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Exercise 8.1 - Chapter 8 Statistics 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Ex $8.1$ : Chapter 8 - Statistics - 10th Maths Guide Samacheer Kalvi Solutions
Question 1.
Find the range and coefficient of range of the following data.
(i) $63,89,98,125,79,108,117,68$
(ii) $43.5,13.6,18.9,38.4,61.4,29.8$
Solution:
Range $\mathrm{R}=\mathrm{L}-\mathrm{S}$.
Co-efficient of range $=\frac{\mathrm{L}-\mathrm{S}}{\mathrm{L}+\mathrm{S}}$
$\mathrm{L}$-Largest value,
$\mathrm{S}$ - Smallest value.
(i) $63,89,98,125,79,108,117,68$.
Here $L=125$
$\mathrm{S}=63$
$\therefore \mathrm{R}=\mathrm{L}-\mathrm{S}=125-63=62$
$
\begin{aligned}
\text { Co-efficient of range } & =\frac{\mathrm{L}-\mathrm{S}}{\mathrm{L}+\mathrm{S}} \\
& =\frac{125-63}{125+63} \\
& =\frac{62}{188}=0.33
\end{aligned}
$
Question 2.
If the range and the smallest value of a set of data are $36.8$ and $13.4$ respectively, then find the largest value.
Answer:

Range $=36.8$
Smallest value $(\mathrm{S})=13.4$
Range $=\mathrm{L}-\mathrm{S}$
$36.8=\mathrm{L}-13.4$
$\mathrm{L}=36.8+13.4=50.2$
Largest value $=50.2$
Question 3.
Calculate the range of the following data.

Solution:
Here the largest value $=650$
The smallest value $=400$
$\therefore$ Range $=\mathrm{L}-\mathrm{S}=650-400$
$=250$
Question 4.
A teacher asked the students to complete 60 pages of a record notebook. Eight students have completed only $32,35,37,30,33,36,35$ and 37 pages. Find the standard deviation of the pages yet to be completed by them.

Solution:
$
\frac{\Sigma x}{n}=\frac{275}{8}=34.3
$

Standard deviation:
$
\begin{aligned}
\sigma & =\sqrt{\frac{\Sigma d^2}{n}} \\
& =\sqrt{\frac{43.92}{8}} \\
& =2.34
\end{aligned}
$
Question 5.
Find the variance and standard deviation of the wages of 9 workers given below: $\square 310, \square 290, \square 320, \square 280, \square 300, \square 290, \square 320, \square 310, \square 280$.
Solution:

$
\bar{x}=\frac{\Sigma x}{n}=\frac{2700}{9}=300
$

$
\begin{aligned}
\text { Variance } & =\frac{\Sigma d^2}{n} \\
& =\frac{2000}{9} \\
& =222.22
\end{aligned}
$
Standard deviation
$
\begin{aligned}
\sigma & =\sqrt{\frac{\Sigma d^2}{n}} \\
& =\sqrt{222.22} \\
& \cong 14.91
\end{aligned}
$
Question 6.
A wall clock strikes the bell once at 1 o'clock, 2 times at 2 o'clock, 3 times at 3 o'clock and so on. How many times will it strike in a particular day? Find the standard deviation of the number of strikes the bell make a day.
Solution:

$
\bar{x}=\frac{\Sigma x}{n}=13
$
$
\begin{aligned}
\sigma & =\sqrt{\frac{\Sigma d^2}{n}} \\
& =\sqrt{\frac{572}{12}} \\
& =\sqrt{47.66} \\
& \cong 6.9
\end{aligned}
$
Question 7.
Find the standard deviation of the first 21 natural numbers.

Solution:

$\bar{x}=\frac{\sum x}{n}=\frac{231}{21}=11$

Standard deviation
$
\begin{aligned}
\sigma & =\sqrt{\frac{\Sigma d^2}{n}} \\
& =\sqrt{\frac{770}{21}} \\
& =\sqrt{36.66} \\
& \cong 6.05
\end{aligned}
$
Question 8.
If the standard deviation of a data is $4.5$ and if each value of the data is decreased by 5 , then find the new standard deviation.
Answer:
The standard deviation of the data $=4.5$
Each data is decreased by 5
The new standard deviation $=4.5$
Question 9.
If the standard deviation of a data is $3.6$ and each value of the data is divided by 3 , then find the new variance and new standard deviation.
Solution:
If the standard deviation of a data is $3.6$, and each of the data is divided by 3 then the new standard deviation is also divided by 3 .
$\therefore$ The new standard deviation $=\frac{3.6}{3}$
$=1.2$
The new variance $=(\text { standard deviation })^2$ $=\sigma^2=1.2^2=1.44$
Question 10.

The rainfall recorded in various places of five districts in a week are given below.

