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clairetwe

  • 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.

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  1. perl
    • one year ago
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    notice that if you add up the decimals they equal to 1

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

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

  4. carson889
    • one year ago
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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.

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

  6. perl
    • one year ago
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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

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

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

  9. reemii
    • one year ago
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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)

  10. clairetwe
    • one year ago
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    i checked at one is wrong

  11. clairetwe
    • one year ago
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    it*

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

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

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

  15. carson889
    • one year ago
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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.

  16. clairetwe
    • one year ago
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    2.32

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

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

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

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

  21. carson889
    • one year ago
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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.

  22. clairetwe
    • one year ago
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    thank you very much!

  23. clairetwe
    • one year ago
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    i grateful!

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

  25. clairetwe
    • one year ago
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    =)

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

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

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