Question

Let S={1,2,3,4,5,6,7,8} be a sample space withP(x)=k2x where x is a member of S,and k is a positive constant. Compute E(S). Round your answer to the nearest hundredths.

Answer 1

Given:

`\begin{gathered} S=\lbrace1,2,3,4,5,6,7,8\rbrace \\ P(x)=k^2x \end{gathered}`
`\begin{gathered} Total\text{ outcome}=1+2+3+4+5+6+7+8 \\ Total\text{ outcome}=36 \end{gathered}`
`\begin{gathered} P(x)=k^2x \\ \\ P(1)=k^2*1 \\ P(1)=k^2 \end{gathered}`

`\begin{gathered} P(x)=k^2x \\ \\ P(2)=k^2*2 \\ P(2)=2k^2 \end{gathered}`

`\begin{gathered} P(x)=k^2x \\ \\ P(3)=k^2*3 \\ P(3)=3k^2 \end{gathered}`

Repeating the procedure, other values of P(x) are;

`\begin{gathered} P(4)=4k^2 \\ P(5)=5k^2 \\ P(6)=6k^2 \\ P(7)=7k^2 \\ P(8)=8k^2 \end{gathered}`

Recall the probability must add up to 1.

Hence;

`\begin{gathered} k^2+2k^2+3k^2+4k^2+5k^2+6k^2+7k^2+8k^2=1 \\ 36k^2=1 \\ k^2=(1)/(36) \\ k=\sqrt{(1)/(36)} \\ k=(1)/(6) \end{gathered}`

Hence,

`P(x)=(1)/(36)x`

The expected value E(x) is;

`\begin{gathered} E(S)=xP(x) \\ E(S)=x.((x)/(36)) \\ E(S)=1((1)/(36))+2((2)/(36))+3((3)/(36))+4((4)/(36))+5((5)/(36))+6((6)/(36))+7((7)/(36))+8((8)/(36)) \\ E(S)=(1)/(36)+(4)/(36)+(9)/(36)+(16)/(36)+(25)/(36)+(36)/(36)+(49)/(36)+(64)/(36) \\ E(S)=(204)/(36) \\ E(S)=(17)/(3) \\ E(S)=5.6667 \\ \\ To\text{ the nearest hundredth,} \\ E(S)=5.67 \end{gathered}`

Therefore, to the nearest hundredth, E(S) = 5.67