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Home  >>  CBSE XII  >>  Math  >>  Probability
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The probability distribution of a random variable x is given as under:\[P(X=x)=\left \{\begin{array}{1 1}kx^2 & for\;x=1,2,3\\2kx & for\;x=4,5,6\\0 & otherwise\end{array}\right.\]Where k is a constant.Calculate\[(i)\;E(X)\quad(ii)\;E(3X^2)\quad(iii)\;P(X\geq 4)\]

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  • \(P(X)=kx^2\:1\;,2\;,3\)
  • \(2kx\;4\;,5\;,6\)
  • \(P(1)=k1^2\:,P(2)\;=k2^2\;,P(3)\;=k3^2\;P(4)=2k.4\;P(5)=2.k.5\;P(6)=2k.6\)
  • \(\sum\:p_i=1\)
  • \(E(X)=\sum\;p_i\;X_i\)
  • \((3x^2)=\sum\;3\;P_i\;X_i^2=3\sum\;P_i\;X_I^2\)
  • \(P(X>4)=P(X=4)+P(X=5)+P(X=6)\)
\(\sum\;P_i\;=1k\;+4k\;+\;9k\;+\:8k\;+\:10k\;+\:12k\:=1\)
44k=1
k=\(\Large\frac{1}{44}\)
\(E(X)=\sum\;p_i\;X_i\)
=\(\Large\frac{1}{44}\times1\;+\;\frac{4}{44}\times2+\;\frac{9}{44}\times3+\;\frac{8}{44}\times4+\;\frac{10}{44}\times5+\;\frac{12}{44}\times6\)
=\(\Large\frac{1}{40}[190]=4.32\)
\(E(x^2)=\Large\frac{1}{44}\times1^2\;+\frac{4}{44}\times2^2\;+\frac{9}{44}\times3^2\;+\frac{8}{44}\times4^2\;+\frac{10}{44}\times5^2\;+\frac{12}{44}\times6^2\;\)
=\(\Large\frac{908}{44}\)
=\(20.63\)
\(E(3x^2)=3\times20.63\)
=\(61.9\)
\(P(x>4)=\Large\frac{8}{44}+\frac{10}{44}+\frac{12}{44}=\frac{30}{44}\)
=\(\Large\frac{15}{22}\)

 

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

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