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anonymous
 3 years 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.
anonymous
 3 years 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.

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perl
 3 years ago
Best ResponseYou've already chosen the best response.0notice that if you add up the decimals they equal to 1

perl
 3 years ago
Best ResponseYou've already chosen the best response.0Let X = # of games will last in a best of 7 games contest X is a random variable.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What is the mean of the probability distribution

carson889
 3 years ago
Best ResponseYou've already chosen the best response.1The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i still dont understand ifi add the games is one?

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

carson889
 3 years ago
Best ResponseYou've already chosen the best response.1Your writting in black would be correct @perl.

reemii
 3 years ago
Best ResponseYou've already chosen the best response.0If 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i checked at one is wrong

reemii
 3 years ago
Best ResponseYou've already chosen the best response.0My computer calculated: 0.0503 + 0.1027 + 0.2499 + 0.5971 = 1.0

SnuggieLad
 3 years ago
Best ResponseYou've already chosen the best response.1\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\) \(\Large \color{red}{~\:\:\:\:\:\:\mathbb Don’t\:\:\mathbb Forget\:\:\mathbb To~~\mathbb Fan~~\mathbb And~~\mathbb Best~~\mathbb The~~\mathbb Answers!}\) \(\Huge \color{red}{~\:\:\:\:\:\:\:\:\:\:\:\:\mathbb Welcome\:\:\mathbb To\:\:\mathbb Open\mathbb Study}\) \(\large \color{red}{~\:\:\:\:\:\:\:\:\:\:\:\: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\mathbb Snuggie\mathbb Lad }\) \(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)

reemii
 3 years ago
Best ResponseYou've already chosen the best response.0Do you have the formulas for mean and variance?

carson889
 3 years ago
Best 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.

carson889
 3 years ago
Best ResponseYou've already chosen the best response.1Pardon me the three games probability should have been for the five games

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm sorry i'm just still confused on this problem its hard for me

perl
 3 years ago
Best ResponseYou've already chosen the best response.0@carson889 please check your mail

carson889
 3 years ago
Best 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.

carson889
 3 years ago
Best ResponseYou've already chosen the best response.1You are very welcome!

perl
 3 years ago
Best ResponseYou've already chosen the best response.0@carson889 please check my answer

perl
 3 years ago
Best ResponseYou've already chosen the best response.0http://openstudy.com/study#/updates/517c3ed0e4b0be6b54aaf02e
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