Consider a set of 3n numbers having variance 4. In this set, the mean of first 2n numbers is 6 and the mean of the remaining n numbers is 3. A new…

Let, the number be ${a}_{1},{a}_{2},{a}_{3},...,{a}_{2n},{b}_{1},{b}_{2},{b}_{3},...,{b}_{n}.$a1​,a2​,a3​,...,a2n​,b1​,b2​,b3​,...,bn​.

Given, the mean of the first $2n$2n numbers is $6$6 and that of the remaining $n$n numbers is $3,$3, then by using the formula for combined mean, we get, the mean of the complete set as $\mu =\frac{2n\times 6+n\times 3}{2n+n}=\frac{15n}{3n}=5.$μ=2n+n2n×6+n×3​=3n15n​=5.

Also, the variance is defined as $\displaystyle {\sigma }^{2}=\frac{\sum {({x}_{i})}^{2}}{n}-{(\mu )}^{2}$σ2=n∑(xi​)2​−(μ)2

$$4=\frac{\sum {({a}_{i})}^{2}+\sum {({b}_{i})}^{2}}{3n}-(5{)}^{2}$$4=3n∑(ai​)2+∑(bi​)2​−(5)2

$$\Rightarrow \frac{\sum {({a}_{i})}^{2}+\sum {({b}_{i})}^{2}}{3n}=29$$⇒3n∑(ai​)2+∑(bi​)2​=29

$$\Rightarrow \sum {({a}_{i})}^{2}+\sum {({b}_{i})}^{2}=87n...(i)$$⇒∑(ai​)2+∑(bi​)2=87n...(i)

Now, as per the given information, the distribution becomes

$${a}_{1}+1,{a}_{2}+1,{a}_{3}+1,...,{a}_{2n}+1,{b}_{1}-1,{b}_{2}-1,...,{b}_{n}-1$$a1​+1,a2​+1,a3​+1,...,a2n​+1,b1​−1,b2​−1,...,bn​−1

Thus, the new variance is

$$k=\frac{\sum ({a}_{i}+1{)}^{2}+\sum ({b}_{i}-1{)}^{2}}{3n}-{(\frac{(12n+2n)+(3n-n)}{3n})}^{2}$$k=3n∑(ai​+1)2+∑(bi​−1)2​−(3n(12n+2n)+(3n−n)​)2

$$\Rightarrow k=\frac{(\sum {({a}_{i})}^{2}+2n+2\sum {a}_{i})+(\sum {({b}_{i})}^{2}+n-2\sum {b}_{i})}{3n}-{(\frac{16}{3})}^{2}$$⇒k=3n(∑(ai​)2+2n+2∑ai​)+(∑(bi​)2+n−2∑bi​)​−(316​)2

$$\Rightarrow k=\frac{\sum {({a}_{i})}^{2}+\sum {({b}_{i})}^{2}+3n+2\sum {a}_{i}-2\sum {b}_{i}}{3n}-{(\frac{16}{3})}^{2}$$⇒k=3n∑(ai​)2+∑(bi​)2+3n+2∑ai​−2∑bi​​−(316​)2

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

$$\Rightarrow k=\frac{87n+3n+2(12n)-2(3n)}{3n}-{(\frac{16}{3})}^{2}$$⇒k=3n87n+3n+2(12n)−2(3n)​−(316​)2

$$\Rightarrow k=\frac{108}{3}-{(\frac{16}{3})}^{2}$$⇒k=3108​−(316​)2

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

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