Baseball, the game where a batter tries to hit a pitched ball before getting 3 s

Question

Baseball, the game where a batter tries to hit a pitched ball before getting 3 strikes. The number of earned runs that a team allows to score against them a 9-inning game is called the earned run average (ERA). Suppose that we have a baseball team’s pitching staff that has an ERA of 4.50. Asumming that the team’s pitching performance on any day is normally distributed, and fractions of a run are possible (it’s possible to allow 6.43 runs)….

1) If the standard deviation is 1.5 runs, what is the probability that the team will allow less than 5 runs in a 9-inning game? Between 4 and 6 runs?
2) What does the standard deviation have to be to have an 80% chance of giving up less than 6 runs?
3) Now, suppose the salary paid to the staff is a function of the ERA
salary= (10-ERA) *20 + 6
what is the expected value of the pitching staff salary? What is the standard deviation of the salary?
4) Now, suppose the team’s hitters ave 4.5 runs/game, and this statistic is also normally distributed. Assume runs scored has a standard deviation of 1.5. How often will the team score more than the ave number of runs allowed (which is 4.5)?

Explanation / Answer

1) The z-score for 5 runs is (5-4.5)/1.5=.3333
technically, the z-score for 0 runs is (0-4.5)/1.5=-3
Using a z-table or calculator software, the area under the normal curve from -3 to .333 z-scores is .6292, or about 63%
[**normally we would find the area from negative infinity z-score away from the mean to the z-score of 1/3, but because we can't have negative runs scored, the lowest possible z-score is -3 (Its only off by about .135% anyway.]

2. We need to find the z-score at which 80% of the normal curve is below, so we can use an invNorm function on a TI graphing caluclator (or simply use a z-table) to get a z-score of .8416. Now, if we set this equal to the z-score for 6, which is (6-4.5)/ we can solve for and get

.8416=(6-4.5)/

1.78==standard deviation

3. Salary will be (10-4.5)*20 + 6=116, and because the standard deviation is only affected by multiplication/division, we can multiply the standard devaition given to us in the problem by 20 to get 1.5*20=30.

4. Because both averages are 4.5 and normally distributed, the runner will score more than the average number of runs given up exactly 50% of the time.

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