Mathematics - Statistics Question with Solution | TestHub

Let$1^{st}5$students are

$x_1,x_2,x_3,x_4$and$x_5$

$\therefore 18=E\left(x^2\right)-{\left(E\left(x\right)\right)}^2$

$=\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-{\left(150\right)}^2$

$\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=112590$

$\therefore$new variance$=\frac{x_1^2+x_2^2+....+x_5^2+{156}^2}{6}-{\left(151\right)}^2$$\left[\mathrm{S}\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{e}\mathrm{w}\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{n}=\frac{150\times 5+156}{6}=151\right]$

$=\frac{112590+24336}{6}-22801$

$=20$