## clairetwe 2 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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1. perl

notice that if you add up the decimals they equal to 1

2. perl

Let X = # of games will last in a best of 7 games contest X is a random variable.

3. clairetwe

What is the mean of the probability distribution

4. carson889

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

i still dont understand ifi add the games is one?

6. perl

carson are you good at probability , there is a question there was disagreement on. would you like to referee the solutions

7. perl

|dw:1367094037874:dw|

8. carson889

Your writting in black would be correct @perl.

9. reemii

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

i checked at one is wrong

11. clairetwe

it*

12. reemii

My computer calculated: 0.0503 + 0.1027 + 0.2499 + 0.5971 = 1.0

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14. reemii

Do you have the formulas for mean and variance?

15. carson889

@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

2.32

17. carson889

Pardon me the three games probability should have been for the five games

18. clairetwe

I'm sorry i'm just still confused on this problem its hard for me

19. perl

|dw:1367094601534:dw|

20. perl

21. carson889

@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

thank you very much!

23. clairetwe

i grateful!

24. carson889

You are very welcome!

25. clairetwe

=)

26. perl