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Home  >>  CBSE XII  >>  Math  >>  Probability
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Let X be a discrete random variable whose probability distribution is defined as follows:\[P(X=x)=\left \{\begin{array}{1 1}k(x+1) & for\;x=1,2,3,4\\2kx & for\;x=5,6,7\\0 & otherwise\end{array} \right.\]Where k is a constant.Calculate\[(i)\;the\;value\;of\;k\quad(ii)\;E(X)\quad(iii)\;standard\;deviation\;of\;X\]

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  • since P(x)=k(x+1)forx{\(1\;2\;3\;4\)}
  • =\(2\;k\;x\; for\; x=5\;6\;7\)
  • We have \(P(X)=k(1+1)=2k\;p(2)=k(2+1)=3k\)
  • \(\sum\)\(P_i\)=1
  • E(X)=\(\sum\)\(P_i\)\(X_i\)
  • \(\sigma\)=\(\sqrt{variance}\)=\(\sqrt{\sum\;P_iX_i-(\sum\;P_iX_i)^2}\)
since \(\sum\)\(P_i\)=1
\(2k+3k+4k+5k+6k+10k+12k+14k=1\)
\(50k=1\)
k=\(\Large\frac{1}{50}\)
\(\sum\)\(P_i\)\(X_i\)=\(1\times2k+2\times3k+3\times4k+4\times5k+5\times10k+6\times12k+7\times14k\)
=\(\Large\frac{1}{50}\)\([2+6+12+50+72+98]\)
=\(\Large\frac{260}{50}\)=\(5.2\)
\(E(X)=5.2\)
\(\sum\;P_iX_i^2\)=\(\Large\frac{2}{50}\times1^2\;+\;\frac{3}{50}\times2^2\;+\;\frac{4}{50}\times3^2\;+\;\frac{5}{50}\times4^2\;+\;\frac{10}{50}\times5^2\;+\;\frac{12}{50}\times6^2\;+\;\frac{14}{50}\times49\)
=\(\Large\frac{1498}{50}\)=\(29.93\)
variance=\(\sum\;P_iX_i^2\)(\(\sum\;P_iX_i\))\(^2\)
=\(29,93-(5.2)^2\)
=\(2.84\)
\(\sigma\)=\(\sqrt{variance}\)=\(\sqrt{2.89}\)
=1.7

 

answered Mar 2, 2013 by poojasapani_1
edited Jun 4, 2013 by poojasapani_1
 

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