In 1998, Mark McGwire of the St. Louis cardinals hit 70 home runs, a new Major League record. Was this feat as surprising as most of us thought/ In the three seasons before 1998, McGwire hit a home run in 11.6% of his times at bat. He went to bat 509 times in 1998. If he continues his past performance, McQwire's home run count in 509 times at bat has approximately the binomial distribution with n = 509 and p = 0.116.

a) What is the mean number of home runs McQwire will hit in 509 times at bat?
b) What is the probability that he hits 70 or more home runs?
c) How does the probability that p falls within 0.01 of the true p change as p gets closer to 0?

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Part A
The mean of a binomial distribution with parameters n and p is calculated as n*p. In this case, we have that n = 509 and that p = 0.116. Therefore, McGwire will hit an average of 509*0.116 = 59.044 home runs in 509 times at the bat.

Part B
In order to answer this question, we can either use a binomial probability calculator, or use the normal approximation to the binomial distribution. I'll use the latter, as it only requires the standard normal probability table, which can be found in most Statistics books.

This ...

Solution Summary

The solution uses binomial distribution to determine the likelihood of a record setting number of home runs in a season for Mark McGwire.

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