7. The random variable X has a probability density function \( \mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}P & ; 01) \)

Let's analyze the given probability density function (pdf) of the random variable \( X \): \[ f(x) = \begin{cases} P & \text{for } 0 < x < 2, \\ P(2x - 3) & \text{for } 2 < x < 3, \\ 0 & \text{otherwise}. \end{cases} \] ### (i) Finding the Value of \( P \) To ensure that \( f(x) \) is a valid pdf, the total area under the curve must be equal to 1: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1. \] Calculating the integral over the defined intervals: \[ \int_0^2 P \, dx + \int_2^3 P(2x - 3) \, dx = 1. \] 1. **First Integral:** \[ \int_0^2 P \, dx = P \cdot (2 - 0) = 2P. \] 2. **Second Integral:** \[ \int_2^3 P(2x - 3) \, dx = P \left[ \int_2^3 2x \, dx - \int_2^3 3 \, dx \right] = P \left[ \left( x^2 \bigg|_2^3 \right) - 3(x \bigg|_2^3) \right] = P \left[ (9 - 4) - (9 - 6) \right] = P \cdot 2. \] Adding both integrals: \[ 2P + 2P = 4P = 1 \quad \Rightarrow \quad P = \frac{1}{4}. \] ### (ii) Calculating the Variance of \( X \) Variance is given by: \[ \text{Var}(X) = E[X^2] - (E[X])^2. \] **First, find \( E[X] \):** \[ E[X] = \int_0^2 x \cdot \frac{1}{4} \, dx + \int_2^3 x \cdot \frac{1}{4}(2x - 3) \, dx. \] 1. **First Expectation Integral:** \[ \frac{1}{4} \int_0^2 x \, dx = \frac{1}{4} \left[ \frac{x^2}{2} \bigg|_0^2 \right] = \frac{1}{4} \cdot 2 = \frac{1}{2}. \] 2. **Second Expectation Integral:** \[ \frac{1}{4} \int_2^3 x(2x - 3) \, dx = \frac{1}{4} \left[ \frac{2x^3}{3} - \frac{3x^2}{2} \bigg|_2^3 \right] = \frac{1}{4} \cdot \frac{31}{6} = \frac{31}{24}. \] So, \[ E[X] = \frac{1}{2} + \frac{31}{24} = \frac{43}{24}. \] **Next, find \( E[X^2] \):** \[ E[X^2] = \int_0^2 x^2 \cdot \frac{1}{4} \, dx + \int_2^3 x^2 \cdot \frac{1}{4}(2x - 3) \, dx. \] 1. **First \( E[X^2] \) Integral:** \[ \frac{1}{4} \int_0^2 x^2 \, dx = \frac{1}{4} \left[ \frac{x^3}{3} \bigg|_0^2 \right] = \frac{1}{4} \cdot \frac{8}{3} = \frac{2}{3}. \] 2. **Second \( E[X^2] \) Integral:** \[ \frac{1}{4} \int_2^3 x^2(2x - 3) \, dx = \frac{1}{4} \cdot \frac{81}{6} = \frac{97}{24}. \] So, \[ E[X^2] = \frac{97}{24}. \] **Finally, compute the variance:** \[ \text{Var}(X) = \frac{97}{24} - \left( \frac{43}{24} \right)^2 = \frac{97}{24} - \frac{1849}{576} = \frac{479}{576}. \] ### (iii) Calculating \( P(X > 1) \) \[ P(X > 1) = \int_1^2 \frac{1}{4} \, dx + \int_2^3 \frac{1}{4}(2x - 3) \, dx. \] 1. **First Probability Integral:** \[ \frac{1}{4} \int_1^2 dx = \frac{1}{4} \cdot 1 = \frac{1}{4}. \] 2. **Second Probability Integral:** \[ \frac{1}{4} \int_2^3 (2x - 3) \, dx = \frac{1}{4} \cdot 2 = \frac{1}{2}. \] Adding both probabilities: \[ P(X > 1) = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}. \] ### **Final Answers:** 1. **Value of \( P \):** \[ P = \frac{1}{4}. \] 2. **Variance of \( X \):** \[ \text{Var}(X) = \frac{479}{576}. \] 3. **Probability \( P(X > 1) \):** \[ P(X > 1) = \frac{3}{4}. \] **Answer Summary:** (i) \( P = \dfrac{1}{4} \); (ii) Variance of \( X \) is \( \dfrac{479}{576} \); (iii) \( \mathrm{P}(X > 1) = \dfrac{3}{4} \).