In Text Questions Try These (Text Book Page No. 96, 97, 99, 101, 103, 106) - Chapter 5 Statistics Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Text Questions : Chapter 5 Statistics  Term 3 Class 7th std Maths Guide Samacheer Kalvi Solutions
Exercise $5.1$
Try These (Text book Page No. 96)
Question $1 .$
Collect the height of students of your class. Organise the data in ascending order.
Solution:
Height of 15 students in our class.
$130 \mathrm{~cm}, 150 \mathrm{~cm}, 155 \mathrm{~cm}, 142 \mathrm{~cm}, 138 \mathrm{~cm}, 145 \mathrm{~cm}, 148 \mathrm{~cm}, 147 \mathrm{~cm}, 148 \mathrm{~cm}, 143 \mathrm{~cm}, 141 \mathrm{~cm}, 152$ $\mathrm{cm}, 147 \mathrm{~cm}, 139 \mathrm{~cm}, 155 \mathrm{~cm}$.
Ascending order:
$130 \mathrm{~cm}, 138 \mathrm{~cm}, 139 \mathrm{~cm}, 141 \mathrm{~cm}, 142 \mathrm{~cm}, 143 \mathrm{~cm}, 145 \mathrm{~cm}, 147 \mathrm{~cm}, 147 \mathrm{~cm}, 148 \mathrm{~cm}, 148 \mathrm{~cm}, 150 \mathrm{~cm}$, $152 \mathrm{~cm}, 155 \mathrm{~cm}, 155 \mathrm{~cm}$.

Try These (Text book Page No. 97)
Find the Arithmetic Mean or average of the following data.
Question $1 .$
The study time spent by Kathir in a week is 3 hrs, 4 hrs, 5 hrs, 3 hrs, 4 hrs, $3: 45$ hrs; $4: 15$ hrs.
Solution:

$\begin{aligned} \text { Arithmetic Mean } &=\frac{\text { Sum of all observations }}{\text { Number of observations }} \\ &=\frac{3+4+5+3+4+3.45+4.15}{7} \\ &=\frac{27}{7}=3 \text { hrs } 52 \mathrm{~min} \\ \text { mean } &=3: 52 \mathrm{hrs} . \end{aligned}$

Question $2 .$
The marks scored by Muhil in five subjects are $75,91,48,63,51$. Solution:
$=\frac{75+91+48+63+51}{5}=\frac{328}{5}$
$\text { Arithmetic Mean }=\frac{\text { Sum of all observations }}{\text { Number of observations }}$
Arithmetic Mean $=65.6$
 

Question $3 .$
Money spent on vegetables for five days is $₹ 120$, ₹ 80 , ₹ 75 , ₹ 95 and $₹ 86$.
Solution:

Arithmetic Mean $=\frac{\text { Sum of all observations }}{\text { Number of observations }}$
$=\frac{120+80+75+95+86}{5}=\frac{456}{5}$
Arithmetic Mean $=91.2$
Arithmetic Mean $=91.2$

Think (Text book Page No. 99)

Check the properties of arithmetic mean for the example given below

Question $1 .$
If the mean is increased by 2 , then what happens to the individual observations.
Solution:
Given number are $3,6,9,12,15$
Arithmetic Mean $=\frac{\text { Sum of all observations }}{\text { Number of observations }}$
$=\frac{3+6+9+12+15}{5}=\frac{45}{5}=9$
If mean is increased by 2 then,
$9+2=\frac{\text { Sum of observations }}{5}$
Sum of observations $=5 \times 11=55$
Difference in sum $=55-45=10$
$\therefore$ Each number is increased by 2 if the mean is increased by 2 .
 

Question $2 .$
If first two items are increased by 3 and last two items are reduced by 3 , then what will be the new mean?
Solution:
If the first two items is increased by 3 , then the numbers will be $3+3,6+3 \Rightarrow 6,9$.
If last two numbers are decreased by 3 , then the numbers will be $12-3,15-3 \Rightarrow 9,12$.

