Question

In college basketball, a turnover is defined as losing possession of the basketball to the opposing team. During a recent basketball season, the Villanova Wildcats averaged 13 turnovers per game. Assume the number of turnovers per game follows the Poisson distribution. a. What is the probability that Villanova will have eight turnovers during a game? b. What is the probability that Villanova will have less than 10 turnovers during a game? (I suggest using Excel for this problem.) c. What is the probability that Villanova will have more than 15 turnovers during a game? ( suggest using Excel for this problem.) d. What is the probability that Villanova will have four turnovers during the first half of a game?

Answer 1

be the number of turnovers Villanova Wildcats will have during a game. Villanova Let Wildcats averaged 13 turnovers per game. It is given that number of turnovers per game follows the Poisson distribution and it will have 13. P.M.F. of Poisson distribution is given by ke- Р(X3D k) %3 k 0, 1, 2,. k! Probability that Villanova will have eight turnovers (k 8) during a game a 138e-13 P(X = 8) 815730721 * 0.00000226 40320 0.0457 Note: We can use =POISSON. DIST(x, mean, cumulative) excel function to calculate Poisson probabilities. For P(X 8) it would be POSSON.DIST(8, 13, FALSE) Probability that Villanova will have less than 10 (k < 10) turnovers during a game i.e. 0 or 1 or 2. or 8 or 9 equivalent to k s 9 Р(X < 9) — Р(X — 0) + P(X %3D 1) + ...+ P(X —D 9) b. Note: For cumulative probabilities put cumulative as TRUE i.e. for P(X9) use =POSSON.DIST(9, 13, TRUE) Hence, P(X 9) 0.1658 Probability that Villanova will have more than 15 (k > 16) turnovers during a game Р(X > 15) %3D1— Р(X < 15) C. P(Xs 15) can be calculated by using -POISSON.DIST(15, 13, TRUE) which gives 0.7636 So, P(X 15) 1 - 0.7636 = 0.2364 Probability that Villanova will have four (k = 4) turnovers during the first half of a game For first half of the game the l =-= 6.5 as rate of Poisson is equally spaced on time interval For P(X 4) it would be POSSON.DIST(4, 6.5, FALSE) gives 0.1118 P(X 8) 0.1118