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clairetwe Group Title

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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  1. perl Group Title
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    notice that if you add up the decimals they equal to 1

    • one year ago
  2. perl Group Title
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    Let X = # of games will last in a best of 7 games contest X is a random variable.

    • one year ago
  3. clairetwe Group Title
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    What is the mean of the probability distribution

    • one year ago
  4. carson889 Group Title
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    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
  5. clairetwe Group Title
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    i still dont understand ifi add the games is one?

    • one year ago
  6. perl Group Title
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    carson are you good at probability , there is a question there was disagreement on. would you like to referee the solutions

    • one year ago
  7. perl Group Title
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    |dw:1367094037874:dw|

    • one year ago
  8. carson889 Group Title
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    Your writting in black would be correct @perl.

    • one year ago
  9. reemii Group Title
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    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
  10. clairetwe Group Title
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    i checked at one is wrong

    • one year ago
  11. clairetwe Group Title
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    it*

    • one year ago
  12. reemii Group Title
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    My computer calculated: 0.0503 + 0.1027 + 0.2499 + 0.5971 = 1.0

    • one year ago
  13. SnuggieLad Group Title
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    \(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\) \(\Large \color{red}{~\:\:\:\:\:\:\mathbb Don’t\:\:\mathbb Forget\:\:\mathbb To~~\mathbb Fan~~\mathbb And~~\mathbb Best~~\mathbb The~~\mathbb Answers!}\) \(\Huge \color{red}{~\:\:\:\:\:\:\:\:\:\:\:\:\mathbb Welcome\:\:\mathbb To\:\:\mathbb Open\mathbb Study}\) \(\large \color{red}{~\:\:\:\:\:\:\:\:\:\:\:\: ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-\mathbb Snuggie\mathbb Lad }\) \(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)\(\large \color{orange}{*}\)\(\small \color{purple}{*}\)\(\Tiny \color{pink}{*}\)\(\tiny \color{violet}{*}\)\(\Tiny \color{pink}{*}\)\(\small \color{purple}{*}\)\(\large \color{orange}{*}\)\(\Large \color{yellow}{*}\)\(\LARGE \color{green}{*}\)\(\huge \color{blue}{*}\)\(\Huge \color{red}{*}\)\(\huge \color{blue}{*}\)\(\LARGE \color{green}{*}\)\(\Large \color{yellow}{*}\)

    • one year ago
  14. reemii Group Title
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    Do you have the formulas for mean and variance?

    • one year ago
  15. carson889 Group Title
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    @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
  16. clairetwe Group Title
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    2.32

    • one year ago
  17. carson889 Group Title
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    Pardon me the three games probability should have been for the five games

    • one year ago
  18. clairetwe Group Title
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    I'm sorry i'm just still confused on this problem its hard for me

    • one year ago
  19. perl Group Title
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    |dw:1367094601534:dw|

    • one year ago
  20. perl Group Title
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    @carson889 please check your mail

    • one year ago
  21. carson889 Group Title
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    @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
  22. clairetwe Group Title
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    thank you very much!

    • one year ago
  23. clairetwe Group Title
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    i grateful!

    • one year ago
  24. carson889 Group Title
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    You are very welcome!

    • one year ago
  25. clairetwe Group Title
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    =)

    • one year ago
  26. perl Group Title
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    @carson889 please check my answer

    • one year ago
  27. perl Group Title
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    http://openstudy.com/study#/updates/517c3ed0e4b0be6b54aaf02e

    • one year ago
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