Exercise 5.3 - Chapter 5 Statistics Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $5.3$
Question $1 .$
Fill in the blanks.
(i) The median of the data $12,14,23,25,34,11,42,45,32,22,44$ is
(ii) The median of first ten even natural numbers is
Answers:
(i) 25
(ii) 11

Question $2 .$
Find the median of the given data: $35,25,34,36,45,18,28$.
Solution:
Arranging the given data in ascending order $18,25,28,34,35,36,45$.
Here the number of observations $n=7$, which is odd.
$\therefore$ Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ term $=\left(\frac{7+1}{2}\right)^{\text {th }}$ term
$=\left(\frac{8}{2}\right)^{\text {th }} \text { term }=4^{\text {th }} \text { term }$
Hence Median $=34$

Question $3 .$
The weekly sale of motor bikes in a showroom for the past 14 weeks given below.
$10,6,8,3,5,6,4,7,12,13,16,10,4,7$.
Find the median of the data.
Solution:

Arranging the given data in ascending order $3,4,4,5,6,6,7,7,8,10,10,12,13,16$. Here number of data $n-14$, which is cven
$=\frac{1}{2}\left\{\left(\frac{14}{2}\right)^{t h} \text { term }+\left(\frac{14}{2}+1\right)^{t h} \text { term }\right\}$
$\begin{aligned}
&\therefore \text { Median }=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text {th }} \text { term }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{\left(\frac{14}{2}\right)^{\text {th }} \text { term }+\left(\frac{14}{2}+1\right)^{\text {th }}\right. \text { terme } \\
&=\frac{1}{2}\left\{7^{\text {th }} \text { term }+8^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\{7+7\}=\frac{1}{2}(14)=7 \\
&\therefore \text { Median }=7
\end{aligned}$

Question $4 .$
Find the median of the 10 observations $36,33,45,28,39,45,54,23,56,25$.
If another observation 35 is added to the above data, what would be the new median?
Solution:
Arranging the given 10 observations in ascending order $23,25,28,33,36,39,45,45,54,56$. Here number of data $\mathrm{n}=10$, which is even
$\begin{aligned}
\therefore \text { Median } &=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text {th }} \text { term }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{\left(\frac{10}{2}\right)^{\text {th }} \text { term }+\left(\frac{10}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }\right\}=\frac{1}{2}\{36+39\}
\end{aligned}$

$=\frac{1}{2}(75)=37.5$
$\therefore$ Median $=37.5$
If 35 is added to the above data then it will be the 5 th term then number of data $\mathrm{n}=11$, odd
$\therefore$ Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ term $=\left(\frac{11+1}{2}\right)^{\text {th }}$ term
$=\left(\frac{12}{2}\right)^{\text {th }}$ term $=6$ th term
New median $=36$
 

Objective Type Questions

Question $1 .$
If the median of $a, 2 a, 4 a, 6 a, 9 a$ is 8 , then find the value of a is
(i) 8
(ii) 6
(iii) 2
(iv) 10
Answer:
(iii) 2

Question $2 .$
The median of the data $24,29,34,38,35$ and 30 , is
(i) 29
(ii) 30
(iii) 34
(iv) 32
Answer:
(iv) 32

Question $3 .$
The median first 6 odd natural numbers is
(i) 6
(ii) 7
(iii) 8
(iv) 14
Answer:
(i) 6