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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions and Answers
Exercise 5.1
Question $1 .$
The class scores on an exam are $97,85,92,78$ and 90 . What is the average score?
Solution:
$\begin{aligned}
\text { Average score } &=\frac{\text { Sum of scores }}{\text { Number of scores }} \\
&=\frac{97+85+92+78+90}{5} \\
&=\frac{442}{5}=88.4
\end{aligned}$
$=\frac{442}{5}=88.4$

Question $2 .$
Mean of 20 observations is 27 . Find the sum of 20 observations.
Solution:
$\begin{aligned}
\text { Mean } &=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\
27 &=\frac{\text { Sum of observation }}{20}
\end{aligned}$
Sum of observations $=27 \times 20$
Sum of observations $=540$

Question $3 .$
Mean of same observations is 15 . Their sum is 405 . Find the number of values. Solution:
$\begin{aligned}
15 &=\frac{405}{\text { Number of }} \\
\text { vaes } &=\frac{405}{15}=27
\end{aligned}$
$\text { Mean }=\frac{\text { Sum of all observations }}{\text { Number of values }}$
Number of values $=\frac{4}{15}$ Number of values $=\frac{405}{15}=27$
 

Exercise $5.2$

Question $1 .$
Find the mode of the following data. $1,2,5,7,3,4,2,5,7,6,2,3$.
Solution:
Arranging the data in ascending order: $1,2,2,2,3,3,4,5,5,6,7,7$
Here 2 occurs maximum number of times.
$\therefore$ Mode is 2 .

Exercise $5.3$
Question $1 .$
Find the median of the data $1,3,7,16,0,19,7,4,3$.
Solution:

Arranging the given data in ascending order: $0,1,3,3,4,7,7,16,19$
Number of terms $\mathrm{n}=9$, which is odd.
$\therefore$ Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ term
$=\left(\frac{9+1}{2}\right)^{\text {th }}$ term $=\left(\frac{10}{2}\right)^{\text {th }}$ term
$=5^{\text {th }}$ term
Hence Median $=4$
 

Question $2 .$
Find the median of first 10 even number.
Solution:
First 10 even numbers are $2,4,6,8,10,12,14,16,18,20$.
Here number of data $\mathrm{n}=10$, which is even
$\begin{aligned} \therefore \text { Median } &=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text {th }} \text { term }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }\right\} \\ &=\frac{1}{2}\left\{\left(\frac{10}{2}\right)^{t h} \text { term }+\left(\frac{10}{2}+1\right)^{t h} \text { term }\right\} \\ &=\frac{1}{2}\left\{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }\right\}=\frac{1}{2}\{10+12\}=\frac{1}{2}(22) \\ \therefore \text { Median } &=11 \end{aligned}$
$\begin{aligned}
&=\frac{1}{2}\left\{\left(\frac{10}{2}\right)^{\text {th }} \text { term }+\left(\frac{10}{2}+1\right)^{t h} \text { term }\right\} \\
&=\frac{1}{2}\left\{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }\right\}=\frac{1}{2}\{10+12\}=\frac{1}{2}(22) \\
\therefore \text { Median } &=11
\end{aligned}$