### Understanding the Problem First, I need to understand what's given and what's being asked: 1. Given: A continuous random variable \( X \) with a probability density function (p.d.f.) \( f(x) \) defined piecewise: \[ f(x) = \begin{cases} Ax(6 - x)^2, & 0 < x < 6 \\ 0, & \text{otherwise} \end{cases} \] 2. Tasks: - Find the mean (expected value) of \( X \). - Find the standard deviation of \( X \). - Construct the cumulative distribution function (C.D.F.) \( F(x) \). - Evaluate \( P(X > \frac{1}{4}) \). ### Step 1: Find the Normalization Constant \( A \) Before anything else, I recall that a p.d.f. must satisfy the condition that the total probability over all possible values is 1. That is: \[ \int_{-\infty}^{\infty} f(x) \, dx = 1 \] Given \( f(x) \) is non-zero only between \( 0 \) and \( 6 \), the condition becomes: \[ \int_{0}^{6} Ax(6 - x)^2 \, dx = 1 \] Calculating the Integral: First, I need to expand \( (6 - x)^2 \): \[ (6 - x)^2 = 36 - 12x + x^2 \] Now, multiply by \( x \): \[ x(6 - x)^2 = 36x - 12x^2 + x^3 \] Now, the integral becomes: \[ A \int_{0}^{6} (36x - 12x^2 + x^3) \, dx = 1 \] Let's break it down: 1. \( \int 36x \, dx = 18x^2 \) 2. \( \int -12x^2 \, dx = -4x^3 \) 3. \( \int x^3 \, dx = \frac{x^4}{4} \) Now, evaluate from 0 to 6: \[ A \left[ 18x^2 - 4x^3 + \frac{x^4}{4} \right]_{0}^{6} = A \left( 18(36) - 4(216) + \frac{1296}{4} \right) = A \left( 648 - 864 + 324 \right) = A (108) \] Set this equal to 1: \[ 108A = 1 \implies A = \frac{1}{108} \] Conclusion: The p.d.f. is: \[ f(x) = \begin{cases} \frac{x(6 - x)^2}{108}, & 0 < x < 6 \\ 0, & \text{otherwise} \end{cases} \] ### Step 2: Calculate the Mean (Expected Value) \( E[X] \) The mean of a continuous random variable is given by: \[ E[X] = \int_{-\infty}^{\infty} x f(x) \, dx \] Again, since \( f(x) \) is non-zero only between \( 0 \) and \( 6 \): \[ E[X] = \int_{0}^{6} x