Exercise 5.4 - 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 5.4
Miscellaneous Practice problems
Question $1 .$
Arithmetic mean of 15 observations was calculated as 85 . In doing so an observation was wrongly taken as 73 for 28 . What would be correct mean?
Solution:
Arithmetic mean $=\frac{\text { Sum of all observations }}{\text { Number of observations }}$
$85=\frac{\text { Sum of } 15 \text { observation }}{15}$
$85 \times 15=$ sum of 15 observations
$1275=$ sum of 15 observations
Wrong observation $=73$
Correct observation $=28$
$\begin{aligned}
\therefore \text { Correct Mean } &=\frac{\text { Sum }-\text { Wrong value }+\text { Correct value }}{\text { Number of obervation }} \\
&=\frac{1275-73+28}{15}=\frac{1202+28}{15}=\frac{1230}{15}=82
\end{aligned}$
Correct mean $=82$
 

Question $2 .$
The median of $25,16,15,10,8,30$.

Solution:
Arranging is ascending order : $8,10,15,16,25,30$
Here $n=6$, even
$\begin{aligned}
\therefore \text { Median } &=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text {th }} \text { term }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{\left(\frac{6}{2}\right)^{\text {th }} \text { term }+\left(\frac{6}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{3^{\text {rd }} \text { term }+4^{\text {th }} \text { term }\right\}=\frac{1}{2}\{15+16\}=\frac{1}{2}(31)=15.5
\end{aligned}$
$\therefore$ Median $=15.5$
 

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

Question $4 .$
The marks scored by the students in social test out of 20 marks are as follows.
$12,10,8,18,14,16$. Find the mean and the median?
Solution:
Arranging the given data in ascending order: $8,10,12,14,16,18$.
$\begin{aligned}
\text { Mean } &=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\
&=\frac{8+10+12+14+16+18}{6}=\frac{78}{6} \\
\text { Mean } &=13
\end{aligned}$

There are $\mathrm{n}=6$ observations, which is even
$\begin{aligned}
\therefore \text { Median } &=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text {th }} \text { term }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{\left(\frac{6}{2}\right)^{\text {th }} \text { term }+\left(\frac{6}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{3^{\text {th }} \text { term }+4^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\{8+18\}=\frac{1}{2}(26)=13
\end{aligned}$

Question $5 .$
The number of goals scored by a football team is given below.
Find the mode and median for the data of $2,3,2,4,6,1,3,2,4,1,6$.
Solution:
Arranging the given data in ascending order: $1,1,2,2,2,3,3,4,4,6,6$
Clearly 2 occurs at the maximum of 3 times and so mode $=2$
Here number of data of data $\mathrm{n}=11$, odd.
$\therefore$ Median $=\left(\frac{n+1}{2}\right)^{\text {th }}$ term
$\begin{aligned}
&=\left(\frac{11+1}{2}\right)^{\text {th }} \\
&\text { term }=\left(\frac{12}{2}\right)^{\text {th }} \text { term }
\end{aligned}$
$=6^{\text {th }} \text { term }$
$\text { Median }=3$
 

Question $6 .$
Find the mean and mode of $6,11,13,12,4,2$.
Answer:
Arranging is ascending order : $2,4,6,11,12,13$

$\begin{aligned}
\text { Mean } &=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\
&=\frac{2+4+6+11+12+13}{6} \\
\text { Mean } &=\frac{48}{6}=8
\end{aligned}$
Mean $=\frac{48}{6}=8$
All observation occurs only once and so there is no mode for this date.
 

Challenge Problems
Question 1.
The average marks of six students is 8 . One more student mark is added and the mean is still 8 . Find the student mark that has been added.
Solution:
$\text { Average }=\frac{\text { Sum of all observations }}{\text { Number of observations }}$

$8=\frac{\text { Sum of observation }}{6}$
Sum of observation $=6 \times 8=48$
If one more mark is added then number of observations $=6+1=7$
Let the number be $x$
Still average $=8$
$\begin{aligned}
&\therefore 8=\frac{48+x}{7} \\
&48+x=7 \times 8 \\
&48+x=56 \\
&48+x=56-48 \\
&x=8
\end{aligned}$
$\therefore$ The number that is added $=8$
 

Question $2 .$
Calculate the mean, mode and median for the following data: $22,15,10,10,24,21$.
Solution:
Arranging in ascending order: $10,10,15,21,22,24$
$\begin{aligned}
\text { Mean } &=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\
&=\frac{10+10+15+21+22+24}{6} \\
&=\frac{102}{6}=17
\end{aligned}$
Here $n=6$, even

$\begin{aligned}
\therefore \text { Median } &=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text {th }} \text { term }+\left(\frac{n}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{\left(\frac{6}{2}\right)^{t h} \text { term }+\left(\frac{6}{2}+1\right)^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\left\{3^{\text {th }} \text { term }+4^{\text {th }} \text { term }\right\}=\frac{1}{2}\{15+21\} \\
&=\frac{1}{2}(36)
\end{aligned}$
$\therefore$ Median $=18$
Clearly the data 10 occurs maximum number of times and so 10 is the mode.
$\therefore$ Mode $=10$

Question $3 .$
Find the median of the given data: $14,-3,0,-2,-8,13,-1,7$.
Solution:
Arranging the data is ascending order: $-8,-3,-2,-1,0,7,13,14$
Here number of data $\mathrm{n}=8$, even
$\therefore$ Median
$\begin{aligned}
&=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{t h} \text { term }+\left(\frac{n}{2}+1\right)^{t h} \text { term }\right\} \\
&=\frac{1}{2}\left\{\left(\frac{8}{2}\right)^{t h} \text { term }+\left(\frac{8}{2}+1\right)^{t h} \text { term }\right\} \\
&=\frac{1}{2}\left\{4^{\text {th }} \text { term }+5^{\text {th }} \text { term }\right\} \\
&=\frac{1}{2}\{-1+0\}=\frac{1}{2}(-1)=-0.5
\end{aligned}$
$=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{\text {th }}\right.$ terr $=\frac{1}{2}\left\{\left(\frac{8}{2}\right)^{t}\right.$ $=\frac{1}{2}\left\{4^{\text {th }}\right.$ tern $=\frac{1}{2}\{-1\}$ $\therefore$ Median $=-0.5$
 

Question $4 .$
Find the mean of first 10 prime numbers and first 10 composite numbers.
Solution:
First 10 prime numbers are $2,3,5,7,11,13,17,19,23,29$

$\begin{aligned}
&\begin{aligned}
\text { Mean } &=\frac{\text { Sum of all data }}{\text { number of data }} \\
&=\frac{2+3+5+7+11+13+17+19+23+29}{10} \\
&=\frac{129}{10} \\
\text { Mean }=12.9 \\
\text { Mean of first } 10 \text { prime numbers }=12.9 \\
\text { First } 10 \text { composite numbers are } 4,6,8,9,10,12,14,15,16,18 \\
\text { Mean } &=\frac{4+6+8+9+10+12+14+15+16+18}{10} \\
&=\frac{112}{10} \\
&=11.2
\end{aligned}
\end{aligned}$
Mean of first 10 composite numbers $=11.2$