Question

Find the mean variance and standard deviation of the following probability distribution

xi :	a	b
pi :	p	q

where p + q = 1

Answer 1

xi	pi	pixi	pixi2
a	p	ap	a2p
b	q	bq	b2q
		${\sum }_{}^{}$pixi = ap + bq	${\sum }_{}^{}$pixi2= a2p + b2q

$$\begin{gathered} \mathrm{Now}, \\ \mathrm{Mean}=\sum p_i{x}_i=ap+bq \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2=a^{2}p+b^{2}q-{\left(ap+bq\right)}^2 \\ =a^{2}p+b^{2}q-a^2{p}^2-b^2{q}^2-2abpq \\ =a^{2}p-a^2{p}^2+b^{2}q-b^2{q}^2-2abpq \\ =a^{2}p\left(1-p\right)+b^{2}q\left(1-q\right)-2abpq \\ =a^{2}pq+b^{2}qp-2abpq\left(\because p+q=1\right) \\ =pq\left(a^2+b^2-2ab\right) \\ =pq{\left(a-b\right)}^2 \\ \mathrm{Step}\mathrm{Deviation}=\sqrt{\mathrm{Variance}} \\ =\sqrt{pq{\left(a-b\right)}^2} \\ =\left|a-b\right|\sqrt{pq} \end{gathered}$$