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?

Solution Preview

Hello!
Here are your answers.

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.

... when he set the record for the most home runs in a ... is a huge skew towards the left of the distribution. ... This is a question of Binomial probability, with n = 20 ...

... 3 During the month of July, a home improvement store sold a ... at random, find the probability that This is a binomial probability distribution, N=15 ...

... in what time interval would you have run the race ... 9. Total homes: 50 Homes with both skylights and attached ... 3. f) The variance of the binomial distribution is np ...

... The probability mass function binomial variable is given by P ... 7. Keyes Home Furnishings ran six local newspaper ... The following frequency distribution resulted: ...

... 0:32), right=F, freq=F, + main="Binomial distribution, p=0.2 ... contributions, each with the same distribution of errors ... Fits many observed distributions of errors ...

... Are the data from a binomial distribution? ... the distribution of the data follows a Weibull distribution. ... fit test for at least some of the common distributions. ...

... was done regarding the number of home runs scored by ... b) Make a histogram for the r probability distribution. ... 3.2498=3.25 Or since r is binomial random variable ...

... ing three discrete probability distributions: the binomial, the hypergeometric ... There is not one t distribution, but rather a family of t distributions. ...

... probability distributions: the binomial, the hypergeometric ... We use probability distributions to evaluate ... F. Yates published the normal probability distribution. ...