Statistics

  $\frac{x_1+x_2+x_3+x_4}{4}=\frac{7}{2}$

$x_1+x_2+x_3+x_4=14$

and $\frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}$

x₅ = 10

Variance $\frac{{\displaystyle \sum _{i=1}^4{x}_i^2}}{4}-{\left(\frac{\Sigma xi}{4}\right)}^2=a$

$\frac{x_1^2+x_2^2+x_3^2+x_4^2}{4}-\frac{49}{4}=a$

$x_1^2+x_2^2+x_3^2+x_4^2=4a+49$

and $\frac{x_1^2+x_2^2+x_5^2+x_4^2+x_5^2}{5}-{\left(\frac{24}{5}\right)}^2=\frac{194}{25}$

$\frac{4a+49+x_5^2}{5}=\frac{576+194}{25}$

49 + 49 + $x_5^2=\frac{770}{5}$

49 $+x_5^2+49=154$

4a + 149 = 154

4a = 5

now 4a + x₅ = 15

 $\overline{x}=\frac{\sum xi}{40}=30\Rightarrow \sum xi=1200$ ………. (i)

${\alpha }^2=\frac{1}{40}\sum x{i}_i^2-{\left(30\right)}^2=25$

$\Rightarrow \sum x_i^2=37000$

after omitting two wrong observations

$\sum y_i^2=37000-144-100=36756$

$\therefore 38a^2=36756-36158=238$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat{\sigma }_C=5\\ WeknowthatC=\frac{5}{9}\left(F-32\right)\Rightarrow F=\frac{9C}{5}+32\\ \therefore {\sigma }_F=\frac{9}{5}{\sigma }_C=\frac{9}{5}*5=9\\ \therefore {\sigma }_F^2={\left(9\right)}^2=81\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Here,wehaveC{V}_1=50,C{V}_2=60\\ \overline{x_1}=30and\overline{x_2}=25\\ \therefore C{V}_1=\frac{{\sigma }_1}{\overline{x_1}}*100\Rightarrow 50=\frac{{\sigma }_1}{30}*100\Rightarrow {\sigma }_1=\frac{50*30}{100}=15\\ andC{V}_1=\frac{{\sigma }_2}{\overline{x_2}}*100\Rightarrow 60=\frac{{\sigma }_2}{25}*100\Rightarrow {\sigma }_2=\frac{60*25}{100}=15\\ \therefore Difference{\sigma }_1-{\sigma }_2=15-15=0\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}WeknowthatVariance\left({\sigma }^2\right)=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\right)}^2\\ =\frac{18000}{60}-{\left(\frac{960}{60}\right)}^2=300-256=44\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}First10positiveintegersare1,2,3,4,5,6,7,8,9,10\\ onmultiplyingeachnumberby-1,weget\\ -1,-2,-3,-4,-5,-6,-7,-8,-9,-10\\ onadding1toeachofthenumber,weget\\ 0,-1,-2,-3,-4,-5,-6,-7,-8,-9\\ \therefore {\displaystyle \sum _{}^{}{x}_i}=0-1-2-3-4-5-6-7-8-9\\ =-45\\ and{\displaystyle \sum _{}^{}{x}_i^2}=0^2+{\left(-1\right)}^2+{\left(-2\right)}^2+{\left(-3\right)}^2+{\left(-4\right)}^2+\dots +{\left(-9\right)}^2\\ =\frac{9*10*19}{6}=285\left[?{\displaystyle \sum _{}^{}{n}^2=}\frac{n\left(n+1\right)\left(2n+1\right)}{6}\right]\\ \therefore S.D=\sqrt[]{\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{N}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{N}\right)}^2}=\sqrt[]{\frac{285}{10}-{\left(\frac{-45}{10}\right)}^2}\\ =\sqrt[]{\frac{285}{10}-\frac{2025}{100}}=\sqrt[]{\frac{2850-2025}{100}}=\sqrt[]{8.25}\\ \therefore Variance={\left(SD\right)}^2={\left(\sqrt[]{8.25}\right)}^2=8.25\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Givennumbersare1,2,3,4,5,6,7,8,9,10\\ Numbersobtainedwhen1isaddedtotheabovenumbersis2,3,4,5,6,7,8,9,10and11.\\ \therefore {\displaystyle \sum _{}^{}{x}_i}=2+3+4+5+6+7+8+9+10+11\\ =\frac{10}{2}\left[2*2+\left(10-1\right).1\right]=5\left[4+9\right]=5*13=65\\ Now{\displaystyle \sum _{}^{}{x}_i^2}=2^2+3^2+4^2+\dots +11^2\\ =\left(1^2+2^2+3^2+4^2+\dots +11^2\right)-{\left(1\right)}^2\\ =\frac{n\left(n+1\right)\left(2n+1\right)}{6}-1=\frac{11*12*23}{6}-1\\ =22*23-1=506-1=505\\ \therefore Variance\left({\sigma }^2\right)=\frac{{\displaystyle \sum _{}^{}{x}_i^2}}{n}-{\left(\frac{{\displaystyle \sum _{}^{}{x}_i}}{n}\right)}^2=\frac{505}{10}-{\left(\frac{65}{10}\right)}^2\\ =50.5-{\left(6.5\right)}^2=50.5-42.25=8.25\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}WeknowthatSDoffirstnnaturalnumbers\sqrt[]{\frac{n^2-1}{12}}\\ Heren=10\\ \therefore SD=\sqrt[]{\frac{{\left(10\right)}^2-1}{12}}=\sqrt[]{\frac{99}{12}}=\sqrt[]{8.25}=2.87\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l}Giventhat{w}_i=x_i+k,\overline{x_i}=48,SD\left(x_i\right)=12,\\ w_i=55andSD\left(w_i\right)=15\\ then\overline{w_i}=\overline{x_i}+k\left(\overline{w_i}=meanof{w}_i\text{'}sand\overline{x_i}isthemeanof{x}_i\text{'}s\right)\\ \Rightarrow 55=48+k\dots \left(i\right)\\ SDof{w}_i=SDof{x}_i\\ 15=l*12\\ \Rightarrow l=\frac{15}{12}=1.25\dots \left(ii\right)\\ fromeqn.\left(i\right)and\left(ii\right)wehave\\ k=\overline{w_i}-\overline{x_i}=55-1.25*48=55-60=-5\\ Hence,thecorrectoptionis\left(a\right).\end{array}$