Construct a probability distribution for the sum shown on the faces when two dice, each with 7 faces, are rolled.  Find the mean

To solve this, let’s go through each part:

1. **Constructing the Probability Distribution**:
   - Since each die has 7 faces, numbered from 1 to 7, the possible sums when two 7-sided dice are rolled range from \(2\) (1+1) to \(14\) (7+7).
   - The number of outcomes for each sum can be calculated by counting the pairs that produce each sum.

2. **Counting Outcomes for Each Sum**:
   - There are \(49\) total outcomes (since each die has 7 faces, and \(7 \times 7 = 49\) combinations).
   - Let's count the occurrences for each sum.

### Probability Distribution Table
Below is the distribution, where each entry shows the possible sum and the number of ways that sum can occur:

| Sum | Occurrences | Probability (Occurrences/49) |
|-----|-------------|------------------------------|
| 2   | 1           | \( \frac{1}{49} \)           |
| 3   | 2           | \( \frac{2}{49} \)           |
| 4   | 3           | \( \frac{3}{49} \)           |
| 5   | 4           | \( \frac{4}{49} \)           |
| 6   | 5           | \( \frac{5}{49} \)           |
| 7   | 6           | \( \frac{6}{49} \)           |
| 8   | 7           | \( \frac{7}{49} \)           |
| 9   | 6           | \( \frac{6}{49} \)           |
| 10  | 5           | \( \frac{5}{49} \)           |
| 11  | 4           | \( \frac{4}{49} \)           |
| 12  | 3           | \( \frac{3}{49} \)           |
| 13  | 2           | \( \frac{2}{49} \)           |
| 14  | 1           | \( \frac{1}{49} \)           |

### Calculating the Mean
The mean \( \mu \) of a probability distribution is calculated by:
\[
\mu = \sum (x \cdot P(x))
\]
where \( x \) is the sum and \( P(x) \) is its probability.

Calculating each product:
\[
\mu = 2 \cdot \frac{1}{49} + 3 \cdot \frac{2}{49} + 4 \cdot \frac{3}{49} + 5 \cdot \frac{4}{49} + 6 \cdot \frac{5}{49} + 7 \cdot \frac{6}{49} + 8 \cdot \frac{7}{49} + 9 \cdot \frac{6}{49} + 10 \cdot \frac{5}{49} + 11 \cdot \frac{4}{49} + 12 \cdot \frac{3}{49} + 13 \cdot \frac{2}{49} + 14 \cdot \frac{1}{49}
\]

Evaluating this gives:
\[
\mu = \frac{2 + 6 + 12 + 20 + 30 + 42 + 56 + 54 + 50 + 44 + 36 + 26 + 14}{49} = \frac{392}{49} = 8
\]

### Conclusion
- **Mean (Expected Value)**: The mean of the sum when two 7-sided dice are rolled is \(8\).

Would you like further details on the calculations? 

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Here are some related questions to explore further:
1. What is the variance of this probability distribution?
2. How would the distribution change if one die had only 6 faces?
3. What is the probability of rolling an odd sum with these dice?
4. How does the mean compare if we used standard 6-sided dice instead?
5. Can we find the mode of this distribution?

**Tip:** Remember, when constructing a probability distribution for dice, consider each unique outcome carefully, as non-standard dice change typical patterns.

Construct a probability distribution for the sum shown on the faces when two dice, each with 7 faces, are rolled. Find the mean.

Explore how to construct the probability distribution for the sum of two 7-sided dice, and learn how to calculate the mean (expected value) of the sum. This detailed solution covers probability concepts and the use of the expected value formula, suitable for students in Grades 10-12.

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