Question

Find the Variance and standard deviation of the probability distribution of the random Variable X which can take only the values 1, 2, and 3, given
that P(1)=10/33, P(2)=1/3, and P(3)=12/33​

Answer 1

Find the Variance and standard deviation of the probability distribution of the random Variable X which can take only the values 1, 2, and 3, given

that P(1)=10/33, P(2)=1/3, and P(3)=12/33

Answer 2

The Variance and Standard deviation of the random Variable X  which take  the values 1, 2, and 3  are   2.0606   and 0.8142 respectively .

Step-by-step explanation:

Given:

The probability distribution of the random Variable X  is  following.

 X             1               2           3

P(X)        $\frac{10}{33}$               $\frac{1}{3}$            $\frac{12}{33}$

To Find:  The Variance and Standard deviation of the random Variable X which take  the values 1, 2, and 3.

Formula Used:

   μ= ∑ x.P(x)      

   μ= The mean  of   random variable.

    б^2  =  ∑ X^2 .P(X) -  μ^2    

    б^2  = The variance of random variable.

    б  = $\sqrt{Variance }$

    б =  Standard deviation of random variable.

Solution:

μ= ∑ X .P(X)  $= 1.\frac{10}{33}+2.\frac{1}{3}+3.\frac{12}{33}$  

                   $= \frac{10}{33}+\frac{2}{3} +\frac{36}{33}$

                   $= \frac{10+22+36}{33}$  $=\frac{68}{33}$

                   $=2.0606$

   ∑ X^2 .P(X)$= 1^2.\frac{10}{33}+2^2.\frac{1}{3}+3^2.\frac{12}{33}$

                      $= \frac{10}{33} +\frac{4}{3}+\frac{108}{33}$

                      $=\frac{10+44+108}{33}$

                      $=\frac{162}{33}$

Variance = б^2 =   ∑ X^2 .P(X) -  μ^2                          

                           $=\frac{162}{33} - \ (\frac{68}{33} ) ^{2}$    

                            $=\frac{722}{1089}$

  Standard deviation б $=\sqrt{\frac{722}{1089} }$  

                                        = 0.8142    

Thus,  The Variance and Standard deviation of the random Variable X are  2.0606   and 0.8142 respectively