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Home  >>  CBSE XII  >>  Math  >>  Probability
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A discrete random variable X has the probability distribution given as below:

X 0.5 1 1.5 2 P(X) k $k^2$ $2k^2$ k \begin{array}{1 1}(i)\quad Find\;the\;value\;of\;k \\(ii)\quad Determine\;the\;mean\;of\;the\;distribution\end{array}
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1 Answer

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Toolbox:
  • X-random variable
  • X={\(0.5\;1\;1.5\;2\)}
  • \(\sum^4_{\substack{I=1 }} P(X_I)\)
  • Mean=\(X_1P(x_1)\)+\(X_2P(X_2)\)+\(X_3P(X_3)\)+\(X_4P(X_4)\)
\(p(x_1)\)=K\(p(x_2)=K^2\ p(x_3)=2K^2 p(x_3)\)=K
\(K+K^2+2K^2+K=1\)
\(2K+3K^2=1\)
\(3K^2+2K-1=0\)
\(3R(K+1)-1(K+1)=0\)
\((3K-1)(K+1)=0\)
3K-1=0
3K=1
K=\(\Large\frac{1}{3}\)
\(X_1\)=\(.5\)\(X_2\)=1\(X_3\)=1.5\(X_4\)=2
\(mean=\Large\;\frac{1}{3}\;\times\;.5\;+\;(\frac{1}{3})^2 +2(\frac{1}{3})^2\times1.5+\frac{1}{3}\times2\)
=\(\Large\frac{1}{6}\)+\(\Large\frac{1}{9}\)+\(\Large\frac{1}{3}\)+\(\Large\frac{2}{3}\)
=\(\Large\frac{23}{18}\)

 

answered Feb 26, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 
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