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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Unit Exercise 8
Question 1.
The mean of the following frequency distribution is $62.8$ and the sum of all frequencies is 50 . Compute the missing frequencies $f_1$ and $f_2$.

Solution:
Mean $\bar{x}=62.8$
$\Sigma x=50$

$
\begin{gathered}
\bar{x}=\frac{\sum f_i x_i}{\sum f_i}=\frac{2060+30 f_1+70 f_2}{30+f_1+f_2} \\
\frac{2060+30 f_1+70 f_2}{30+f_1+f_2}=62.8
\end{gathered}
$

$
\begin{aligned}
30+f_1+f_2 & =50 \text { (given) } \\
f_1+f_2 & =20 \\
2060+30 f_1+70 f_2 & =3140 \\
30 f_1+70 f_2 & =3140-2060 \\
30 f_1+70 f_2 & =1080
\end{aligned}
$
Solving (2) \& (3) we get,
$
\begin{aligned}
30 f_1+70 f_2 & =1080 \\
(2) \times 30 \Rightarrow 30 f_1+30 f_2 & =600 \\
\hline 40 f_2 & =480 \\
f_2 & =12
\end{aligned}
$
Sub. $f_2=12$ in (2), we get
$
\begin{aligned}
& f_1+12 \leq 20 \Rightarrow f_1=8 \\
& f_1=8, f_2=12
\end{aligned}
$
Question 2.
The diameter of circles (in $\mathrm{mm}$ ) drawn in a design are given below.

Claculate the standard deviation.

Solution:

$
\begin{array}{r}
\sigma=\sqrt{\frac{\Sigma f_i\left(x_i-\bar{x}\right)^2}{\Sigma f}} \\
=\sqrt{\frac{3184}{100}}=\sqrt{31.84}=5.64
\end{array}
$
Question 3.
The frequency distribution is given below.

In the table, $\mathrm{k}$ is a positive integer, has a varience of 160 . Determine the value of $\mathrm{k}$.

Solution:

$
\begin{aligned}
& \sigma^2=\frac{\Sigma f_i d_1^2}{\Sigma f} \\
& =\frac{k^2}{7^2} \frac{\left[1^2+6^2+8^2+13^2+15^2+20^2\right]}{7} \\
& 160=\frac{k^2}{7^3} \times 1120 \\
& k^2=\frac{160 \times 7^3}{1120} \\
& =49 \Rightarrow k=\pm 7 \\
& k=7 \text { since } k \text { is a +ve number. } \\
&
\end{aligned}
$
Question 4.
The standard deviation of some temperature data in degree Celsius $\left({ }^{\circ} \mathrm{C}\right)$ is 5 . If the data were converted into degree Fahrenheit $\left({ }^{\circ} \mathrm{F}\right)$ then what is the variance?
Answer:
Standard deviation $(\sigma)=5$
Variance $=5^2=25$
We know the formula, $\mathrm{F}=\frac{9}{5} \mathrm{C}+32$
Variance $(\mathrm{F})=$ Vanance $\frac{9}{5} \mathrm{C}^{\circ}+32$
[Variance of $\mathrm{ax}+\mathrm{b}=\mathrm{a}^2$ (variance of $\left.\mathrm{x}\right)$ ]
$
\begin{aligned}
& =\left(\frac{9}{5}\right)^2 \text {. variance } \\
& =\frac{81}{25} \times 25 \\
& =81^{\circ} \mathrm{F}
\end{aligned}
$
New variance $=81^{\circ} \mathrm{F}$
Question 5.
If for a distribution, $\Sigma(x-5)=3, \Sigma(x-5)^2=43$ and total number of observations is 18 , find the mean and standard deviation.
Solution:

$
\begin{aligned}
& \text { Mean } \bar{x}=\frac{\sum x}{n}, \Sigma(x-5)=3, n=18 \\
& \qquad \Sigma x-5 \times 18=3 \\
& \Sigma x=3+90=93 \\
& \quad \Sigma=\frac{93}{18}=5.17 \\
& \Rightarrow \Sigma(x-5)^2=43 \\
& \Rightarrow \Sigma x^2-10 \Sigma x+\Sigma 25=43 \\
& \Rightarrow \Sigma x^2-10 \times 93+25 \times 18=43 \\
& \therefore \Sigma x^2=43+930-450=523 \\
& \text { Standard Deviation } \sigma=\sqrt{\frac{\sum d^2}{n}}=\sqrt{\frac{\sum(x-\bar{x})^2}{n}} \\
& \text { ie } \sqrt{\frac{\sum(x-5.17)^2}{18}} \\
& =\sqrt{\frac{\sum\left(x^2-2 \times 5.17 x+5.17^2\right)}{18}} \\
& \sigma=\sqrt{\frac{523-961.62+481.12}{18}=\sqrt{\frac{42.5}{18}}}=\sqrt{2.36} \\
& =\sqrt{\frac{\sum x^2-2 \times 5.17 \sum x+5.17^2 \times 18}{18}} \\
& =\sqrt{\frac{523-2 \times 5.17 \times 93+5.17^2 \times 18}{18}}
\end{aligned}
$
Question 6.
Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?

