Question 73841: When a pair of dice is rolled, the total will range from 2 (1,1) to 12 (6,6). It is a fact that some numbers will occur more frequently than others as the dice are rolled over and over.
Why will some numbers come up more frequently than others?
Each die has six sides numbered from 1 to 6. How many possible ways can a number be rolled? In other words, we can roll (2,3) or (3,2) or (6,1) and so on. What are the total (x,y) outcomes that can occur?
How might you then estimate the percentage of the time a particular number will come up if the dice are rolled over and over?
Once these percentages have been calculated, how might the mean value of the all the numbers thrown be determined?
If you have completed the Discussion Board assignment, you have an idea of what a population distribution is. There is a very famous distribution that describes the frequency of the number of times a number comes up in a series of dice rolls. Use the Library or the Internet to see if you can find its name.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! When a pair of dice is rolled, the total will range from 2 (1,1) to 12 (6,6). It is a fact that some numbers will occur more frequently than others as the dice are rolled over and over.
Why will some numbers come up more frequently than others?
Because combinations of two die sometimes give few occurrences of some numbers
and more frequest occurrence of others.
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Each die has six sides numbered from 1 to 6. How many possible ways can a number be rolled? In other words, we can roll (2,3) or (3,2) or (6,1) and so on.
2: 1,1
3: 1,2 or 2,1
4: 1,3 or 2,4 or 3,1
5: 1,4 or 2,3 or 3,2, or 4,1
6: 1,5 or 2,4 or 3,3 or 4,2 or 5,1
7: 1,6 or 2,5 or 3,4 or 4,3 or 5,2 or 6,1
etc
What are the total (x,y) outcomes that can occur?
6*6 or 36 outcomes
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How might you then estimate the percentage of the time a particular number will come up if the dice are rolled over and over?
Divide the number of times a sum can occur by 36
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Once these percentages have been calculated, how might the mean value of the all the numbers thrown be determined?
Expected value is the sum of the product of each of the sums with the probability that sum will occur.
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Cheers,
Stan H.
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