# 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