Find its standard deviation.
Solution:

$
\text { mean, } \begin{aligned}
\bar{x} & =\frac{\Sigma x_i f_i}{\mathrm{~N}} \\
& =\frac{2240}{40} \\
& =56
\end{aligned}
$
$\therefore$ Standard deviation
$
\begin{aligned}
\sigma & =\sqrt{\frac{\Sigma f_i d_i^2}{\mathrm{~N}}} \\
& =\sqrt{\frac{2410}{40}} \\
& =\sqrt{60.25} \\
& =7.76
\end{aligned}
$
Question 11.
In a study about viral fever, the number of people affected in a town were noted as

Find its standard deviation.
Solution:
Let the assumed mean $\mathrm{A}=35, \mathrm{C}=10$

$
\begin{aligned}
& \text { Standard deviation } \sigma=c \times \sqrt{\frac{\Sigma f_i d^2}{\mathrm{~N}}-\left(\frac{\Sigma f_i d_i}{\mathrm{~N}}\right)^2} \\
= & 10 \times \sqrt{\frac{139}{65}-\left(\frac{3}{65}\right)^2} \\
= & 10 \times \sqrt{2.138-(0.046)^2} \\
= & 10 \times \sqrt{2.138-0.002116} \\
= & 10 \times \sqrt{2.136} \\
= & 10 \times 1.46 \\
= & 14.6
\end{aligned}
$
Question 12.
The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find its standard deviation.

Solution:

Assumed mean $\mathrm{A}=30.5, \mathrm{C}=4$
$
\begin{aligned}
\text { Standard deviation } \sigma & =c \times \sqrt{\frac{\Sigma f_i d^2}{\mathrm{~N}}-\left(\frac{\Sigma f_i d_i}{\mathrm{~N}}\right)^2} \\
& =4 \times \sqrt{\frac{189}{84}-\left(\frac{5}{84}\right)^2} \\
& =4 \times \sqrt{2.25-(0.059)^2} \\
& =4 \times \sqrt{2.25-0.0035} \\
& =4 \times \sqrt{2.2465} \\
& =4 \times 1.498 \\
& \cong 5.99 \\
& =6
\end{aligned}
$

Question 13.
The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

Solution:
Assumed mean $\mathrm{A}=11, \mathrm{C}=1$

Standard deviation
$
\begin{aligned}
\sigma & =c \times \sqrt{\frac{\Sigma f_i d^2}{\mathrm{~N}}-\left(\frac{\Sigma f_i d_i}{\mathrm{~N}}\right)^2} \\
& =1 \times \sqrt{\frac{78}{50}-\left(\frac{8}{50}\right)^2} \\
& =1 \times \sqrt{1.56-(0.16)^2} \\
& =1 \times \sqrt{1.56-0.0256} \\
& =1 \times \sqrt{1.534} \\
& =1 \times 1.213 \\
& =1.213 \\
& =1.2
\end{aligned}
$
Question 14.
For a group of 100 candidates the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.
Solution:

$\begin{aligned}
& n=100 \\
& \bar{x}=\frac{\Sigma x}{n} \\
& \bar{x}=60, \sigma=15 \\
& \therefore \quad \Sigma x=\bar{x} \times n=60 \times 100=6000 \\
& \text { Correct } \Sigma x=6000+45+72-40-27 \\
& =6117-67 \\
& \Sigma x=6050 \\
& n=100 \\
& \text { Correct } \bar{x}=\frac{\Sigma x}{n}=\frac{6050}{100}=60.5 \\
& \text { Standard deviation } \sigma=\sqrt{\left(\frac{\Sigma x^2}{n}\right)-\left(\frac{\Sigma x}{n}\right)^2} \\
& \text { Incorrect value of } \sigma=15=\sqrt{\frac{\Sigma x^2}{100}-60^2} \\
& 225=\frac{\Sigma x^2}{100}-3600 \text { gives, } \\
& \frac{\Sigma x^2}{100}=3825 \\
& \Sigma x^2=382500 \\
&
\end{aligned}$

$
\text { Correct value of } \begin{aligned}
& \Sigma x^2= 382500+45^2+72^2 \\
& \quad-40^2-27^2 \\
&= 382500+2025+ \\
& 5184-1600-729 \\
&= 389709-2329 \\
&= 387380
\end{aligned}
$
Correct standard deviation
$
\begin{aligned}
\sigma & =\sqrt{\frac{387380}{100}-(60.5)^2} \\
& =\sqrt{3873.80-3660.25} \\
& =\sqrt{213.55} \\
& =14.61
\end{aligned}
$
Question 15.
The mean and variance of seven observations are 8 and 16 respectively. If five of these are $2,4,10$, 12 and 14, then find the remaining two observations.
Solution:
$\bar{x}=8, \sigma^2=16, n=7$. If five of these are $2,4,10$,
12,14
$
\begin{aligned}
\bar{x}=\frac{\Sigma x}{n} & =\frac{2+4+10+12+14+a+b}{7} \\
8 & =\frac{42+a+b}{7} \\
42+a+b & =56 \\
a+b & =56-42=14
\end{aligned}
$
The given 5 number are 2,4 , a, b, $10,12,14$. a, $b$ are 6 and 8 .