To find the mean of the sum of the numbers when two dice are rolled, we can follow these steps:
### Step 1: Identify the possible sums
When two dice are rolled, the possible sums (denoted as \(X\)) range from 2 (1+1) to 12 (6+6). Therefore, the values of \(X\) are:
\[ X = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \]
### Step 2: Calculate the probabilities of each sum
Next, we need to determine the probability of each possible sum:
- \(P(X = 2) = \frac{1}{36}\) (only 1+1)
- \(P(X = 3) = \frac{2}{36} = \frac{1}{18}\) (1+2, 2+1)
- \(P(X = 4) = \frac{3}{36} = \frac{1}{12}\) (1+3, 2+2, 3+1)
- \(P(X = 5) = \frac{4}{36} = \frac{1}{9}\) (1+4, 2+3, 3+2, 4+1)
- \(P(X = 6) = \frac{5}{36}\) (1+5, 2+4, 3+3, 4+2, 5+1)
- \(P(X = 7) = \frac{6}{36} = \frac{1}{6}\) (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)
- \(P(X = 8) = \frac{5}{36}\) (2+6, 3+5, 4+4, 5+3, 6+2)
- \(P(X = 9) = \frac{4}{36} = \frac{1}{9}\) (3+6, 4+5, 5+4, 6+3)
- \(P(X = 10) = \frac{3}{36} = \frac{1}{12}\) (4+6, 5+5, 6+4)
- \(P(X = 11) = \frac{2}{36} = \frac{1}{18}\) (5+6, 6+5)
- \(P(X = 12) = \frac{1}{36}\) (6+6)
### Step 3: Calculate the mean (expected value)
The mean (or expected value) \(E(X)\) is calculated using the formula:
\[
E(X) = \sum_{i=2}^{12} i \cdot P(X = i)
\]
Substituting the values we calculated:
\[
E(X) = 2 \cdot \frac{1}{36} + 3 \cdot \frac{1}{18} + 4 \cdot \frac{1}{12} + 5 \cdot \frac{1}{9} + 6 \cdot \frac{5}{36} + 7 \cdot \frac{1}{6} + 8 \cdot \frac{5}{36} + 9 \cdot \frac{1}{9} + 10 \cdot \frac{1}{12} + 11 \cdot \frac{1}{18} + 12 \cdot \frac{1}{36}
\]
Calculating each term:
- \(2 \cdot \frac{1}{36} = \frac{2}{36}\)
- \(3 \cdot \frac{1}{18} = \frac{3}{18} = \frac{6}{36}\)
- \(4 \cdot \frac{1}{12} = \frac{4}{12} = \frac{12}{36}\)
- \(5 \cdot \frac{1}{9} = \frac{5}{9} = \frac{20}{36}\)
- \(6 \cdot \frac{5}{36} = \frac{30}{36}\)
- \(7 \cdot \frac{1}{6} = \frac{7}{6} = \frac{42}{36}\)
- \(8 \cdot \frac{5}{36} = \frac{40}{36}\)
- \(9 \cdot \frac{1}{9} = 1 = \frac{36}{36}\)
- \(10 \cdot \frac{1}{12} = \frac{10}{12} = \frac{30}{36}\)
- \(11 \cdot \frac{1}{18} = \frac{11}{18} = \frac{22}{36}\)
- \(12 \cdot \frac{1}{36} = \frac{12}{36}\)
Now, summing these fractions:
\[
E(X) = \frac{2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12}{36} = \frac{252}{36} = 7
\]
### Final Answer
Thus, the mean of the sum when two dice are rolled is:
\[
\boxed{7}
\]
