1. **Stating the problem:** Find the constant $c$ such that $f(x) = c x e^{-2x}, x > 0$ is a valid pdf. 2. **Total probability must be 1:** $$\int_0^{\infty} c x e^{-2x} dx = 1$$ 3. **Calculate the integral:** $$\int_0^{\infty} x e^{-2x} dx = \frac{1}{2^2} = \frac{1}{4}$$ 4. **Substitute the integral result:** $$c \times \frac{1}{4} = 1 \implies c = 4$$ --- 5. **Find the harmonic mean of $X$:** $$HM = \frac{1}{E\left(\frac{1}{X}\right)}$$ 6. **Calculate $E\left(\frac{1}{X}\right)$:** $$E\left(\frac{1}{X}\right) = \int_0^{\infty} \frac{1}{x} \cdot 4 x e^{-2x} dx = 4 \int_0^{\infty} e^{-2x} dx = 4 \times \frac{1}{2} = 2$$ 7. **Hence, harmonic mean:** $$HM = \frac{1}{2}$$ --- 8. **Find the moment generating function (m.g.f.) of $X$:** $$M_X(t) = E(e^{tX}) = \int_0^{\infty} e^{t x} \cdot 4 x e^{-2x} dx = 4 \int_0^{\infty} x e^{-(2 - t) x} dx$$ 9. **Integral converges if $t < 2$. Recall:** $$\int_0^{\infty} x e^{-a x} dx = \frac{1}{a^2}$$ 10. **Thus:** $$M_X(t) = \frac{4}{(2 - t)^2}, \quad t < 2$$ --- 11. **Find m.g.f. of $Y = \frac{1}{X} + 9$:** $$M_Y(t) = E\left(e^{tY}\right) = e^{9t} E\left(e^{\frac{t}{X}}\right) = e^{9t} \int_0^{\infty} 4 x e^{\frac{t}{x}} e^{-2x} dx$$ 12. **Note:** This integral does not simplify to elementary functions. --- 13. **Find the second factorial moment $E[X(X-1)]$:** Recall: $$E[X(X-1)] = E[X^2] - E[X]$$ 14. **Calculate $E[X]$:** $$E[X] = 4 \int_0^{\infty} x^2 e^{-2x} dx = 4 \times \frac{2!}{2^3} = 4 \times \frac{2}{8} = 1$$ 15. **Calculate $E[X^2]$:** $$E[X^2] = 4 \int_0^{\infty} x^3 e^{-2x} dx = 4 \times \frac{3!}{2^4} = 4 \times \frac{6}{16} = 1.5$$ 16. **Hence:** $$E[X(X-1)] = 1.5 - 1 = 0.5$$ --- 17. **Calculate $P(X < 3)$ by CDF:** $$F_X(3) = \int_0^3 4 x e^{-2x} dx$$ 18. **Integrate by parts:** $$\int x e^{-2x} dx = -\frac{x}{2} e^{-2x} - \frac{1}{4} e^{-2x} + C$$ 19. **Evaluate definite integral:** $$\int_0^3 4 x e^{-2x} dx = 4 \left[-\frac{3}{2} e^{-6} - \frac{1}{4} e^{-6} + \frac{1}{4} \right] = 4 \times \frac{1 - 7 e^{-6}}{4} = 1 - 7 e^{-6}$$ --- 20. **Find the second non-central moment around 3, $E[(X-3)^2]$:** Recall: $$E[(X-3)^2] = E[X^2] - 2 \times 3 \times E[X] + 3^2$$ 21. **Substitute values:** $$E[(X-3)^2] = 1.5 - 6 + 9 = 4.5$$