Question #112929

5. Let X represents the number of computers in an Australian household, for those that own a

computer.


a. Find and interpret the expected number of computers in a randomly selected Australian

household. (2 marks)

b. What is the probability that a randomly selected Australian household will have more than 2

computers?

c. Find V (4X +2). (1 mark)

d. Find E(30X + 20]. (1 mark)

Expert's answer

x12345p(x)0.250.330.170.150.10\begin{matrix} x & 1 & 2 & 3 & 4 & 5 \\ p(x) & 0.25 & 0.33 & 0.17 & 0.15 & 0.10 \end{matrix}


a. The expected number of computers in a randomly selected Australian household


E(X)=1(0.25)+2(0.33)+3(0.17)+4(0.15)+5(0.10)=E(X)=1(0.25)+2(0.33)+3(0.17)+4(0.15)+5(0.10)==2.52=2.52

The expexted number of computers in an Australian household, for those that own a computer is

2.522.52 computers.


b. The probability that a randomly selected Australian household will have more than 2 computers is


P(X>2)=0.17+0.15+0.10=0.42P(X>2)=0.17+0.15+0.10=0.42

c.


V(X)=(1−2.52)2(0.25)+(2−2.52)2(0.33)+V(X)=(1-2.52)^2(0.25)+(2-2.52)^2(0.33)++(3−2.52)2(0.17)+(4−2.52)2(0.15)++(3-2.52)^2(0.17)+(4-2.52)^2(0.15)++(5−2.52)2(0.10)=1.6496+(5-2.52)^2(0.10)=1.6496

V(4X+2)=42⋅V(X)=16⋅1.6496=26.3936V(4X+2)=4^2\cdot V(X)=16\cdot1.6496=26.3936

d.


E(30X+20)=30E(X)+20=30⋅2.52+20=95.6E(30X+20)=30E(X)+20=30\cdot2.52+20=95.6


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