$\therefore$ Mean of new numbers $=\frac{6+9+9+9+12}{5}$
$=\frac{45}{5}=9$
$=\frac{45}{5}=9$
There is no change in the mean.
 

Exercise $5.2$
Try These (Text book Page No. 101)
Question 1.
Find the mode of the following data. 2, 6, 5, 3, 0, 3, 4,3,2,4,5,2
Solution:
Arranging the numbers in ascending order we get $0,2,2,2,3,3,3,4,4,5,5,6$ Since 2 and 3 occurs the maximum of 3 times. So mode of this data is 2 and 3 .
 

Question $2 .$
Find the mode of the following data set. 3, 12,15,3,4,12,11,3,12, 9, 19 .
Solution:
Arranging the given data in ascending order : 3,3,3,4, 9,11, 12, 12,12,15, 19 .
The data 3 and 12 occurs the maximum of 3 times.
So mode of this data is 3 and 12 .
 

Question $3 .$
Find the mode of even numbers within 20 .
Solution:
Even numbers within 20 are 2, $4,6,8,10,12,14,16,18$.
There is no mode for this data.

Think (Text book Page No. 102)
Question $1 .$
A toy factory making variety of toys for kids, wants to know the most popular toy liked by all the kids. Which average will be the most appropriate for it?
Solution:
Mode.
 

Question $2 .$
Is there a mode exists between the odd numbers from 20 to 40 ? Discuss.
Solution:
Odd number between 20 to 40 are $21,23,25,27,29,31,33,35,37,39$.
As all numbers occurs only once there is no mode for this data.
 

Think (Text book Page No. 103)
Question $3 .$
Which average will be most aprropriate for the companies producing the following goods? why?
(i) Diaries and notebooks
(ii) School bags
(iii) Jeans and T-shirts.
Solution:

for all the above data mode will be more appropriate.
 

Exercise $5.3$
Try These (Text book Page No. 106)
Question $1 .$
Find the median of $3,8,7,8,4,5,6$.
Solution:
Arranging in ascending order: $3,4,5,6,7,8,8$.
Here $\mathrm{n}=7$, which is odd.
$\therefore \quad$ Median $=\left(\frac{n+1}{2_{t h}}\right)^{\text {th }}$ term $=\left(\frac{7+1}{2}\right)^{\text {th }}$ term
$=\left(\frac{8}{2}\right)^{t h} \text { term }=4^{\text {th }} \text { term }=6$
Hence the median is 6 .
 

Question $2 .$
Find the median: $11,14,10,9,14,11,12,6,7,7$.
Solution:
Arranging in ascending order: $6,7,7,9,10,11,11,12,14,14$

Here $\mathrm{n}=10$, which is even.
$\begin{aligned}
\therefore \text { Median } &=\frac{1}{2}\left\{\left(\frac{n}{2}\right)^{t h} \text { term }+\left(\frac{n}{2}+1\right)^{t h} \text { term }\right\} \\
&=\frac{1}{2}\left\{\left(\frac{10}{2}\right)^{\text {th }} \text { term }+\left(\frac{10}{2}+1\right)^{t h} \text { term }\right\} \\
&=\frac{1}{2}\left\{5^{\text {th }} \text { term }+6^{\text {th }} \text { term }\right\}=\frac{1}{2}\{10+11\} \\
&=\frac{1}{2}(21)=10.5 \\
\therefore \text { Median } &=10.5
\end{aligned}$

Think (Text book Page No. 108)
Complete the table given below and observe it to answer the following questions.

Solution:
Question $1 .$
Which are all the series having common mean and median?
Solution:
$A, B$ and $C$.
 

Question $2 .$
Why median is same for all the 4 series?
Solution:
Since the middle value is 100 .

Question $3 .$
How mean is unchanged in the series $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ ?
Solution:
The difference between the given numbers are equal.
 

Question $4 .$
What change is to be made in the data, so that mean and median of ' $D$ ' series is equal to other series?
Solution:
If 99 becomes 0 or 200 becomes 101 then mean becomes 100 .