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# 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
Can you answer this question?

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

answered Mar 15, 2013
edited Jun 6, 2013