A random variable X has probability density function given by \[ \mathrm{f}(x)=\left\{\begin{array}{ll}k(3-x) & 1 \leqslant x \leqslant 2 \\ 0 & \text { otherwise, }\end{array}\right. \] where k is a constant. a) Show that \( \mathrm{k}=\frac{2}{3} \). b) Find the median of X .

To solve the problem, we will first find the value of the constant \( k \) in the probability density function (PDF) defined as: \[ f(x)=\left\{\begin{array}{ll}k(3-x) & 1 \leqslant x \leqslant 2 \\ 0 & \text{otherwise, }\end{array}\right. \] ### Part (a): Show that \( k=\frac{2}{3} \) To determine \( k \), we need the total area under the curve of the PDF to equal 1, since the total probability must equal 1. Thus, we set up the integral: \[ \int_{1}^{2} f(x) \, dx = 1 \] Substituting in the expression for \( f(x) \): \[ \int_{1}^{2} k(3-x) \, dx = 1 \] Now we compute the integral: \[ \int_{1}^{2} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{1}^{2} \] Calculating the limits: 1. For \( x=2 \): \[ 3(2) - \frac{(2)^2}{2} = 6 - 2 = 4 \] 2. For \( x=1 \): \[ 3(1) - \frac{(1)^2}{2} = 3 - \frac{1}{2} = 2.5 \] Thus, the value of the integral from 1 to 2 is: \[ 4 - 2.5 = 1.5 \] Substituting back, we have: \[ k(1.5) = 1 \implies k = \frac{1}{1.5} = \frac{2}{3} \] ### Part (b): Find the median of \( X \) The median \( m \) of a continuous random variable \( X \) is defined such that: \[ P(X \leq m) = \frac{1}{2} \] This means we need to find \( m \) such that: \[ \int_{1}^{m} f(x) \, dx = \frac{1}{2} \] Using \( k = \frac{2}{3} \): \[ \int_{1}^{m} \frac{2}{3} (3 - x) \, dx = \frac{1}{2} \] Calculating the integral: \[ \int_{1}^{m} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{1}^{m} = \left(3m - \frac{m^2}{2}\right) - \left(3 - \frac{1}{2}\right) = 3m - \frac{m^2}{2} - 2.5 \] Substituting this into our equation: \[ \frac{2}{3} \left( 3m - \frac{m^2}{2} - 2.5 \right) = \frac{1}{2} \] Clearing the fraction by multiplying both sides by 3: \[ 2(3m - \frac{m^2}{2} - 2.5) = \frac{3}{2} \] This simplifies to: \[ 6m - m^2 - 5 = \frac{3}{2} \] Rearranging gives us the quadratic equation: \[ -m^2 + 6m - \frac{13}{2} = 0 \] Multiplying through by -2 to eliminate the fraction: \[ m^2 - 12m + 13 = 0 \] We can solve for \( m \) using the quadratic formula: \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4(1)(13)}}{2(1)} \] Calculating the discriminant: \[ 144 - 52 = 92 \] So: \[ m = \frac{12 \pm \sqrt{92}}{2} = 6 \pm \sqrt{23} \] Since \( m \) must be within the range [1, 2], we take: \[ m = 6 - \sqrt{23} \] Calculating: We know that \( \sqrt{23} \approx