Power ratings and normal distributions can be used to calculate theprobability t

Question

Power ratings and normal distributions can be used to calculate theprobability that the home team wins an NCAA basketball game. Theaverage amount by which a home team will win a game is givenby

In college basketball the home court advantage is 4 points.

For example, in March 2010 the Jeff Sagarin power ratings for UNCand NC State were 82.34 and 80.50, respectively. So in a gamebetween NC State and UNC played at NC State, the average amount bywhich State wins is

Statisticians have shown that the distribution of the final marginof victory for a home NCAA basketball team can be well-approximatedusing a normal model with mean = (home court advantage + home teampower rating - visiting team power rating) and standard deviation10 points.

Question 1. In a gamebetween NC State and UNC played at NC State, what is theprobability that NC State wins by at least 5 points? (If you usethe Z-table, round your z-score to 2 decimal places).
Use4 decimal places in your answer.

Question 2. In a gamebetween NC State and UNC played at NC State, what is theprobability that UNC wins by at least 4 points? (If you use theZ-table, round your z-score to 2 decimalplaces) Note: ifUNC wins by at least 4 points, then NC State "wins" by -4 points orless.
Use4 decimal places in your answer.

Explanation / Answer

Given =2.16, =10 Question 1: The probability is P(X>5)=P((X-)/ >(5-2.16)/10) =P(Z>0.284) =0.388 (check normal table) Question 2: The probability is P(X

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