1. **Problem statement:** We have a random variable $X$ representing the shelf life of medicine bottles with density function $$f(x) = \frac{20000}{(x+100)^3} \text{ for } x>0, \text{ and } 0 \text{ elsewhere}.$$ We want to find probabilities: (a) $P(X \geq 200)$ (b) $P(80 \leq X \leq 120)$ 2. **Formula and rules:** The probability for continuous variables is found by integrating the density function over the desired interval: $$P(a \leq X \leq b) = \int_a^b f(x) \, dx$$ 3. **Calculate (a) $P(X \geq 200)$:** $$P(X \geq 200) = \int_{200}^\infty \frac{20000}{(x+100)^3} \, dx$$ Use substitution $u = x+100$, so when $x=200$, $u=300$: $$= 20000 \int_{300}^\infty u^{-3} \, du = 20000 \left[-\frac{1}{2u^2}\right]_{300}^\infty = 20000 \left(0 - \left(-\frac{1}{2 \times 300^2}\right)\right) = \frac{20000}{2 \times 90000} = \frac{20000}{180000} = \frac{1}{9} \approx 0.1111$$ 4. **Calculate (b) $P(80 \leq X \leq 120)$:** $$P(80 \leq X \leq 120) = \int_{80}^{120} \frac{20000}{(x+100)^3} \, dx$$ Substitute $u = x+100$, so when $x=80$, $u=180$ and when $x=120$, $u=220$: $$= 20000 \int_{180}^{220} u^{-3} \, du = 20000 \left[-\frac{1}{2u^2}\right]_{180}^{220} = 20000 \left(-\frac{1}{2 \times 220^2} + \frac{1}{2 \times 180^2}\right)$$ Calculate values: $$= 20000 \times \frac{1}{2} \left(\frac{1}{180^2} - \frac{1}{220^2}\right) = 10000 \left(\frac{1}{32400} - \frac{1}{48400}\right)$$ Find common denominator and subtract: $$= 10000 \left(\frac{48400 - 32400}{32400 \times 48400}\right) = 10000 \times \frac{16000}{1,568,160,000} = \frac{160,000,000}{1,568,160,000} \approx 0.102$$ **Final answers:** (a) $P(X \geq 200) = \frac{1}{9} \approx 0.1111$ (b) $P(80 \leq X \leq 120) \approx 0.102$