Discussion Forum

A random variable X takes  the values 1,2,3 and 4 such that 2P(X=1)=3P (X=2)=P(X=3)=5P(X=4) . If $\sigma^{2}$  is the variance and $\mu$  is the mean of X . Then,

 $\sigma^{2}+\mu^{2}$=

Answer:

Explanation:

Given that   ,

$\frac{P(x=1)}{\frac{1}{2}}=\frac{P(x=2)}{\frac{1}{2}}=\frac{P(x=3)}{1}=\frac{P(x=4)}{\frac{1}{5}}=k$   (let)

 $\because$   P(x=1)+P(x=2)+P(x=3)+P(x=4)=1

 $\Rightarrow$      $\frac{k}{2}+\frac{k}{3}+k+\frac{k}{5}=1\Rightarrow k=\frac{30}{61}$

 So,  $P(x=1)=p_{1}=\frac{30}{2 \times 61}=\frac{15}{61}$

       $P(x=2)=p_{2}=\frac{30}{3 \times 61}=\frac{10}{61}$

   $P(x=3) =p_{3}= \frac{30}{61}$

 and $P(x=4)=p_{4}= \frac{30}{5 \times 61}= \frac{6}{61}$

 $\because$    $\sigma^{2}=\left(\sum_{i=1}^4 p_{i}x_{i}^{2}\right)-\mu^{2}$

 $\Rightarrow$       $\sigma^{2}+\mu^{2}=\left(\frac{15}{61}\times 1^{2}\right)+\left(\frac{10}{61} \times2^{2}\right)+\left(\frac{30}{61} \times3^{2}\right)+\left(\frac{6}{61} \times4^{2}\right)$

 = $\frac{15+40+270+96}{61}= \frac{421}{61}$