For the following probability distribution determine standard deviation of the random variable X.

X 2 3 4 P(X) 0.2 0.5 0.3

Toolbox:
• $$variance=\sum\;P_iX^2_i\;-\;(\sum\;P_iX_i)^2$$
• $$\sigma\;=\sqrt{variance}$$
$$\sum\;P_iX_i\;=\;2\;\times\;2\;+\;.5\;\times\;3\;+\;.3\;\times\;4$$
=$$.4\;+\;1.5\;+\;1.2$$
=$$3.1$$
$$(\sum\;P_iX_i\;)^2=\;2^2\;\times\;2\;+\;.5\;\times\;3^2\;+\;.3\;\times\;4^2$$
=$$.8+4.5+4.8$$
=$$10.1$$
$$variance\;X\;v(X)=10.1-(3.1)^2$$
=$$10.1-9.61$$
=$$0.49$$
$$\sigma\;=\sqrt{variance}$$
=$$\sqrt{0.49}$$
=$$0.7$$

edited Jun 6, 2013