xp(x)10.2520.3330.1740.1550.10
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)==2.52The expexted number of computers in an Australian household, for those that own a computer is
2.52 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=0.42
c.
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)++(5−2.52)2(0.10)=1.6496
V(4X+2)=42⋅V(X)=16⋅1.6496=26.3936
d.
E(30X+20)=30E(X)+20=30⋅2.52+20=95.6
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