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A baseball player has a lifetime batting average of 0.349. If each at bat is independent and he comes to bat 6 times in a given game, what is the probability that the following occur? (Round your answers to five decimal places.) (a) he will get 4 hits (b) he will get more than 4 hits

Respuesta :

a) The probability that he will get 4 hits is 0.263.

b) The probability he will get more than 4 hits is 0.271.

The situation is binomial:

  • categorical variable with only 2 categories - hit (success)or no hit (failure)
  • known probability of success - 0.349 batting average
  • fixed number of trials - 4 at bats
  • each at-bat should be independent if the player has more than 70 "at bats."

The random variable is X = # hits in 4 at-bats

X has a binomial distribution with n = 4 & p = 0.349.

We're looking for the probability that X is at least 3: P(X>=3)

You can solve with the formula found on your formula sheet for x = 3 + x = 4 + ... + x = 7. If you have a TI 83 or 84, you can use the function binomcdf:

P(X >=3) = 1 - P(X <= 2) = 1 - binomcdf(n=7, p=0.263, x=2) = 0.271

Learn more about the binomial distribution at

https://brainly.com/question/27999221?referrer=searchResults

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