Question

Find \( k \), if the probability density function is given by: \[ f(x) = kx^2(1 - x), \quad {for } 0<x<1, \] \[ = 0, \quad {otherwise.} \]

Hint:

The total probability for a probability density function (p.d.f.) must always satisfy: \[ \int_{-\infty}^{\infty} f(x) dx = 1. \]

Answer 1

Step 1: Use the Probability Density Function Property
The total probability must be 1: \[ \int_{0}^{1} f(x) dx = 1. \] Step 2: Compute the Integral
\[ \int_{0}^{1} kx^2(1 - x) dx = 1. \] Expanding: \[ \int_{0}^{1} k (x^2 - x^3) dx = 1. \] Step 3: Evaluate the Integrals
\[ k \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_{0}^{1} = 1. \] \[ k \left( \frac{1}{3} - \frac{1}{4} \right) = 1. \] \[ k \left( \frac{4 - 3}{12} \right) = 1. \] \[ k \times \frac{1}{12} = 1. \] \[ k = 12. \]