1. From the given probability distribution, the total probability must equal 1:
\[ \sum P_i = 1 \implies a + 2a + a + b + 2b + 3b = 1. \]
Simplify:
\[ 4a + 6b = 1 \quad \cdots \text{(I)} \]
2. The mean is given by:
\[ E(X) = \sum P_i X_i = \frac{46}{9}. \]
Substitute the probabilities:
\[ E(X) = 0 \times a + 2 \times 2a + 4 \times (a + b) + 6 \times 2b + 8 \times 3b. \]
Simplify:
\[ 4a + 4b + 12b + 24b = \frac{46}{9}, \]
\[ 8a + 40b = \frac{46}{9} \quad \cdots \text{(II)}. \]
3. Solve equations (I) and (II) simultaneously: From (I): \(b = \frac{1}{9} - \frac{2a}{3}\). Substitute \(b = \frac{1}{9}\) and solve to find:
\[ a = \frac{1}{12}, \quad b = \frac{1}{9}. \]
4. Variance is calculated as:
\[ \text{Variance} = E(X^2) - (E(X))^2. \]
Step 1: Find \(E(X^2)\):
\[ E(X^2) = \sum P_i X_i^2 = 0^2 \times a + 2^2 \times 2a + 4^2 \times (a + b) + 6^2 \times 2b + 8^2 \times 3b. \]
Simplify:
\[ E(X^2) = 4a + 16(a + b) + 72b + 192b. \]
Substitute \(a = \frac{1}{12}, b = \frac{1}{9}\):
\[ E(X^2) = \frac{298}{9}. \]
Step 2: Find \((E(X))^2\):
\[ (E(X))^2 = \left(\frac{46}{9}\right)^2 = \frac{2116}{81}. \]
Step 3: Calculate variance:
\[ \text{Variance} = \frac{298}{9} - \frac{2116}{81} = \frac{566}{81}. \]
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32