# 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}$ ## 1 Answer 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$$

edited Jun 4, 2013