Exercise 8.2 - Chapter 8 - Statistics - 9th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Ex 8.2
Question 1.
Find the median of the given values : $47,53,62,71,83,21,43,47,41$.
Solution:
$47,53,62,71,83,21,43,47,41$
Ascending order $=21,41,43,47,47,53,62,71,83$
The number of values 9 , which is odd
$\therefore \text { Median }=\left(\frac{9+1}{2}\right)^{\text {th }} \text { Value }=\left(\frac{10}{2}\right)^{\text {th }} \text { Value }=5^{\text {th }} \text { Value }=47$
 

Question $2 .$
Find the Median of the given data: $36,44,86,31,37,44,86,35,60,51$.
Solution:
$36,44,86,31,37,44,86,35,60,51$
Ascending order $31,35,36,37,44,44,51,60,86,86$
The number of values $=10$, which is an even number
$\therefore$ Median $=$ Average of $\left(\frac{10}{2}\right)^{\text {th }}$ and $\left(\frac{10}{2}+1\right)^{\text {th }}$ Value
$=$ Average of $5^{\text {th }}$ and $6^{\text {th }}$ values $=\frac{44+44}{2}=\frac{88}{2}=44$
 

Question $3 .$
The median of observation $11,12,14,18, x+2, x+4,30,32,35,41$ arranged in ascending order is
24 . Find the values of $x$.
Solution:
$11,12,14,18, x+2,, x+4,30,32,35,41$

The number of values $=10$, which is an even number
Median $=$ Average of $\left(\frac{10}{2}\right)^{\text {th }}$ and $\left(\frac{10}{2}+1\right)^{\text {th }}$ Values
$\begin{aligned}
24 &=\frac{x+2+x+4}{2} \\
24 &=\frac{2 x+6}{2}=\frac{2(x+3)}{2} \\
\therefore x+3 &=24 \\
x &=24-3=21
\end{aligned}$

Question $4 .$
A researcher studying the behaviour of mice has recorded the time (in seconds) taken by each mouse to locate its food by considering 13 different mice as $31,33,63,33,28,29,33,27,27,34$, $35,28,32$. Find the median time that mice spent in searching its food.
Solution:
$31,33,63,33,28,29,33,27,27,34,35,28,32$ Writing in ascending order we get $27,27,28,28$, $29,31,32,33,33,33,34,35,63$
Number of values $=13$ which is an odd number
$\therefore \text { Median }=\left(\frac{13+1}{2}\right) \text { Value }=\left(\frac{14}{2}\right)^{\text {th }} \text { Value }=7^{\text {th }} \text { Value }=32$

Question $5 .$
The following are the marks scored by the students in the Summative Assessment exam

Calculate the median.

Solution

Question $6 .$
The mean of five positive integers is twice their median. If four of the integers are $3,4,6,9$ and median is 6 , then find the fifth integer.
Solution:
The five integers are $3,4,6,9, \mathrm{x}$

$\begin{aligned}
\text { Mean } \bar{x} &=\frac{\sum x}{n}=\frac{3+4+6+9+x}{5}=\frac{22+x}{5} \\
\text { Their Median } &=6 \\
\text { Mean } &=\text { twice of the median } \\
\frac{22+x}{5} &=2 \times 6 \\
22+x &=5 \times 12=60 \\
\therefore x &=60-22=38 \\
\mathrm{~N} &=50 \\
\text { Median Class } &=\left(\frac{\mathrm{N}}{2}\right)^{\text {th }} \text { Value }=\left(\frac{50}{2}\right)^{\text {th }} \\
\text { Value } &=25^{\text {th }} \text { Value } \\
\text { Median value } &=30-40 \\
\frac{\mathrm{N}}{2} &=25, l=30 \\
m &=24, c=10, f=10 \\
\therefore \text { Median } &=\frac{l+\left(\frac{\mathrm{N}}{2}-m\right)}{f} \times c \\
&=\frac{30+25-24}{10} \times 10=31
\end{aligned}$