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

 

answered Mar 15, 2013 by poojasapani_1
edited Jun 6, 2013 by poojasapani_1
 

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