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Home  >>  CBSE XII  >>  Math  >>  Probability
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The probability distribution of a discrete random variable X is given as under:

x 0 2 4 2A 3A 5A P(x) $\Large \frac{1}{2}$ $\Large\frac{1}{5}$ $\Large\frac{3}{25}$ $\Large\frac{1}{10}$ $\Large\frac{1}{25}$ $\Large\frac{1}{25}$ Calculate\[\begin{array}{1 1}(i)\;the\;value\;of\;A\;if\;E(X)=2.94\\(ii)\;variance \;of \;x\end{array}\]  
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1 Answer

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Toolbox:
  • \(E(X)=\sum\;P_i\;X_i\)
  • \(Also\;variance\;=\sum\;P_i\;X_i\;-(\sum\:P_i\;X_i)^2\)
\(E(X)=\Large\frac{1}{2}\;\times1+\;\frac{1}{5}\;\times2+\;\frac{3}{25}\;\times4+\;\frac{1}{10}\;\times2A+\;\frac{1}{25}\;\times3A+\;\frac{1}{25}\;\times5A\)
\(since\; E(X)=\;2.94\)
\(.5+.4+.48+\Large\frac{1A}{0}+\frac{3A}{25}+\frac{5A}{25}=2.94\)
\(\Large\frac{26A}{50}=2.94-1.38\)
\(A=\Large\frac{1.56\times50}{26}\)
A=3
\(variance\; X=\sum\;P_i\;X_i\;-(\sum\:P_i\;X_i)^2\)
\(\sum\;P_i\;X_i^2\)
=\(\Large\frac{1}{2}\;\times1^2\;+\;\frac{1}{5}\;\times2^2\;+\;\frac{3}{5}\;\times4^2\;+\;\frac{1}{10}\;\times6^2\;+\;\frac{1}{25}\;\times9^2\;+\;\frac{1}{25}\;\times15^2\)
\(\frac{1337}{50}\)=\(27.69\)
\(variance=27.69-(2.94)^2\)
=\(19.05\)

 

answered Mar 3, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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