Exercise 13.2 (Revised) - Chapter 15 - Statistics - Ncert Solutions class 11 - Maths

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Chapter 13 - Statistics | NCERT Solutions for Class 11 Maths

Find the mean and variance for each of the data in Exercises 1 to 5 .
Ex 13.2 Question 1.

$6,7,10,12,13,4,8,12$

Answer.

Given: $x=6,7,10,12,13,4,8,12$
$
\therefore \sum x=6+7+10+12+13+4+8+12=72
$

And $n=8 \therefore \bar{x}=\frac{72}{8}=9$
Also $\sum x^2=(6)^2+(7)^2+(10)^2+(12)^2+(13)^2+(4)^2+(8)^2+(12)^2=722$
$\therefore$ Variance $=\sigma^2=\frac{\mathrm{N} \sum x^2-\left(\sum x\right)^2}{\mathrm{~N}^2}$
$
\begin{aligned}
& =\frac{8 \times 722-(72)^2}{(8)^2} \\
& =\frac{5776-5184}{64}=\frac{592}{64}=9.25
\end{aligned}
$
Ex 13.2 Question 2.

First $n$ natural numbers

Answer.

Given: $x=1,2,3,4, \ldots \ldots \ldots \ldots ., n$

$\therefore \sum x=1+2+3+4+\ldots \ldots \ldots+n=\frac{n(n+1)}{2}$

And $\sum x^2=(1)^2+(2)^2+(3)^2+(4)^2+\ldots \ldots \ldots(n)^2=\frac{n(n+1)(2 n+1)}{6}$
$\therefore$ Mean $(\bar{x})=\frac{n(n+1)}{2 n}=\frac{n+1}{2}$
Now, Variance $=\sigma^2=\frac{\mathrm{N} \sum x^2-\left(\sum x\right)^2}{\mathrm{~N}^2}$
$=\frac{n \times \frac{n(n+1)(2 n+1)}{6}-\left(\frac{n(n+1)}{2}\right)^2}{n^2}$
$=\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{(n+1)}{2}\right)^2$
$=\frac{(n+1)(n-1)}{12}$
$=\frac{n^2-1}{12}$
Ex 13.2 Question 3.

First 10 multiples of 3

Answer.

Given: $x=3,6,9,12,15,18,21,24,27,30$
$\therefore \sum x=3+6+9+12+\ldots \ldots \ldots+30=165$

And $\sum x^2=3^2+6^2+9^2+12^2+15^2+\ldots \ldots+30^2=3465$
$\therefore$ Mean $(\bar{x})=\frac{165}{10}=16.5$
Now, Variance $=\sigma^2=\frac{N \sum x^2-\left(\sum x\right)^2}{\mathrm{~N}^2}$
$
=\frac{10 \times 3465-(165)^2}{(10)^2}
$

$
\begin{aligned}
& =\frac{34650-27225}{100} \\
& =\frac{7425}{100}=74.25
\end{aligned}
$
Ex 13.2 Question 4.

Answer.

$\begin{aligned}
& \text { Mean }(\bar{x})=\frac{1}{\mathrm{~N}} \sum f_i x_i=\frac{1}{40} \times 760=19 \\
& \text { Variance }=\sigma^2=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2 \\
& =\frac{1}{40} \times 1736=43.4
\end{aligned}$

Ex 13.2 Question 5. 

Answer.

Mean $(\bar{x})=\frac{1}{\mathrm{~N}} \sum f_i x_i=\frac{1}{22} \times 2200=100$
$
\begin{aligned}
& \text { Variance }=\sigma^2=\frac{1}{\mathrm{~N}} \sum_{i=1}^n f_i\left(x_i-\bar{x}\right)^2 \\
& =\frac{1}{22} \times 640=29.09
\end{aligned}
$
Ex 13.2 Question 6.

Find the mean and standard deviation using short cut method.

Answer.

Mean $(\bar{x})=\mathrm{A}+\frac{\sum f u}{\mathrm{~N}}=64+\frac{0}{100}=64$
$
\begin{aligned}
& \text { Variance }=\sigma^2=\frac{1}{\mathrm{~N}} \sqrt{\mathrm{N} \sum f u^2-\left(\sum f u\right)^2} \\
& =\frac{1}{100} \times \sqrt{100 \times 286-(0)^2} \\
& =\frac{1}{100} \sqrt{28600}=\frac{1}{100} \times 169.1=1.69
\end{aligned}
$

Find the mean and variance for the following frequency distribution in Exercises 7 and 8.
Ex 13.2 Question 7.

Answer.

Mean $(\bar{x})=\mathrm{A}+\frac{\sum f u}{\mathrm{~N}} \times h=105+\frac{2}{30} \times 30=107$
$
\text { Variance }=\sigma^2=\frac{h^2}{\mathrm{~N}^2}\left[\mathrm{~N} \sum f u^2-\left(\sum f u\right)^2\right]=\frac{(30)^2}{(30)^2} \times\left[30 \times 76-(2)^2\right]=2280-4=2276
$
Ex 13.2 Question 8 .

Answer.

$
\begin{aligned}
& \text { Mean }(\bar{x})=\mathrm{A}+\frac{\sum f u}{\mathrm{~N}} \times h \\
& =25+\frac{10}{50} \times 10=25+2=27 \\
& \text { Variance }=\sigma^2=\frac{h^2}{\mathrm{~N}^2}\left[\mathrm{~N} \sum f u^2-\left(\sum f u\right)^2\right] \\
& =\frac{(10)^2}{(50)^2} \times\left[50 \times 68-(10)^2\right] \\
& =\frac{1}{25} \times 3300=132
\end{aligned}
$
Ex 13.2 Question 9.

Find the mean, variance and standard deviation using short-cut method.

Answer.

$
\begin{aligned}
& \text { Mean }(\bar{x})=\mathrm{A}+\frac{\sum f u}{\mathrm{~N}} \times h \\
& =92.5+\frac{6}{60} \times 5=92.5+0.5=93 \\
& \text { Variance }=\sigma^2=\frac{h^2}{\mathrm{~N}^2}\left[\mathrm{~N} \sum f u^2-\left(\sum f u\right)^2\right] \\
& =\frac{(5)^2}{(60)^2} \times\left[60 \times 254-(6)^2\right] \\
& =\frac{25}{3600}[15240-36] \\
& =\frac{25}{3600} \times 15204=105.58
\end{aligned}
$

Standard deviation $(\sigma)=\sqrt{105.58}=10.27$
Ex 13.2 Question 10.

The diameters of circles (in $\mathrm{mm}$ ) drawn in a design are given below:

Calculate the standard deviation and mean diameter of the circles.
[Hint: First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,40.5$ 44.5, 44.5-48.5, 48.5-52.5 and then proceed]

Answer.

$
\begin{aligned}
& \text { Mean }(\bar{x})=\mathrm{A}+\frac{\sum f u}{\mathrm{~N}} \times h \\
& =42.5+\frac{25}{100} \times 4=42.5+1=43.5 \mathrm{~mm}
\end{aligned}
$

Standard deviation $(\sigma)=\frac{h}{\mathrm{~N}} \sqrt{\mathrm{N} \sum f u^2-\left(\sum f u\right)^2}$
$
\begin{aligned}
& =\frac{4}{100} \sqrt{100 \times 199-(25)^2} \\
& =\frac{1}{25} \times 138.83=5.55
\end{aligned}
$