# Continuous Probability (PDF based) Solution ## Given Data - Probability density function (pdf): \( f(x) = kx^2 \), for \( \leq x \leq 1 \) - \( f(x) = \) otherwise ## Concepts and Definitions - A probability density function (pdf) describes the likelihood of a continuous random variable taking a specific value. - For any pdf \( f(x) \) over the interval \( a \leq x \leq b \), the total probability is 1: \( \int_a^b f(x) \, dx = 1 \) - The probability that \( X \) is in an interval \([c, d]\) is: \( P(c \leq X \leq d) = \int_c^d f(x) \, dx \) - The expected value (mean) \( E(X) \) for a continuous random variable is: \( E(X) = \int_{-\infty}^{\infty} x f(x) \, dx \) --- ## 1. Find the value of k Set the integral of the pdf over the possible range equal to 1: \[ \int_^1 kx^2 \, dx = 1 \] \[ k \int_^1 x^2 \, dx = 1 \] \[ k \left[ \frac{x^3}{3} \right]_^1 = 1 \] \[ k \left( \frac{1^3}{3} - \frac{^3}{3} \right) = 1 \] \[ k \left( \frac{1}{3} \right) = 1 \] \[ k = 3 \] --- ## 2. Find \( P(X > .5) \) \[ P(X > .5) = \int_{.5}^{1} 3x^2 \, dx \] \[ = 3 \left[ \frac{x^3}{3} \right]_{.5}^1 \] \[ = \left[ x^3 \right]_{.5}^1 \] \[ = (1^3) - (.5^3) \] \[ = 1 - .125 = .875 \] --- ## 3. Find \( E(X) \) \[ E(X) = \int_^1 x \cdot 3x^2 \, dx \] \[ = 3 \int_^1 x^3 \, dx \] \[ = 3 \left[ \frac{x^4}{4} \right]_^1 \] \[ = 3 \left( \frac{1^4}{4} - \frac{^4}{4} \right) \] \[ = 3 \left( \frac{1}{4} \right) = \frac{3}{4} \] --- ## Summary - \( k = 3 \) - \( P(X > .5) = .875 \) - \( E(X) = \frac{3}{4} \)