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Let $A$ and $B$ be non-empty disjoint sets of real numbers. Suppose that the average of the numbers in the first set is $\mu_{A}$ and the average of the numbers in the second set is $\mu_{B}$; let the corresponding variances be $v_{A}$ and $v_{B}$ respectively. If the average of the elements in $A \cup B$ is $\mu= p.\mu_{A} + (1 - p).\mu_{B}$, what is the variance of the elements in $A \cup B?$

1. $p.v_{A}+ (1 - p).v_{B}$
2. $(1 - p). v_{A}+ p. v_{B}$
3. $p.[v_{A}+(\mu_{A}-\mu)^{2}]+(1 - p). [v_{B}+ (\mu_{B}-\mu)^{2}]$
4. $(1 - p).[v_{A}+(\mu_{A}-\mu)^{2}]+ p. [v_{B}+ (\mu_{B}-\mu)^{2}]$
5. $p.v_{A}+ (1 - p). v_{B} + (\mu_{A}- \mu_{B})^{2}$

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$Var\left ( X+Y \right )=Var\left ( X \right )+Var\left ( Y \right )$

According to this formula $C)$ should be the ans.
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what is p ?
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$p$ is any value between $0$ and $1.$

We have $v =\displaystyle \frac{1}{n} \sum_{i=1}^n (x_i - \mu)^2$

$\qquad =\displaystyle \frac{1}{n} \sum_{i=1}^n \left[x_i^2 - 2x_i\mu+ \mu^2\right]$

$\qquad = \displaystyle \frac{1}{n} \sum_{i=1}^n x_i^2 - 2\mu\frac{1}{n}\sum_{i=1}^n x_i+ \mu^2$

$\qquad = \displaystyle \frac{1}{n} \sum_{i=1}^n x_i^2 - 2\mu^2+ \mu^2$

$\qquad = \displaystyle \frac{1}{n} \sum_{i=1}^n x_i^2 - \mu^2$

Thus,

• $v_A = \displaystyle \frac{1}{n_A} \left(\sum_{i=1}^{n_A} x_i^2\right) - \mu_A^2 \quad \to (1)$
• $v_B = \displaystyle \frac{1}{n_B} \left(\sum_{i=1}^{n_B} x_i^2\right) - \mu_B^2 \quad \to (2)$

$v = \displaystyle \frac{1}{n} \left(\sum_{i=1}^n x_i^2\right) - \mu^2$

$\qquad = \displaystyle \frac{1}{n} \left(\sum_{i=1}^{n_A} x_i^2+\sum_{i=1}^{n_B} x_i^2\right) - \mu^2$

$\qquad = \displaystyle \frac{1}{n} \left(n_A( v_A + \mu_A^2)+n_B( v_B + \mu_B^2)\right) - \mu^2$ (From $(1)$ and $(2))$

$\qquad = \displaystyle \frac{n_A}{n} \left(v_A + \mu_A^2 - \mu^2\right)+\frac{n_B}{n}\left( v_B + \mu_B^2-\mu^2\right)$ (Since, $n_A+n_B = n)$

$\qquad = \displaystyle p[v_A + \mu_A^2 - \mu^2] +(1-p)[v_B +\mu_B^2 - \mu^2]$ (Since, $\frac{n_A}{n} = p)$

$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 + 2\mu\mu_A -2\mu^2] +(1-p)[v_B +(\mu_B - \mu)^2+2\mu\mu_B-2\mu^2]$

$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 ] +(1-p)[v_B +(\mu_B - \mu)^2] + 2p\mu\mu_A + 2(1-p)\mu\mu_B -2\mu^2$

$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 ] +(1-p)[v_B +(\mu_B - \mu)^2] + 2\mu\left[p\mu_A + (1-p)\mu_B\right]-2\mu^2$

$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 ] +(1-p)[v_B +(\mu_B - \mu)^2] + 2\mu^2-2\mu^2$

$\qquad = \displaystyle p[v_A + (\mu_A - \mu)^2 ] +(1-p)[v_B +(\mu_B - \mu)^2]$

So, Correct option C.

by Veteran (431k points)

Say, there are $N_{1}$ elements in set $A$.

So, $\mu _{A}$ will be $\frac{X_{1}+X_{2}+......+X_{N_{1}}}{N_{1}}$

Similarly $\mu _{B}$ will be $\frac{X_{1}+X_{2}+......+X_{N_{2}}}{N_{2}}$

Now, suppose $A\cup B$ has two elements   {$\mu _{A},\mu _{B}$}

Therefore mean of $A\cup B$ is $\frac{\mu _{A}+\mu _{B}}{2}$

Now, if $p=1/2$ then mean of $A\cup B$ is   $p.\mu _{A}+\left ( 1-p \right ).\mu _{B}$

Now, find variance of $A$

$v_{A}=\frac{1}{N_{1}}\sum_{i=1}^{N_{1}}\left [ \left ( X_{1}-\mu _{A} \right ) +\left ( X_{2}-\mu _{A} \right )+\left ( X_{3}-\mu _{A} \right )+\left ( X_{4}-\mu _{A} \right )+..........+\left ( X_{N_{1}}-\mu _{A} \right )\right ]$

and,

$v_{B}=\frac{1}{N_{2}}\sum_{i=1}^{N_{2}}\left [ \left ( X_{1}-\mu _{B} \right ) +\left ( X_{2}-\mu _{B} \right )+\left ( X_{3}-\mu _{B} \right )+\left ( X_{4}-\mu _{B} \right )+..........+\left ( X_{N_{1}}-\mu _{B} \right )\right ]$

Now ,mean of $\left ( A\cup B \right )=\mu$

$Variance\left ( A\cup B \right )=Variance\left ( A-\left ( A\cap B \right ) \right )+Variance\left ( B-\left ( A\cap B \right ) \right )+Variance\left ( A\cap B \right )$

$\Rightarrow Variance\left ( A\cup B \right )=Variance\left ( A\right)+Variance\left ( B \right )+Variance\left ( A\cap B \right )$

$=p.v_{A}+\left ( 1-p \right ).v_{B}+p\left ( \mu _{A}-\mu \right )^{2}+\left ( 1-p \right )\left ( \mu_{B} -\mu \right )^{2}$

Ans will be C)

by Veteran (119k points)
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