Answer to Question #112733 in Statistics and Probability for warsha prasad

Question

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? (1 mark)
c. Find V (4X +2). (1 mark)
d. Find E(30X + 20]. (1 mark)

Answer 1

$P\{X=1\}=0.25\\ P\{X=2\}=0.33\\ P\{X=3\}=0.17\\ P\{X=4\}=0.15\\ P\{X=5\}=0.1\\ a)M(X)=1\cdot (0.25)+2\cdot (0.33)+3\cdot (0.17)+\\ +4\cdot (0.15)+5\cdot (0.1)=2.52.\\ \text{If we randomly select Australian household (among those}\\ \text{that own a computer) we will get } 2.52 \text{ computers on average}.\\ b)P=P\{X=3\}+P\{X=4\}+P\{X=5\}=0.42.\\ c)V(4X+2)=M((4X+2)^2)-(M(4X+2))^2\\ Y=4X+2\\ P\{Y=6\}=0.25\\ P\{Y=10\}=0.33\\ P\{Y=14\}=0.17\\ P\{Y=18\}=0.15\\ P\{Y=22\}=0.1\\ Y^2=(4X+2)^2\\ P\{Y^2=36\}=0.25\\ P\{Y^2=100\}=0.33\\ P\{Y^2=196\}=0.17\\ P\{Y^2=324\}=0.15\\ P\{Y^2=484\}=0.1\\ V(Y)=36(0.25)+100(0.33)+196(0.17)+\\ +324(0.15)+484(0.1)-(6(0.25)+10(0.33)+\\ +14(0.17)+18(0.15)+22(0.1))^2=26.3936.\\ d)Z=30X+20\\ P\{Z=50\}=0.25\\ P\{Z=80\}=0.33\\ P\{Z=110\}=0.17\\ P\{Z=140\}=0.15\\ P\{Z=170\}=0.1\\ E(Z)=50(0.25)+80(0.33)+110(0.17)+\\ +140(0.15)+170(0.1)=95.6$