Statistics

  $LetG.P.be{a}_1=a,a_2=ar,a_3=a{r}^2$ ${}_{}{}_{}{}_{}{}^{}$ , ….

$?3a_2+a_3=2a_4$

^{$\Rightarrow 3ar+a{r}^2=2a{r}^3$}

$\Rightarrow 2a{r}^2-r-3=0$

$\therefore a_2+a_4+2a_5=a\left(r+r^3+2r^4\right)$

$\frac{}{}$ $=\frac{8}{3}\left(\frac{3}{2}+\frac{27}{8}+\frac{81}{8}\right)$  = 40

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

$x_1+x_2+x_3+x_4=14$

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

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