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# Let X be a discrete random variable whose probability distribution is defined as follows:$P(X=x)=\left \{\begin{array}{1 1}k(x+1) & for\;x=1,2,3,4\\2kx & for\;x=5,6,7\\0 & otherwise\end{array} \right.$Where k is a constant.Calculate$(i)\;the\;value\;of\;k\quad(ii)\;E(X)\quad(iii)\;standard\;deviation\;of\;X$

Toolbox:
• since P(x)=k(x+1)forx{$1\;2\;3\;4$}
• =$2\;k\;x\; for\; x=5\;6\;7$
• We have $P(X)=k(1+1)=2k\;p(2)=k(2+1)=3k$
• $\sum$$P_i$=1
• E(X)=$\sum$$P_i$$X_i$
• $\sigma$=$\sqrt{variance}$=$\sqrt{\sum\;P_iX_i-(\sum\;P_iX_i)^2}$
since $\sum$$P_i$=1
$2k+3k+4k+5k+6k+10k+12k+14k=1$
$50k=1$
k=$\Large\frac{1}{50}$
$\sum$$P_i$$X_i$=$1\times2k+2\times3k+3\times4k+4\times5k+5\times10k+6\times12k+7\times14k$
=$\Large\frac{1}{50}$$[2+6+12+50+72+98]$
=$\Large\frac{260}{50}$=$5.2$
$E(X)=5.2$
$\sum\;P_iX_i^2$=$\Large\frac{2}{50}\times1^2\;+\;\frac{3}{50}\times2^2\;+\;\frac{4}{50}\times3^2\;+\;\frac{5}{50}\times4^2\;+\;\frac{10}{50}\times5^2\;+\;\frac{12}{50}\times6^2\;+\;\frac{14}{50}\times49$
=$\Large\frac{1498}{50}$=$29.93$
variance=$\sum\;P_iX_i^2$($\sum\;P_iX_i$)$^2$
=$29,93-(5.2)^2$
=$2.84$
$\sigma$=$\sqrt{variance}$=$\sqrt{2.89}$
=1.7

edited Jun 4, 2013