Given that the total probability sums to 1:
\[ \frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1. \]
Simplifying:
\[ K = \frac{1}{4}. \]
Step 1: Calculate the Mean \(\mu\) The mean \(\mu\) is given by:
\[ \mu = \alpha \cdot \frac{1}{3} + 1 \cdot K + 0 \cdot \frac{1}{6} + (-3) \cdot \frac{1}{4}. \]
Substituting the values:
\[ \mu = \frac{\alpha}{3} + \frac{1}{4} \cdot 1 + 0 - \frac{3}{4} = \frac{\alpha}{3} - \frac{1}{2}. \]
Step 2: Calculate the Variance \(\sigma^2\) The variance \(\sigma^2\) is given by:
\[ \sigma^2 = \left(\alpha^2 \cdot \frac{1}{3} + 1^2 \cdot K + 0^2 \cdot \frac{1}{6} + (-3)^2 \cdot \frac{1}{4} \right) - \mu^2. \]
Substituting the values:
\[ \sigma^2 = \frac{\alpha^2}{3} + \frac{1}{4} + \frac{9}{4} - \left(\frac{\alpha}{3} - \frac{1}{2}\right)^2. \]
Simplifying:
\[ \sigma^2 = \frac{\alpha^2}{3} + \frac{1}{4} + \frac{9}{4} - \left(\frac{\alpha^2}{9} - \alpha + \frac{1}{4}\right). \]
Further simplification gives:
\[ \sigma^2 = \frac{2\alpha^2}{9} + \alpha + \frac{9}{4}. \]
Step 3: Given Condition \(\sigma - \mu = 2\) Given:
\[ \sigma = \mu + 2. \]
Substituting this condition and solving for \(\alpha\):
\[ \sigma^2 = (\mu + 2)^2. \]
Equating and simplifying:
\[ \alpha = 6 \quad \text{(since \(\alpha = 0\) is rejected)}. \]
Step 4: Calculate \(\sigma + \mu\) Substitute \(\alpha = 6\) into the expressions for \(\mu\) and \(\sigma\):
\[ \sigma + \mu = 2\mu + 2 = 5. \]
Therefore, the correct answer is $5$.
\(X\) | 0 | 1 | 2 |
\(P(X)\) | \(\frac{25}{36}\) | k | \(\frac{1}{36}\) |
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is: