Find i)k ii) mean iii) variance.

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 05

Year : DEC 2015

We know that total probability is 1.

$$\therefore \sum p=1$$

$\therefore 0.1+k+0.2+2k+0.3+k=1\\ \therefore 4k=1-0.6\\ \therefore k=0.1$

Now the probability distribution is

Now mean $=E(x)=\sum pixi\\ =0.1 \times(-2)+0.1\times(-1)+0.2\times0+ 0.2 \times1+ 0.3 \times 2+0.1 \times3\\ =-0.2-0.1+0.2+0.6+0.3\\ E(x)=0.8\\ E(x^2)=\sum pi(xi)^2\\E(x^2)=0.1 \times (-2)^2+0.1\times (-1)^2 + 0.2\times(0)^2 +(0.2)\times1^2+0.3 \times2^2+0.1\times(3^2)\\ E(x^2)=0.4+0.1+0.2+1.2+0.9\\ E(x^2)=2.8 $

Variance $=E(x^2)-[E(x)]^2 \\ =2.8-0.8^2 \\ =2.8-0.64 $

Variance $=2.16$