Question 1083606: Hello, I'm having problems with these and I need sort of explaination on how to solve them.
I figured out the first one, but I can't figure out the last two. Also, my textbook taught me about combinations, the binomial thereom and how to construct it, and normal distribution. Therefore, I just can't use any method to solve this or it would be incorrect. Any help or a point in the right direction would be much appreciated.
Thank you in advance for your time and help
Michael and Angela run an animal shelter that houses only cats and dogs. The ration of cats to dogs is 5:4. Of the cats, 40% are kittens, while 3/7 of the dogs are puppies. The shelter houses a total of 63 animals.
2.) A group of 8 adult animals is chosen at random from the shelter. What is the most likely number of cats and dogs in this group and what is the probability of that arrangement happening? Give your probability to 4 decimal places.
___ cats and ___ dogs with a probability of _____
3. A group of 10 baby animals is chosen from the shelter. Is the probability distribution for the makeup of this group symmetric or skewed? Justify your answer.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 5C:4D
therefore, because 9 goes into 63 7 times
35 cats and 28 dogs
21 cats are adults, 14 kittens
16 dogs are adults, 12 puppies
If 8 adults are chosen, the number of ways is 37C8
The expected value is 8*21/37=168/37 or 4 20/37 rounding to 5 for cats
and 8*16/37 or 3 17/37 rounding to 3.
the fraction is 21C5*16C3/37C8=0.2952
I will check with 21C4*16C4/37C8=0.2821. It is close but not quite as much.
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For 10 babies, there are 26, 14 kittens and 12 puppies
The probability is skewed slightly, because the kittens are a few more than the puppies. They would have to be equal to have a symmetric probability distribution.
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