\( \alpha - \beta \) | Case | \( P \) |
---|---|---|
5 | (6, 1) | \(\frac{1}{36}\) |
4 | (6, 2), (5, 1) | \(\frac{2}{36}\) |
3 | (6, 3), (5, 2), (4, 1) | \(\frac{3}{36}\) |
2 | (6, 4), (5, 3), (4, 3), (3, 1) | \(\frac{4}{36}\) |
1 | (6, 5), (5, 4), (4, 3), (3, 2), (2, 1) | \(\frac{5}{36}\) |
0 | (6, 6), (5, 5), ..., (1, 1) | \(\frac{6}{36}\) |
-1 | --- | \(\frac{5}{36}\) |
-2 | --- | \(\frac{4}{36}\) |
-3 | --- | \(\frac{3}{36}\) |
-4 | (2, 6), (1, 5) | \(\frac{2}{36}\) |
-5 | (1, 6) | \(\frac{1}{36}\) |
Step 1: Calculation of \( \Sigma(x^2) \)
\( \Sigma(x^2) = \Sigma x^2 P(x) = 2 \left( \frac{25}{36} + \frac{32}{36} + \frac{27}{36} + \frac{16}{36} + \frac{5}{36} \right) \)
Simplify:
\( \Sigma(x^2) = \frac{105}{18} = \frac{35}{6} \)
Step 2: Mean (\( \mu \))
As the data is symmetric:
\( \mu = \Sigma(x) = 0 \)
Step 3: Variance (\( \sigma^2 \))
\( \sigma^2 = \Sigma(x^2 P(x)) = \frac{35}{6} \)
The total probability \( P \) is:
\( P = 35 = 5 \times 7 \)
Step 4: Sum of divisors of \( P \)
The sum of divisors of 35 is:
\( (5^0 + 5^1)(7^0 + 7^1) = 6 \times 8 = 48 \)