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

a)

There are 119 games where he did not hit a home run out of 161 games.

Hence,

P(0) = 119/161 = 0.739130435 [ANSWER]

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b)

Note that the probability of x successes is          
          
P(x) = u^x e^(-u) / x!          
          
where          
          
u = the mean number of successes = 47/161 =   0.291925466      
          
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.746824198

As this is close to the answer in part a), YES, IT APPROXIMATELY FOLLOWS A POISSON DISTRIBUTION WITH PARAMETER LAMDA = 47/161.

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