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For the following probability distribution:
X 1 2 3 4 P(X) $\Large\frac{1}{10}$ $\Large\frac{1}{5}$ $\Large\frac{3}{10}$ $\Large\frac{2}{5}$ $E(X^2)$ is equal to
\[(A)\;3\quad(B)\;5\quad(C)\;7\quad(D)\;10\]
cbse
class12
ch13
q89
p284
objective
exemplar
sec-a
easy
math
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asked
Jan 10, 2013
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sreemathi.v
retagged
Apr 8, 2014
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For a given probability distribution with random variable $X$ taking values $x_{i},i=0,1,\dots$ and $p_{i}$ the corresponding probabilities.
$E(X^{2})=\sum p_{i}x^{2}_{2}.$
$ \begin{array} {llllllll} \textbf{X:}& 1& 2& 3& 4 \\ \textbf{P(X):}& \large\frac{1}{10}&\large \frac{1}{5} &\large\frac{3}{10}& \large\frac{2}{5} \end{array}$
$E(X^{2})=\Large\sum p_{i}x_{i}^{2}=\frac{1}{10}\times 1+\frac{1}{5}\times 2^{2}+\frac{3}{10}\times 3^{2}+\frac{2}{5}\times 4^{2}$
$\Large\frac{1+8+27+64}{10}$
$=\Large\frac{100}{10}$
=$10$
$'D'$ option is correct
answered
Jun 10, 2013
by
poojasapani_1
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