To find the mean and variance of the random variable \( X \), which represents the sum of the numbers when two dice are rolled, we can follow these steps:
### Step 1: Determine the Possible Values of \( X \)
When rolling two dice, the minimum sum \( X \) can be is 2 (when both dice show 1), and the maximum sum is 12 (when both dice show 6). Therefore, the possible values of \( X \) are \( 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 \).
### Step 2: Calculate the Number of Outcomes for Each Value of \( X \)
Next, we need to calculate the number of ways to achieve each possible sum \( X \):
- \( X = 2 \): 1 way (1,1)
- \( X = 3 \): 2 ways (1,2), (2,1)
- \( X = 4 \): 3 ways (1,3), (2,2), (3,1)
- \( X = 5 \): 4 ways (1,4), (2,3), (3,2), (4,1)
- \( X = 6 \): 5 ways (1,5), (2,4), (3,3), (4,2), (5,1)
- \( X = 7 \): 6 ways (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
- \( X = 8 \): 5 ways (2,6), (3,5), (4,4), (5,3), (6,2)
- \( X = 9 \): 4 ways (3,6), (4,5), (5,4), (6,3)
- \( X = 10 \): 3 ways (4,6), (5,5), (6,4)
- \( X = 11 \): 2 ways (5,6), (6,5)
- \( X = 12 \): 1 way (6,6)
### Step 3: Calculate the Probability for Each Value of \( X \)
Since there are a total of 36 outcomes when rolling two dice, the probabilities \( P(X = x) \) for each sum \( x \) can be calculated as follows:
\[
P(X = x) = \frac{\text{Number of ways to get } x}{36}
\]
### Step 4: Calculate the Mean \( \mu \)
The mean \( \mu \) is calculated using the formula:
\[
\mu = \sum (x_i \cdot P(X = x_i))
\]
Calculating this gives:
\[
\mu = 2 \cdot \frac{1}{36} + 3 \cdot \frac{2}{36} + 4 \cdot \frac{3}{36} + 5 \cdot \frac{4}{36} + 6 \cdot \frac{5}{36} + 7 \cdot \frac{6}{36} + 8 \cdot \frac{5}{36} + 9 \cdot \frac{4}{36} + 10 \cdot \frac{3}{36} + 11 \cdot \frac{2}{36} + 12 \cdot \frac{1}{36}
\]
Calculating each term:
\[
= \frac{2}{36} + \frac{6}{36} + \frac{12}{36} + \frac{20}{36} + \frac{30}{36} + \frac{42}{36} + \frac{40}{36} + \frac{36}{36} + \frac{30}{36} + \frac{22}{36} + \frac{12}{36}
\]
Summing these gives:
\[
= \frac{252}{36} = 7
\]
### Step 5: Calculate the Variance \( \sigma^2 \)
The variance \( \sigma^2 \) is calculated using the formula:
\[
\sigma^2 = \sum (x_i^2 \cdot P(X = x_i)) - \mu^2
\]
Calculating \( \sum (x_i^2 \cdot P(X = x_i)) \):
\[
= 2^2 \cdot \frac{1}{36} + 3^2 \cdot \frac{2}{36} + 4^2 \cdot \frac{3}{36} + 5^2 \cdot \frac{4}{36} + 6^2 \cdot \frac{5}{36} + 7^2 \cdot \frac{6}{36} + 8^2 \cdot \frac{5}{36} + 9^2 \cdot \frac{4}{36} + 10^2 \cdot \frac{3}{36} + 11^2 \cdot \frac{2}{36} + 12^2 \cdot \frac{1}{36}
\]
Calculating each term:
\[
= \frac{4}{36} + \frac{18}{36} + \frac{48}{36} + \frac{100}{36} + \frac{180}{36} + \frac{294}{36} + \frac{320}{36} + \frac{324}{36} + \frac{300}{36} + \frac{242}{36} + \frac{144}{36}
\]
Summing these gives:
\[
= \frac{1974}{36}
\]
Now, substituting back into the variance formula:
\[
\sigma^2 = \frac{1974}{36} - 7^2 = \frac{1974}{36} - 49
\]
Calculating this gives:
\[
= \frac{1974 - 1764}{36} = \frac{210}{36} \approx 5.83
\]
### Final Results
- Mean \( \mu = 7 \)
- Variance \( \sigma^2 \approx 5.83 \)
