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Home  >>  CBSE XII  >>  Math  >>  Probability
0 votes

For the following probability distribution:

X -4 -3 -2 -1 0 P(X) 0.1 0.2 0.3 0.2 0.2 E(X) is equal to $(A)\;0\quad(B)\;-1\quad(C)\;-2\quad(D)\;-1.8$
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1 Answer

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Toolbox:
  • For a given probability distribution.
  • With random variable $X$ taking value $x_{i}\; i=1,2,\dots$
  • $E(X)=\sum p_{i}x_{i}$
  • and $p_{i}$ are the corresponding probability.
$ \begin{array} {llllllll} \textbf{X:}& -4& -3& -2& -1& 0 \\ \textbf{P(X):}& 0.1& 0.2 &0.3& 0.2& 0.2 \end{array}$
$E(X)=(-4)(0.1)+(-3)(0.2)+(-2)(0.3)+(-1)(0.2)+0(0.2)$
$=-0.4-0.6-0.6-0.2-0$
$=-1.8$
$'D'$ option is correct.
answered Jun 10, 2013 by poojasapani_1
 

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