Question

gambler plays roulette 100 times betting $1 on the numbers 1, 2, and 3 each time. If the ball lands on 1, 2 or 3, the gambler wins $11 and if the ball lands on any of the other numbers, the gambler loses $1. The roulette wheel has 38 slots number 1-36, 0, and 00. 5. Which is the appropriate box model? The box has 38 tickets, 3 marked "11" and 35 marked "-1" The box has 100 tickets, half marked "1" and half marked "0" The box has 100 tickets, some marked "11" and some marked "-1", but the exact percentages are unknown The box has 38 tickets: 1 marked "1", 1 marked "2", 1 marked "3", and 35 marked "1" Submit Answer Tries 0/99 draws replacement b. This corresponds to taking Submit Answer Tries 0/99 (round to 3 decimal places) c. What is the average of the box? Submit Answer Tries 0/99 (round to 3 decimal places) d. What is the SD of the box? Sun Answer Tries 0/99 e. Use the normal approximation and the fact that the EV is about $5 and the SE is about $32 to figure out the chance that the gambler will win more than $11 in 100 plays First, calculate the score: (round to 2 decimal places) Su Are Tries 0/99 (round to the nearest integer) then chance is (use the Z score in the normal table closest to the one calculated above): Sum Answer Tries 0/99

Answer 1

a) The box has 38 tickets, 3 marked 11 and 35 marked -1 6) This corresponds to taking too draws with replacement c) box model: X: 11 - 1 - EX= P(X=11) × 11 + P(x=-1)×(-1) = 11x3 + (-1) 835 = 3 = -0.05263 since we drow 100 times with replacement (er independently) overage drow = = 1/2x100=-5.263 d) SD 2 boxe model: so = Jex- cena = (2x 3 + 352) - (ye = 3.236 o For 100 draws (independently) S.D fu drow = 5100 *3.236 = 32.36 Given u=-5 0=32 h=100 - scone is given by ze xoll = 11-6-21 2=5 then chance is P(EX>10) = P(275) =