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# The probability distribution of a discrete random variable X is given as under:

x 0 2 4 2A 3A 5A P(x) $\Large \frac{1}{2}$ $\Large\frac{1}{5}$ $\Large\frac{3}{25}$ $\Large\frac{1}{10}$ $\Large\frac{1}{25}$ $\Large\frac{1}{25}$ Calculate$\begin{array}{1 1}(i)\;the\;value\;of\;A\;if\;E(X)=2.94\$ii)\;variance \;of \;x\end{array}$ Can you answer this question? ## 1 Answer 0 votes Toolbox: • \(E(X)=\sum\;P_i\;X_i$
• $Also\;variance\;=\sum\;P_i\;X_i\;-(\sum\:P_i\;X_i)^2$
$E(X)=\Large\frac{1}{2}\;\times1+\;\frac{1}{5}\;\times2+\;\frac{3}{25}\;\times4+\;\frac{1}{10}\;\times2A+\;\frac{1}{25}\;\times3A+\;\frac{1}{25}\;\times5A$
$since\; E(X)=\;2.94$
$.5+.4+.48+\Large\frac{1A}{0}+\frac{3A}{25}+\frac{5A}{25}=2.94$
$\Large\frac{26A}{50}=2.94-1.38$
$A=\Large\frac{1.56\times50}{26}$
A=3
$variance\; X=\sum\;P_i\;X_i\;-(\sum\:P_i\;X_i)^2$
$\sum\;P_i\;X_i^2$
=$\Large\frac{1}{2}\;\times1^2\;+\;\frac{1}{5}\;\times2^2\;+\;\frac{3}{5}\;\times4^2\;+\;\frac{1}{10}\;\times6^2\;+\;\frac{1}{25}\;\times9^2\;+\;\frac{1}{25}\;\times15^2$
$\frac{1337}{50}$=$27.69$
$variance=27.69-(2.94)^2$
=$19.05$

answered Mar 3, 2013
edited Jun 4, 2013