The probability distribution of a random variable $X$  is given below

 

 

 

Question

Find  variance
 

we know that sum of all probabilities is = 1

$\sum P\left(xi\right)=1$

$P\left(1\right)+P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)=1$

$k+2k+3k+4k+5k=1$

$15\mathrm{k}=1$

$k=\frac{1}{15}$

variance= ${\sigma }^2=\sum x_i^{2}p\left(x_i\right)-{\mu }^2$

${\sigma }^2=\left(1{)}^2\right(k)+(2{)}^2\left(2k\right)+\left(3{)}^2\right(3k)+(4{)}^2\left(4k\right)+\left(5{)}^2\right(5k)-{\left(\frac{11}{3}\right)}^2$

${\sigma }^2=\left(1{)}^2\right(k)+(4\left)\right(2k)+(9\left)\right(3k)+(16\left)\right(4k)+(25\left)\right(5k)-\frac{121}{9}$

$=225k-\frac{121}{9}\left(\because k=\frac{1}{15}\right)$

$=225k\left(\frac{1}{15}\right)-\frac{121}{9}=15-\frac{121}{9}$

$=\frac{\left(15\right)\left(9\right)-121}{9}=\frac{135-121}{9}=\frac{14}{9}$