Consider a set of 3n numbers having variance 4 . In this set, the...

Let, the number be ${a}_{1}, {a}_{2}, {a}_{3}, . . . , {a}_{2n}, {b}_{1}, {b}_{2}, {b}_{3}, . . . , {b}_{n} .</p><p>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 $\mu =$\dfrac{2n \times 6 + n \times 3}{2n + n} = \dfrac{15n}{3n} =$5 .</p><p>Also, the variance is defined as${\sigma}^{2} = \dfrac{∑ {\left({x}_{i}\right)}^{2}}{n} - {\left(\mu\right)}^{2}$</p><p>$4 = \dfrac{∑ {\left({a}_{i}\right)}^{2} + ∑ {\left({b}_{i}\right)}^{2}}{3n} - ( 5{)}^{2}$</p><p>$⇒ \dfrac{∑ {\left({a}_{i}\right)}^{2} + ∑ {\left({b}_{i}\right)}^{2}}{3n} = 29$</p><p>$⇒ ∑ {\left({a}_{i}\right)}^{2} + ∑ {\left({b}_{i}\right)}^{2} = 87n . . . \left(i\right)$</p><p>Now, as per the given information, the distribution becomes</p><p>${a}_{1} + 1, {a}_{2} + 1, {a}_{3} + 1, . . . , {a}_{2n} + 1, {b}_{1} - 1, {b}_{2} - 1, . . . , {b}_{n} - 1$</p><p>Thus, the new variance is</p><p>$k = \dfrac{∑ ( {a}_{i} + 1{)}^{2} + ∑ ( {b}_{i} - 1{)}^{2}}{3n} - {\left(\dfrac{\left(12n + 2n\right) + \left(3n - n\right)}{3n}\right)}^{2}$</p><p>$⇒ k = \dfrac{\left(∑ {\left({a}_{i}\right)}^{2} + 2n + 2 ∑ {a}_{i}\right) + \left(∑ {\left({b}_{i}\right)}^{2} + n - 2 ∑ {b}_{i}\right)}{3n} - {\left(\dfrac{16}{3}\right)}^{2}$</p><p>$⇒ k = \dfrac{∑ {\left({a}_{i}\right)}^{2} + ∑ {\left({b}_{i}\right)}^{2} + 3n + 2 ∑ {a}_{i} - 2 ∑ {b}_{i}}{3n} - {\left(\dfrac{16}{3}\right)}^{2}$</p><p>On putting the given values and value from the equation \left(i\right),$

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

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

$⇒ 9k = 3 ( 108 ) - ( 16{)}^{2}$

$⇒ 9k = 324 - 256 = 68 .</p>$