7. From the following tree, find the probability that a randomly-chosen person w

Question

7. From the following tree, find the probability that a randomly-chosen person will get the 10 Getflu Yes .70 Don't get .90 Received flu vaccine? 40 Get flu .30 No Don't get flu 60 A. .0700 B. .1900 C..8100 D. .7000 8. A carnival has a game of chance: a fair coin is tossed. If it lands heads you win S1.00 and if it lands tails you lose S0.50. How much should a ticket to play this game cost if the carnival wants to break even? A·S.25 B. S.50 C. S.75 D. $1.00 9. The expected value of a random variable X is 140 and the standard deviation is 14. The standard deviation of the random variable Y=3X-10 is: A. 14 B. 6.48 C. 42 D. 32 10. The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. The first quartile for the lengths of brook trout would be A. 16.01 inches B. 11.00 inches C. 11.98 inches D. 10.65 inches $75) = 11. Assume that X is normally distributed with a mean = $64. Given that P(X 0.2981, we can calculate that the standard deviation of X is approximately A. $20.76 B. S13.17 C. $5.83 D. $7.05

Explanation / Answer

7) probability of flu=get vaccine and flu+have not received vaccine and flu =0.7*0.1+0.3*0.4=0.19

option B

8) expected cost =expected win =1*(1/2)-0.5*(1/2)=0.25

option A

9)std deviation of Y =3*SD(X) =3*14=42

option C

10) for first quartile ; zscore =-0.6745

hence corresponding value =mean +z*std deviaiton =14-0.6745*3 =11.98

option C

11) for P(X>75) =0.2981

P(X<75) =1-0.2981

P(X<75)=0.5299

(X-mean)/std deviation =0.5299

std deviation =(75-64)/0.5299=20.76

option A

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