Exercise 8.1 - Chapter 8 Statistics 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $8.1$ : Chapter 8 - Statistics - 9th Maths Guide Samacheer Kalvi Solutions
Question 1.
In a week, temperature of a certain place is measured during winter are as follows $26^{\circ} \mathrm{C}, 24^{\circ} \mathrm{C}, 28^{\circ} \mathrm{C}, 31^{\circ} \mathrm{C}$, $30^{\circ} \mathrm{C}, 26^{\circ} \mathrm{C}, 24^{\circ} \mathrm{C}$. Find the mean temperature of the week.
Solution:
$\begin{aligned}
\operatorname{Mean} \bar{x} &=\frac{\sum x}{n} \\
&=\frac{26+24+28+31+30+26+24}{7}=\frac{189}{7}
\end{aligned}$
Mean temperature of the week $=27^{\circ} \mathrm{C}$

Question $2 .$
The mean weight of 4 members of a family is $60 \mathrm{~kg}$. Three of them have the weight $56 \mathrm{~kg}, 68 \mathrm{~kg}$ and $72 \mathrm{~kg}$ respectively. Find the weight of the fourth member.
Solution:

$\begin{aligned}
\bar{x} &=60 \mathrm{~kg} \\
\bar{x} &=\frac{\sum x}{n}=\frac{56+68+72+x}{4}=60 \\
196+x &=240 \\
x &=240-196
\end{aligned}$
$\therefore$ The weight of the fourth member $=44 \mathrm{~kg}$

Question 3.
In a class test in mathematics, 10 students scored 75 marks, 12 students scored 60 marks, 8 students scored 40 marks and 3 students scored 30 marks. Find the mean of their score.
Solution:
Total number of students $=10+12+8+3=33$

Total score of 33 students $=10 \times 75+12 \times 60+8 \times 40+3 \times 30$
$=750+720+320+90=1880$
Mean of their score $=\frac{\text { Total Marks }}{\text { number of students }}=\frac{1880}{33}$
$=56.96$ or 57 approximately
 

Question $4 .$
In a research laboratory scientists treated 6 mice with lung cancer using natural medicine. Ten days later, they measured the volume of the tumor in each mouse and given the results in the table.

Find the mean.
Solution:
$\begin{aligned}
&\bar{x}=\frac{\sum x}{n}=\frac{145+148+142+141+139+140}{6}=\frac{855}{6} \\
&x=142.5 \mathrm{~mm}^{3}
\end{aligned}$

Question $5 .$
If the mean of the following data is $20.2$, then find the value of $p$

Solution:
$\begin{aligned}
\bar{x} &=20.2 \\
\bar{x} &=\frac{\sum f x}{\sum f}=\frac{10 \times 6+15 \times 8+20 p+25 \times 10+30 \times 6}{6+8+p+10+6} \\
20.2 &=\frac{60+120+20 p+250+180}{30+p} \\
(30+p) 20.2=610+20 \mathrm{p} \\
606+20.2 \mathrm{p}=610+20 \mathrm{p} \\
20.2 \mathrm{p}-20 \mathrm{p}=610-606=4 \\
0.2=4
\end{aligned}$
$\begin{aligned}
&\Rightarrow \\
&p=\frac{4 \times 10}{0.2 \times 10}=\frac{40}{2} \\
&p=20
\end{aligned}$

Question $6 .$
In the class, weight of students is measured for the class records. Calculate mean weight of the class students using Direct method.

Question $7 .$
Calculate the mean of the following distribution using Assumed Mean Method:

Question $8 .$
Find the Arithmetic Mean of the following data using Step Deviation Method

Solution: