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there is a 0.0503 probability that a best of seven contest that will last four games, a 0.1027 probability that will last five games a 0.2499 probability that it will last six games and a 0.5971 probability that it will last seven games. verify that this is a probability distribution. find its mean and standard deviation.
 one year ago
 one year ago
there is a 0.0503 probability that a best of seven contest that will last four games, a 0.1027 probability that will last five games a 0.2499 probability that it will last six games and a 0.5971 probability that it will last seven games. verify that this is a probability distribution. find its mean and standard deviation.
 one year ago
 one year ago

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perlBest ResponseYou've already chosen the best response.0
notice that if you add up the decimals they equal to 1
 one year ago

perlBest ResponseYou've already chosen the best response.0
Let X = # of games will last in a best of 7 games contest X is a random variable.
 one year ago

clairetweBest ResponseYou've already chosen the best response.0
What is the mean of the probability distribution
 one year ago

carson889Best ResponseYou've already chosen the best response.1
The mean is \[\mu = \sum_{x}^{n} x\times f(x)\] Where x is the number of games and f(x) is the probability that it will last x number of games.
 one year ago

clairetweBest ResponseYou've already chosen the best response.0
i still dont understand ifi add the games is one?
 one year ago

perlBest ResponseYou've already chosen the best response.0
carson are you good at probability , there is a question there was disagreement on. would you like to referee the solutions
 one year ago

carson889Best ResponseYou've already chosen the best response.1
Your writting in black would be correct @perl.
 one year ago

reemiiBest ResponseYou've already chosen the best response.0
If this sum \(0.0503 + 0.1027 + 0.2499 + 0.5971\) equals 1, then you got a probability distribution. (i checked and it does equal 1)
 one year ago

clairetweBest ResponseYou've already chosen the best response.0
i checked at one is wrong
 one year ago

reemiiBest ResponseYou've already chosen the best response.0
My computer calculated: 0.0503 + 0.1027 + 0.2499 + 0.5971 = 1.0
 one year ago

SnuggieLadBest ResponseYou've already chosen the best response.1
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 one year ago

reemiiBest ResponseYou've already chosen the best response.0
Do you have the formulas for mean and variance?
 one year ago

carson889Best ResponseYou've already chosen the best response.1
@clairetwe The first thing to know is that the probability of an event needs to add up to one. Thus as perl mentioned before if we add all the probabilities they would sum to one. Since reemii proved this we can move on. As I stated before the mean of the distribution is equal to the sum of the number of games the contest lasted multiplied by the probability that a game would last that number of games. So for four games: 4 * 0.0503, three games: 3 * 0.1027, and so on. Try to calculate that and if you are still having trouble let me know.
 one year ago

carson889Best ResponseYou've already chosen the best response.1
Pardon me the three games probability should have been for the five games
 one year ago

clairetweBest ResponseYou've already chosen the best response.0
I'm sorry i'm just still confused on this problem its hard for me
 one year ago

perlBest ResponseYou've already chosen the best response.0
@carson889 please check your mail
 one year ago

carson889Best ResponseYou've already chosen the best response.1
@clairetwe Are you familiar with the summation symbol, Sigma, \[\sum_{x}^{n}\] starting at x and summing until n. ie: \[\sum_{x=0}^{5}x\], x starts at 0 and sums till 5: 0 + 1 + 2 + 3 + 4 + 5 = 15. If not, you must understand it to calculate the mean and variance and I can help with that. Mean: \[\mu= \sum_{n}^{x}(x_{n} \times f(x_{n}))\] = (4 * 0.0503) + (5 * 0.1027) + (6 * 0.2499) + (7 * 0.5971) = 6.3938. I'm sort of giving yuou a freebie by answering this one. My hopes are that you will be able to understand the summation equation by seeing it written out. In addition, the variance relies on the mean, so hopefully by giving you the mean you can now calculate the variance and thus the standard deviations. Variance: \[\sigma^{2}=\sum_{x=4}^{n}(x\mu)^{2}f(x)\] Now the standard deviation is just the square root of the variance.
 one year ago

clairetweBest ResponseYou've already chosen the best response.0
thank you very much!
 one year ago

carson889Best ResponseYou've already chosen the best response.1
You are very welcome!
 one year ago

perlBest ResponseYou've already chosen the best response.0
@carson889 please check my answer
 one year ago

perlBest ResponseYou've already chosen the best response.0
http://openstudy.com/study#/updates/517c3ed0e4b0be6b54aaf02e
 one year ago
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