Solution:

$\begin{array}{rlrl}
\bar{x}_1 & =\frac{100}{5}=20 & \bar{x}_2 & =\frac{\sum x}{n}=\frac{75}{5}=15 \\
\sigma & =\sqrt{\frac{\sum d^2}{n}} & \sigma & =\sqrt{\frac{\sum d^2}{n}} \\
& =\sqrt{\frac{30}{5}} & & =\sqrt{\frac{68}{5}} \\
& =\sqrt{6} & & =\sqrt{13.6} \\
& =2.44 & & =3.68 \\
\mathrm{CV} & =\frac{\sigma}{\bar{x}} \times 100 & \mathrm{CV} & =\frac{\sigma}{\bar{x}} \times 10 \\
& =\frac{2.44}{20} \times 100 & & =\frac{3.68}{15} \times 100 \\
& =12.2 & & =\frac{368}{15}=24.53
\end{array}$

C.V. of city $\mathrm{A}<$ C.V. of city $\mathrm{B}$.
$\therefore$ City $\mathrm{A}$ is more consistents.
Question 7.
If the range and coefficient of range of the data are 20 and $0.2$ respectively, then find the largest and smallest values of the data.

Solution:
Range $=\mathrm{L}-\mathrm{S}=20$
$
\begin{aligned}
& \text { Co-efficient of range }=\frac{\mathrm{L}-\mathrm{S}}{\mathrm{L}+\mathrm{S}}=0.2 \\
& \mathrm{~L}-\mathrm{S}=20 \\
& \mathrm{~L}-\mathrm{S}=0.2(\mathrm{~L}+\mathrm{S}) \\
&(\mathrm{L}+\mathrm{S}) 0.2=20 \\
& 0.2 \mathrm{~L}+0.2 \mathrm{~S}=20
\end{aligned}
$
$
\begin{aligned}
(1) \times 0.2 \Rightarrow 0.2 \mathrm{~L}-0.2 \mathrm{~S} & =4 \\
0.4 \mathrm{~L} & =24 \\
\mathrm{~L} & =\frac{24}{0.4}=60
\end{aligned}
$
Substitute
$
\begin{aligned}
\mathrm{L} & =60 \text { in (1) } \\
60-\mathrm{S} & =20 \\
-\mathrm{S} & =20-60=-40 \\
\mathrm{~S} & =40
\end{aligned}
$
$\therefore$ The largest is 60 , the smallest is $40^{\circ}$.

Question 8.
If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5 .
Solution:
Product of face values $6:\{(1,6),(2,3),(6,1),(3,2)\}$
Difference of face value $5:\{(1,6),(6,1)\}$
$
\begin{aligned}
& P(\text { product } 6)=\frac{4}{6 \times 6}=\frac{4}{36}=\frac{1}{9} \\
& P(\text { difference } 5)=\frac{2}{6 \times 6}=\frac{1}{18}
\end{aligned}
$
Question 9.
In a two children family, find the probability that there is at least one girl in a family.
Solution:
$
\begin{aligned}
& \mathrm{S}=\{\mathrm{BB}, \mathrm{BG}, \mathrm{GB}, \mathrm{GG}\} \\
& \mathrm{n}(\mathrm{S})=4
\end{aligned}
$
Event of atleast one girl in a family say A
$
\begin{aligned}
& \mathrm{A}=\{\mathrm{BG}, \mathrm{GB}, \mathrm{GG}\} \\
& \mathrm{n}(\mathrm{A})=3 \\
& \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{3}{4}
\end{aligned}
$
Probability of at least one girl in a family is $\frac{3}{4}$
Question 10.
A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.
Solution:
Let a number of black balls be ' $x$ '.

Number of white balls $=5$.
$\mathrm{P}($ black ball $)=\frac{x}{x+5}$
$\mathrm{P}($ white ball $)=\frac{5}{x+5}$
$\because \mathrm{P}($ black ball $)=2 \times \mathrm{P}($ white ball $)$
$\Rightarrow \quad \frac{x}{x+5}=2 \times \frac{5}{x+5} \Rightarrow x=10$
Number of Black balls $=10$.
Question 11.
The probability that a student will pass the final examination in both English and Tamil is $0.5$ and the probability of passing neither is $0.1$. If the probability of passing the English examination is $0.75$, what is the probability of passing the Tamil examination?
Solution:

$\mathrm{P}($ English $)=0.75$
$\mathrm{P}($ Tamil $)=\mathrm{x}$ (assume $)$
$\mathrm{P}($ English $\mathrm{U}$ Tamil $)=\mathrm{P}($ English $)+\mathrm{P}($ Tamil $)-\mathrm{P}($ English $\cap$ Tamil $)$
$\Rightarrow 1-0.1=0.75+\mathrm{x}-0.5$

$
\begin{aligned}
& \Rightarrow \mathrm{x}=0.9-0.25 \\
& \Rightarrow \quad x=0.65=\frac{13}{20}
\end{aligned}
$
Question 12.
The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5 .
Solution:
King spade, Queen spade, Jack spade are removed
$\therefore$ total number of cards $=52-3=49$.
(i) Prob (diamond) $=\frac{13}{49}$
(ii) Prob(queen) $=\frac{4-1}{49}=\frac{3}{49}$
(iii) Prob(spade) $=\frac{13-3}{49}=\frac{10}{49}$
(iv) Prob(heart bearing number 5 ) $=\frac{1}{49}$