# The probability distribution of a random variable x is given as under:$P(X=x)=\left \{\begin{array}{1 1}kx^2 & for\;x=1,2,3\\2kx & for\;x=4,5,6\\0 & otherwise\end{array}\right.$Where k is a constant.Calculate$(i)\;E(X)\quad(ii)\;E(3X^2)\quad(iii)\;P(X\geq 4)$

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• $$P(X)=kx^2\:1\;,2\;,3$$
• $$2kx\;4\;,5\;,6$$
• $$P(1)=k1^2\:,P(2)\;=k2^2\;,P(3)\;=k3^2\;P(4)=2k.4\;P(5)=2.k.5\;P(6)=2k.6$$
• $$\sum\:p_i=1$$
• $$E(X)=\sum\;p_i\;X_i$$
• $$(3x^2)=\sum\;3\;P_i\;X_i^2=3\sum\;P_i\;X_I^2$$
• $$P(X>4)=P(X=4)+P(X=5)+P(X=6)$$
$$\sum\;P_i\;=1k\;+4k\;+\;9k\;+\:8k\;+\:10k\;+\:12k\:=1$$
44k=1
k=$$\Large\frac{1}{44}$$
$$E(X)=\sum\;p_i\;X_i$$
=$$\Large\frac{1}{44}\times1\;+\;\frac{4}{44}\times2+\;\frac{9}{44}\times3+\;\frac{8}{44}\times4+\;\frac{10}{44}\times5+\;\frac{12}{44}\times6$$
=$$\Large\frac{1}{40}[190]=4.32$$
$$E(x^2)=\Large\frac{1}{44}\times1^2\;+\frac{4}{44}\times2^2\;+\frac{9}{44}\times3^2\;+\frac{8}{44}\times4^2\;+\frac{10}{44}\times5^2\;+\frac{12}{44}\times6^2\;$$
=$$\Large\frac{908}{44}$$
=$$20.63$$
$$E(3x^2)=3\times20.63$$
=$$61.9$$
$$P(x>4)=\Large\frac{8}{44}+\frac{10}{44}+\frac{12}{44}=\frac{30}{44}$$
=$$\Large\frac{15}{22}$$

edited Jun 4, 2013