Mathematics - Statistics Question with Solution | TestHub

Let, the number be $a_1, a_2, a_3,..., a_{2n}, b_1, b_2, b_3,..., b_n.$

Given, the mean of the first $2n$ numbers is $6$ and that of the remaining $n$ numbers is $3,$ then by using the formula for combined mean, we get, the mean of the complete set as $\mu =\frac{2n\times 6+n\times 3}{2n+n}=\frac{15n}{3n}=5.$

Also, the variance is defined as ${\sigma }^2=\frac{\sum {\left(x_i\right)}^2}{n}-{\left(\mu \right)}^2$

$4=\frac{\sum {\left(a_i\right)}^2+\sum {\left(b_i\right)}^2}{3n}-(5{)}^2$

$\Rightarrow \frac{\sum {\left(a_i\right)}^2+\sum {\left(b_i\right)}^2}{3n}=29$

$\Rightarrow \sum {\left(a_i\right)}^2+\sum {\left(b_i\right)}^2=87n ...\left(i\right)$

Now, as per the given information, the distribution becomes

$a_1+1, a_2+1, a_3+1,..., a_{2\mathrm{n}}+1,{ b}_1-1, b_2-1,..., b_{\mathrm{n}}-1$

Thus, the new variance is

$k=\frac{\sum (a_i+1{)}^2+\sum (b_i-1{)}^2}{3n}-{\left(\frac{\left(12n+2n\right)+\left(3n-n\right)}{3n}\right)}^2$

$\Rightarrow k=\frac{\left(\sum {\left(a_i\right)}^2+2n+2\sum a_i\right)+\left(\sum {\left(b_i\right)}^2+n-2\sum b_i\right)}{3n}-{\left(\frac{16}{3}\right)}^2$

$\Rightarrow k=\frac{\sum {\left(a_i\right)}^2+\sum {\left(b_i\right)}^2+3n+2\sum a_i-2\sum b_i}{3n}-{\left(\frac{16}{3}\right)}^2$

On putting the given values and value from the equation $\left(i\right),$ 

$\Rightarrow k=\frac{87n+3n+2\left(12n\right)-2\left(3n\right)}{3n}-{\left(\frac{16}{3}\right)}^2$

$\Rightarrow k=\frac{108}{3}-{\left(\frac{16}{3}\right)}^2$

$\Rightarrow 9k=3\left(108\right)-(16{)}^2$

$\Rightarrow 9k=324-256=68.$