Solution:
Given:
[tex]\begin{gathered} S=\lbrace1,2,3,4,5,6,7,8\rbrace \\ P(x)=k^2x \end{gathered}[/tex][tex]\begin{gathered} Total\text{ outcome}=1+2+3+4+5+6+7+8 \\ Total\text{ outcome}=36 \end{gathered}[/tex][tex]\begin{gathered} P(x)=k^2x \\ \\ P(1)=k^2\times1 \\ P(1)=k^2 \end{gathered}[/tex][tex]\begin{gathered} P(x)=k^2x \\ \\ P(2)=k^2\times2 \\ P(2)=2k^2 \end{gathered}[/tex][tex]\begin{gathered} P(x)=k^2x \\ \\ P(3)=k^2\times3 \\ P(3)=3k^2 \end{gathered}[/tex]Repeating the procedure, other values of P(x) are;
[tex]\begin{gathered} P(4)=4k^2 \\ P(5)=5k^2 \\ P(6)=6k^2 \\ P(7)=7k^2 \\ P(8)=8k^2 \end{gathered}[/tex]Recall the probability must add up to 1.
Hence;
[tex]\begin{gathered} k^2+2k^2+3k^2+4k^2+5k^2+6k^2+7k^2+8k^2=1 \\ 36k^2=1 \\ k^2=\frac{1}{36} \\ k=\sqrt{\frac{1}{36}} \\ k=\frac{1}{6} \end{gathered}[/tex]Hence,
[tex]P(x)=\frac{1}{36}x[/tex]The expected value E(x) is;
[tex]\begin{gathered} E(S)=xP(x) \\ E(S)=x.(\frac{x}{36}) \\ E(S)=1(\frac{1}{36})+2(\frac{2}{36})+3(\frac{3}{36})+4(\frac{4}{36})+5(\frac{5}{36})+6(\frac{6}{36})+7(\frac{7}{36})+8(\frac{8}{36}) \\ E(S)=\frac{1}{36}+\frac{4}{36}+\frac{9}{36}+\frac{16}{36}+\frac{25}{36}+\frac{36}{36}+\frac{49}{36}+\frac{64}{36} \\ E(S)=\frac{204}{36} \\ E(S)=\frac{17}{3} \\ E(S)=5.6667 \\ \\ To\text{ the nearest hundredth,} \\ E(S)=5.67 \end{gathered}[/tex]Therefore, to the nearest hundredth, E(S) = 5.67
