During the 2013 season, a baseball player hit a total of 47 home runs in the 161

Question

During the 2013 season, a baseball player hit a total of 47 home runs in the 161 games he played, distributed as follows: he had 3 games during which he hit 2 or more home runs; he had 39 games during which he hit 1 home run, and he had 119 games during which he did not hit a home run.

(a) On a given day that season, what is the probability that the player did not hit a home run?

(b) Does the observed number of home runs per game approximately follow a Poisson distribution with parameter = 47/161?

Explanation / Answer

the distribution table

x f(x) p(x)

2+ 47 47/161

1 39 39/161

0 119 119/161

a) as per the distribution table

the probability that there will be no home run = 119/161 = 0.739

b) no there is no chance of having a poisson distribution because this distributio has 3 variables or 2+,1 and 0 while the poission will be able to show the mean for only one variable.

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