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Home  >>  CBSE XII  >>  Math  >>  Probability
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Consider the probability distribution of a random variable X.

X 0 1 2 3 4 p(X) 0.1 0.25 0.3 0.2 0.15   Calculate $(i)\;V\bigg(\Large \frac{X}{2}\normalsize \bigg)\quad (ii)\;variance\;of\;\;X$
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1 Answer

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Toolbox:
  • define y=\(\frac{x}{2}\)
  • x=\((0\;1\;2\;3\;4)\)
  • y=\(\frac{x}{2}\)=\((0\;0.5\;1\;1.5\;2)\)
  • P(X)=\((.1\;.25\;.3\;.2\;.15)\)
  • \(\vee\)(\(\large\frac{x}{2}\))=\(\vee\)(y)=\(\sum\)\(P_i\)\(Y^2_i\)_(\(\sum\)\(P_i\)\(Y_i\))\(^2\)
  • \(\vee\)(X)=\(\sum\)\(P_i\)\(X^2_i\)_(\(\sum\)\(P_i\)\(X_i\))\(^2\)
\(\sum\)\(P_i\)\(Y_i\)=\(0\times.1+\large\frac{0.25}{2}\)+1\(\times\).3+\(\large\frac{3}{2}\times.2+2\times.15\)
=\(0+0.125+.3+0.3+0.3\)
=\(1.025\)
=\(\sum\)\(P_i\)\(Y_i\)\(^2\)=\((0\times.1+(\large\frac{1}{2})^2\)\(\times\).25+1\(^2\)\(\times\).3+\((\large\frac{3}{2})^2\)\(\times\).2=2\(^2\)\(\times0.15\)
=\(0.0625+0.3+0.45+.6\)
=\(1.976\)
var(Y)=var(\(\large\frac{x}{2}\))=\(1.976\)-(\(1.025\))\(^2\)
=\(0.1118\)
\(\sum\)\(P_i\)\(x_i\)=\(0\times.1+1\times2.5+2\times.3+3\times.2+4\times1.5\)
=\(2.05\)
=\(\sum\)\(P_i\)\(x_i\)\(^2\)=\(0\times.1+1{^2}\times.25+2{^2}\times.3+3{^2}\times.2+4{^2}\times.15\)
=.25+1.2+1.8+2.4
=5.65
var(X)=5.65-(2.05)\(^2\)
=5.65-4.1025
=0.4475

 

answered Feb 27, 2013 by poojasapani_1
edited Jun 6, 2013 by poojasapani_1
 

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