Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

In roulette, the bet on a “split” pays 17 to 1 and there are 2 chances in 38 to win. The bet on “red” pays 1 to 1 and there are 18 chances in 38 to win. Compare the following two strategies: A: bet $1 200 times (independently) on a split B: bet $1 200 times independently on red In what follows, “making more than $x” means having a net gain of more than $x; “losing more than $x” means having a net gain of less than -$x. Pick all that are correct. The chance of coming out ahead is greater with strategy A than with strategy B The chance of making more than $20 is greater with strategy A than with strategy B. The chance of losing more than $20 is greater with strategy A than with strategy B.

OpenStudy (agent0smith):

I wrote this earlier when solving this, and it'd take a while to explain it all, so interpret it if you can... A: bet $1 200 times (independently) on a split E(x) = 17(2/38) -1(36/38) = -1/19 variance = (2/38)(17+1/19)^2 + (36/38)(-1+1/19)^2 = 5832/361 Population mean = -200/19 SD = sqrt(200*5832/361) = 56.842 P(x > 20.5) = 0.291 z = (20.5 + 200/19)/56.842 = 0.5458 P(x < -20.5) = 0.429 z = (-20.5 + 200/19)/56.842 = -0.1755 P(x>0.5) = 0.425 z = (0.5 + 200/19)/56.842 = 0.194 B: bet $1 200 times independently on red E(x) = 1(18/38) – 1(20/38) = -1/19 variance = (18/38)(1+1/19)^2 + (20/38)(-1+1/19)^2 = 360/361 Population Mean = -200/19 SD = sqrt(200*360/361) = 14.122 P(x > 20.5) = 0.014 z = (20.5 + 200/19)/14.122 = 2.197 P(x < -20.5) = 0.239 z = (-20.5 + 200/19)/14.122 = -0.706 P(x>0.5) = 0.218 z = (0.5 + 200/19)/14.122 = 0.78

OpenStudy (anonymous):

the interpretation see, confusing, help me out with it

OpenStudy (agent0smith):

You first have to find the expected value E(x) for each. Then find the variance. Then find the mean for 200 times, by multiplying the E(x) by 200. Get the standard deviation for 200 times, by multiplying the variance by 200, then take the square root of that. Then you find the z-score and use that to find the probability.

OpenStudy (anonymous):

I have also some problems... When i selected Answers "A and C" together - was Incorect :-(

OpenStudy (agent0smith):

@Oles... it's not answers A and C.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!