product chosen is defective? If X is a continuous random variable having the following probability density function: \[ f(x)=\left\{\begin{array}{ll}2(x-1), & 1

Ask by Ray Dickson. in Pakistan

Jan 22,2025

To find the variance of a continuous random variable \( X \) with the given probability density function (PDF), we first need to calculate the mean \( E[X] \) and then the expected value of \( X^2 \), \( E[X^2] \). 1. **Finding the Mean \( E[X] \)**: \[ E[X] = \int_{1}^{2} x \cdot f(x) \, dx = \int_{1}^{2} x \cdot 2(x - 1) \, dx = \int_{1}^{2} 2x^2 - 2x \, dx \] Evaluating the integral: \[ E[X] = \left[ \frac{2}{3}x^3 - x^2 \right]_{1}^{2} = \left( \frac{2}{3}(2^3) - (2^2) \right) - \left( \frac{2}{3}(1^3) - (1^2) \right) \] \[ = \left( \frac{16}{3} - 4 \right) - \left( \frac{2}{3} - 1 \right) = \left( \frac{16}{3} - \frac{12}{3} \right) - \left( \frac{2}{3} - \frac{3}{3} \right) = \left( \frac{4}{3} \right) - \left( -\frac{1}{3} \right) \] \[ = \frac{4}{3} + \frac{1}{3} = \frac{5}{3} \] 2. **Finding \( E[X^2] \)**: \[ E[X^2] = \int_{1}^{2} x^2 \cdot f(x) \, dx = \int_{1}^{2} x^2 \cdot 2(x - 1) \, dx = \int_{1}^{2} 2x^3 - 2x^2 \, dx \] Evaluating the integral: \[ E[X^2] = \left[ \frac{1}{2}x^4 - \frac{2}{3}x^3 \right]_{1}^{2} = \left( \frac{1}{2}(2^4) - \frac{2}{3}(2^3) \right) - \left( \frac{1}{2}(1^4) - \frac{2}{3}(1^3) \right) \] \[ = \left( \frac{16}{2} - \frac{16}{3} \right) - \left( \frac{1}{2} - \frac{2}{3} \right) = \left( 8 - \frac{16}{3} \right) - \left( \frac{3}{6} - \frac{4}{6} \right) = \left( \frac{24}{3} - \frac{16}{3} \right) + \frac{1}{6} \] \[ = \frac{8}{3} + \frac{1}{6} = \frac{16}{6} + \frac{1}{6} = \frac{17}{6} \] 3. **Calculating Variance \( Var(X) \)**: \[ Var(X) = E[X^2] - (E[X])^2 = \frac{17}{6} - \left( \frac{5}{3} \right)^2 = \frac{17}{6} - \frac{25}{9} \] Finding a common denominator (the least common multiple of 6 and 9 is 18): \[ = \frac{51}{18} - \frac{50}{18} = \frac{1}{18} \] Thus, the variance of \( X \) is \( \frac{1}{18} \).