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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions
Question 1.
Find the standard deviation of $30,80,60,70,20,40,50$ using the direct method.
Solution:
Direct method:

$
\begin{aligned}
\sigma & =\sqrt{\frac{\sum x^2}{n}-\left(\frac{\Sigma x}{n}\right)^2} \\
& =\sqrt{\frac{20300}{7}-\left(\frac{350}{7}\right)^2} \\
& =\sqrt{400}=20
\end{aligned}
$
Question 2.
Find the standard deviation for the following data. 5, 10, 15, 20,25. And also find the new S.D. if three is added to each value.
Solution:

$
\begin{aligned}
& \bar{x}=\frac{\Sigma x}{n}=\frac{75}{5}=15 \\
& d^{\prime}=\frac{x-\bar{x}}{c}=\frac{x-\mathrm{A}}{c}
\end{aligned}
$
$\mathrm{A}$ is assumed mean $\mathrm{c}$ is common factor.
Here $\mathrm{A}=15, \mathrm{C}=5$
$
\sigma=\sqrt{\left(\frac{\Sigma d^{\prime 2}}{n}\right)-\left(\frac{\Sigma d^{\prime}}{n}\right)^2} \times c
$

$
\begin{aligned}
& =\sqrt{\frac{10}{5}-0} \times c \\
& =\sqrt{2} \times 5 \\
& =5 \sqrt{2}
\end{aligned}
$
If 3 is added to each value, we get $8,13,18,23$, 28 as new values.

$
\begin{aligned}
\therefore \quad \sigma & =\sqrt{\left(\frac{\Sigma d^{\prime 2}}{n}\right)-\left(\frac{\Sigma d^{\prime}}{n}\right)^2} \times c \\
& =\sqrt{\frac{10}{5}-0 \times 5} \\
& =\sqrt{2} \times 5=5 \sqrt{2}
\end{aligned}
$
S.D. doesn't change when a number is added or subtracted to the values.
Question 3.
The marks scored by 5 students in a test for 50 marks are 20, 25, 30, 35, 40. Find the S.D for the marks. If the marks are converted for 100 marks, find the S.D. for newly obtained marks.
Solution:
Let assumed mean $\mathrm{A}=30$
$
C=5
$

$\begin{aligned}
\sigma & =\sqrt{\left(\frac{\Sigma d^{\prime 2}}{n}\right)-\left(\frac{\Sigma d^{\prime}}{n}\right)^2} \times c \\
& =\sqrt{\frac{10}{5}-0 \times 5} \\
& =\sqrt{2} \times 5=5 \sqrt{2}
\end{aligned}$

To convert the values for 100 , all the values will be multiplied by 2 . Therefore the new values are $40,50,60,70,80$.
Let $\mathrm{A}=60$,
$
\mathrm{C}=10
$

$
\begin{aligned}
\sigma & =\sqrt{\left(\frac{\Sigma d^{\prime 2}}{n}\right)-\left(\frac{\Sigma d^{\prime}}{n}\right)^2} \times c \\
& =\sqrt{\frac{10}{5}-0 \times 10} \\
& =\sqrt{2} \times 10 \\
& =10 \sqrt{2}
\end{aligned}
$
S.D. also be multiplied by 2 . It is also true for the division also.
Question 4.
$\Sigma x=99, n=9, \Sigma(x-10)^2=79$, then find, (i) $\Sigma x^2$ (ii) $\Sigma(x-\bar{x})^2$
Solution:

$
\begin{aligned}
n=9, \Sigma x & =99, \bar{x}=\frac{\Sigma x}{n}=\frac{99}{9}=11 \\
\Sigma(x-10)^2 & =79=\Sigma x^2-20 x+100=79 \\
& =\Sigma x^2-20 \Sigma x+100 \times 9=79 \\
& =\Sigma x^2-20 \times 99+900=79 \\
\Sigma x^2 & =79+1980-900=1159 \\
\Sigma(x-\bar{x})^2 & =\Sigma(x-11)^2=\Sigma\left(x^2-22 x+121\right) \\
& =\Sigma x^2-22 \Sigma x+121 \times 9 \\
& =1159-22 \times 99+1089=70 \\
\therefore \quad \Sigma x^2 & =1159, \Sigma(x-\bar{x})^2=70
\end{aligned}
$
Question 5.
Find the co-efficient of variation for the following data: $16,13,17,21,18$.

Solution:
$
\text { Mean } \bar{x}=\frac{16+13+17+21+18}{5}=\frac{85}{5}=17
$

$
\begin{aligned}
& \sigma=\sqrt{\frac{\sum d^2}{n}}=\sqrt{\frac{34}{5}}=\sqrt{6.8} \\
& \sigma=2.61
\end{aligned}
$
Co-efficient of variation
$
\begin{aligned}
\mathrm{CV} & =\frac{\sigma}{\bar{x}} \times 100=\frac{2.61}{17} \times 100 \\
& =15.35 \%
\end{aligned}
$
Question 6.
C.V. of a data is $69 \%$, S.D. is $15.6$, then find its mean.
Solution:
$
\begin{aligned}
\mathrm{CV} & =\frac{\sigma}{\bar{x}} \times 100 \Rightarrow \bar{x}=\frac{\sigma}{\mathrm{CV}} \times 100 \\
\bar{x} & =\frac{15.6}{6.9} \times 100=22.6
\end{aligned}
$
Question 7.
S.D. of a data is 2102 , mean is $36.6$, then find its C.V.

Solution:
$
\begin{aligned}
\sigma & =21.2, \bar{x}=36.6 \\
\mathrm{CV}=\frac{\sigma}{\bar{x}} \times 100 & =\frac{21.2}{36.6} \times 100=57.92 \%
\end{aligned}
$
Question 8.

Which team is more consistent?
Solution:

$
\bar{x}_1=\frac{140}{5}=28
$

$\begin{aligned}
\sigma_1 & =\sqrt{\frac{\Sigma d^2}{n}}=\sqrt{\frac{880}{5}} \\
& =\sqrt{176} \\
& =13.27 \\
\mathrm{CV}_1 & =\frac{\sigma_1}{\bar{x}_1} \times 100 \\
\mathrm{CV}_1 & =\frac{13.27}{28} \times 100 \\
& =47.39 \% \\
\mathrm{CV}_1 & <\mathrm{CV}_2
\end{aligned}$

$
\begin{aligned}
\bar{x}_2 & =\frac{150}{5}=30 \\
\sigma_2 & =\sqrt{\frac{\sum d^2}{n}}=\sqrt{\frac{1600}{5}} \\
& =\sqrt{320} \\
& =17.89 \\
\mathrm{CV}_1 & =\frac{\sigma_2}{\bar{x}_2} \times 100 \\
\mathrm{CV}_2 & =\frac{17.89}{30} \times 100 \\
& =59.63 \%
\end{aligned}
$
$\therefore$ Team $\mathrm{A}$ is more consistent.
Question 9.
Final the probability of choosing a spade or a heart card from a deck of cards.
Solution:
Total number of cards $=52$
Event of selecting a spade card $=\mathrm{A}$
Event of selecting a heart card $=B$
$\mathrm{n}(\mathrm{A})=13$
$n(B)=13$