 5000$.
Writing in ascending order $₹ 5000$, ₹ 5000 , ₹ 5000 , ₹ 7000 , ₹ 7000 , ₹ 7000 , ₹ 8000 , ₹ 8000
Number of values $=10$ which is an even number.
Median $=$ Average of $\frac{10^{\text {th }}}{2}$ and $\left(\frac{10}{2}+1\right)^{\text {th }}$ Value
$=$ Average of $5^{\text {th }}$ and $6^{\text {th }}$ Value
$=\frac{7000+7000}{2}=\frac{14000}{2}=₹ 7000$
$5000+5000+5000+7000+7000+7000+7000+7000$
$\operatorname{Mean} \bar{x}=\frac{\sum x}{n}=$
$+8000+8000$
$=\frac{66000}{10}=6600$
In this given data $₹ 7000$ occurs maximum number of 5 times
$\therefore$ Mode $=₹ 7000 /-$
$\therefore$ Mean $=₹$ 6600/-
Mode $=₹ 7000 /-$
 

Question $2 .$
Find the mode of the given data : $3.1,3.2,3.3,2.1,1.3,3.3,3.1$
Solution:
$3.1,3.2,3.3,2.1,1.3,3.3,3.1$

In this given data $3.1,3.3$ occurs twice
$₹$ Mode $=3.1$ and $3.3$ (bimodal)
 

Question $3 .$
For the data $11,15,17, x+1,19, x-2,3$ if the mean is 14 , find the value of $x$. Also find the mode of the data.
Solution:
The data given is $11,15,17, x+1,19, x-2,3$
$x=14$
$\bar{x}=\frac{\sum x}{n}=\frac{11+15+17+x+1+19+x-2+3}{7}=\frac{66+2 x-2}{7}=\frac{64+2 x}{7}$
$\frac{64+2 x}{7}=14$
$64+2 x=98$
$2 x=98-64=34$
$x=\frac{34}{2}=17$
The data $11,15,17,17+1,19,17-2,3=11,15,17,18,19,15,3$ In this given data 15 occurs twice. Hence the mode is 15 .
 

Question 4 .
The demand of track suit of different sizes as obtained by a survey is given below:

Which size is in greater demand ?

Solution:

Question 5.

Find the mode of the following data

Solution:

$\text { Mode }=l+\left[\frac{f-f_{1}}{2 f-f_{1}-f_{2}}\right] \times c$
Here Modal class is $20-30$ since it has the maximum frequency.
$\begin{aligned}
l &=20, f=46, f_{1}=38, f_{2}=34, c=30-20=10 \\
\therefore \text { Mode } &=l+\left[\frac{f-f_{1}}{2 f-f_{1}-f_{2}}\right] \times c=20+\left[\frac{46-38}{2 \times 46-38-34}\right] \times 10 \\
&=20+\left[\frac{8}{92-72}\right] \times 10=20+\frac{\frac{8}{20}}{20} \times 10=20+4=24 \\
\text { Mode } &=24
\end{aligned}$

Question $6 .$
Find the mode of the following distribution:

Solution:

$\begin{aligned}
\frac{\mathrm{N}}{2} &=\frac{25}{\not 2}=25 \\
\text { :. Median class } &=25^{\text {th }} \text { value }=54.5-64.5 \\
l &=54.5, m=22, f=14, \\
c &=10 \\
\text { Median } &=l+\frac{\left(\frac{\mathrm{N}}{2}-m\right)}{f} \times c=54.5+\frac{25-22}{14} \times 10=54.5+\frac{3}{14} \times 10 \\
&=54.5+\frac{30}{14}=54.5+2.14=56.64
\end{aligned}$
Modal class is $54.5-64.5$ since it has the maximum frequency.
$\begin{aligned}
l &=54.5, f=14, f_{1}=10, f_{2}=8, c=10, \\
\therefore \text { Mode } &=l+\left[\frac{f-f_{1}}{2 f-f_{1}-f_{2}}\right] \times c=54.5+\left[\frac{14-10}{2 \times 14-10-8}\right] \times 10 \\
&=54.5+\left(\frac{4}{28-18}\right) \times 10=54.5+\frac{4}{1 Q} \times 1 Q=58.5
\end{aligned}$