Exercise 5.1 - 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 5.1
Question $1 .$
Fill in the blanks.
1. The mean of first ten natural numbers is
2. If the average selling price of 15 books is $₹ 235$, then the total selling price is
3. The average of the marks $2,9,5,4,4,8,10$ is
4. The average of integers between $-10$ to 10 is
Answers:
1. $5.5$
2. 3,525
3. 6
4. 0

Question $2 .$
Ages of 15 students in 8th standard is $13,12,13,14,12,13,13,14,12,13,13,14,13,12,14$. Find the mean age of the students.
Solution:
$\begin{gathered}
\text { Arithmetic Mean }=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\
=\frac{13+12+13+14+12+13+13+14+12+13+13+14+13+12+14}{15} \\
=\frac{195}{15}=13
\end{gathered}$
 

Question $3 .$
The marks of 14 students in a science test out of 50 are given below. $34,23,10,45,44,47,35,37$, $41,30,28,32,45,39$ Find
(i) the mean mark.
(ii) the maximum mark obtained.
(iii) the minimum mark obtained.
Solution:
$\begin{aligned}
\text { Mean marks } &=\frac{\text { Sum of all marks }}{\text { Total number of marks }} \\
=\frac{34+23+10+45+44+47+35+37+41+30+28+32+45+39}{14} \\
&=\frac{490}{14}
\end{aligned}$
Mean marks $=35$
(ii) Maximum mark obtained $=47$
(iii) Minimum mark obtained $=10$

Question $4 .$
The mean height of 11 students in a group is $150 \mathrm{~cm}$. The heights of the students are $154 \mathrm{~cm}, 145$ $\mathrm{cm}, \mathrm{Ycm}, \mathrm{Y}+4 \mathrm{~cm}, 160 \mathrm{~cm}, 151 \mathrm{~cm}, 149 \mathrm{~cm}, 149 \mathrm{~cm}, 150 \mathrm{~cm}, 144 \mathrm{~cm}$ and $140 \mathrm{~cm}$. Find the value of $Y$ and the heights of two students.
Solution:

$\begin{gathered}
\text { Mean Height }=\frac{\text { Sum of heights of all students }}{\text { Number of students }} \\
150=\frac{154+145+Y+(Y+4)+160+151+149+149+150+144+140}{11}
\end{gathered}$
$\begin{aligned}
&\text { Mean Height }=\underline{\text { Sum }} \\
&150=\frac{154+145+\mathrm{Y}+(\mathrm{Y}+4)+160+}{150} \\
&150=\frac{1342+Y+Y+4}{11} \\
&150=\frac{1346+2 \mathrm{Y}}{11} \\
&150 \times 11=1346+2 \mathrm{Y} \\
&1650=1346+2 \mathrm{Y} \\
&2 \mathrm{Y}=1650-1346=304 \\
&\mathrm{Y}=\frac{304}{2}=152 \\
&\text { Height of two students are } \mathrm{Y} \text { and } \mathrm{Y}+4 \\
&\Rightarrow 152 \text { and } 152+4 \\
&\Rightarrow 152 \mathrm{~cm} \text { and } 156 \mathrm{~cm}
\end{aligned}$

Question $5 .$
The mean of runs scored by a cricket team in the last 10 innings is 276 . If the scores are 235,400 , $351, \mathrm{x}, 100,315,410,165,260,284$, then find the runs scored in the fourth innings.
Solution:
Let the runs scored in the fourth innings be $x$.

$\begin{aligned}
\text { Mean runs scored } &=\frac{\text { Total runs of all innings }}{\text { number of innings }} \\
276 &=\frac{235+400+351+x+100+315+410+165+260+284}{10} \\
276 &=\frac{2520+x}{10}
\end{aligned}$
$\therefore \text { Number of runs scored in the fourth innings }=240$
$\begin{aligned}
&276 \times 10=2520+x \\
&2760=2520+x \\
&x=2760-2520=240
\end{aligned}$

Question 6 .

Find the mean of the following data. $5.1,4.8,4.3,4.5,5.1,4.7,4.5,5.2,5.4,5.8,4.3,5.6$,
Solution:

Mean $=\frac{\text { Sum of all numbers }}{\text { Number of values }}$
$\frac{5.1+4.8+4.3+4.5+5.1+4.7+4.5+5.2+5.4+5.8+4.3+5.6+5.2+5.5}{14}$
$=\frac{70.0}{14}=5$
Mean $=5$

Question $7 .$
Arithmetic mean of 10 observations was found to be 22 . If one more observation 44 was to be added to the data, what would be the new mean?
Solution:
Arithmetic mean of 10 observation is 22 .
$\begin{aligned}
\text { Arithmetic mean } &=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\
22 &=\frac{\text { sum of } 10 \text { observations }}{10}
\end{aligned}$
Sum of 10 observations $=22 \times 10=220$ Now if new number is added, then
Now if new number is added, then Mean of 11 observations $=\frac{\text { Sum of } 10 \text { observation }+44}{11}$
$=\frac{220+44}{11}=\frac{264}{11}=24$
New mean $=24$

Objective Type Questions
Question 1.

___ is a representative value of the entire data.
(i) Mean
(ii) range
(iii) minimum value
(iv) maximum value
Answer:
(i) Mean

Question $2 .$
The mean of first fifteen even numbers is
(i) 4
(ii) 16
(iii) 5
(iv) 10
Answer:
(ii) 16

Question $3 .$
The average of two numbers are 20 . One number is 24 , another number is
(i) 16
(ii) 26
(iii) 20
(iv) 40
Ans:
(i) 16
Hint:
$\frac{x+y}{2}=20$
$x+y=40$
$24+y=40$
$y=40-24=16$

Question $4 .$
The mean of the data $12, x, 28$ is 18 . Find the value of $x$.
(i) 18
(ii) 16
(iii) 14
(iv) 22
Answer:
(iii) 14
Hint:
$\begin{aligned}
&\frac{12+x+28}{3}=18 \\
&x+40=54 \\
&x=14
\end{aligned}$