1. **State the problem:** We have a continuous random variable $T$ representing the lifetime of a battery with probability density function (pdf) $$f(t) = \begin{cases} ce^{-3t}, & t > 0 \\ 0, & \text{otherwise} \end{cases}$$ We need to find: (a) The constant $c$. (b) The median lifetime. (c) The mean lifetime. (d) The variance $\mathrm{Var}(T)$. 2. **Find $c$:** The total probability must be 1, so $$\int_0^\infty ce^{-3t} dt = 1.$$ Calculate the integral: $$c \int_0^\infty e^{-3t} dt = c \left[-\frac{1}{3} e^{-3t} \right]_0^\infty = c \left(0 + \frac{1}{3} \right) = \frac{c}{3}.$$ Set equal to 1: $$\frac{c}{3} = 1 \implies c = 3.$$ 3. **Find the median $m$:** The median satisfies $$P(T \leq m) = \int_0^m 3 e^{-3t} dt = 0.5.$$ Calculate the integral: $$\int_0^m 3 e^{-3t} dt = \left[-e^{-3t} \right]_0^m = 1 - e^{-3m}.$$ Set equal to 0.5: $$1 - e^{-3m} = 0.5 \implies e^{-3m} = 0.5.$$ Take natural log: $$-3m = \ln(0.5) = -\ln(2) \implies m = \frac{\ln(2)}{3}.$$ 4. **Find the mean $E(T)$:** $$E(T) = \int_0^\infty t \cdot 3 e^{-3t} dt.$$ Use integration by parts: Let $u = t$, $dv = 3 e^{-3t} dt$, then $du = dt$, $v = -e^{-3t}$. So, $$E(T) = \left. -t e^{-3t} \right|_0^\infty + \int_0^\infty e^{-3t} dt = 0 + \left[-\frac{1}{3} e^{-3t} \right]_0^\infty = \frac{1}{3}.$$ 5. **Find the variance $\mathrm{Var}(T)$:** First find $E(T^2)$: $$E(T^2) = \int_0^\infty t^2 \cdot 3 e^{-3t} dt.$$ Using the gamma function or repeated integration by parts, $$E(T^2) = \frac{2}{9}.$$ Then, $$\mathrm{Var}(T) = E(T^2) - (E(T))^2 = \frac{2}{9} - \left(\frac{1}{3}\right)^2 = \frac{2}{9} - \frac{1}{9} = \frac{1}{9}.$$ **Final answers:** (a) $c = 3$ (b) Median $m = \frac{\ln(2)}{3}$ (c) Mean $E(T) = \frac{1}{3}$ (d) Variance $\mathrm{Var}(T) = \frac{1}{9}$