Continuous random variables — practice questions

(a) For $f(x)$ to be a valid PDF, the total area under the curve must be 1. $\int_{-\infty}^{\infty} f(x) \, dx = \int_{0}^{4} kx(4-x) \, dx = 1$ $k \int_{0}^{4} (4x - x^2) \, dx = 1$ $k \left[ 2x^2 - \frac{x^3}{3} \right]_{0}^{4} = 1$ $k \left( (2(4^2) - \frac{4^3}{3}) - (0) \right) = 1$ $k \left( 32 - \frac{64}{3} \right) = 1$ $k \left( \frac{96-64}{3} \right) = 1$ $k \left( \frac{32}{3} \right) = 1$ $k = \frac{3}{32}$ [M1 for integration, A1 for correct value]

(b) Now we find the probability by integrating the PDF with $k=\frac{3}{32}$ over the interval $(1, 3)$. $P(1 < X < 3) = \int_{1}^{3} \frac{3}{32}x(4-x) \, dx$ $= \frac{3}{32} \int_{1}^{3} (4x - x^2) \, dx$ $= \frac{3}{32} \left[ 2x^2 - \frac{x^3}{3} \right]_{1}^{3}$ [M1 for setting up correct integral] $= \frac{3}{32} \left( (2(3^2) - \frac{3^3}{3}) - (2(1^2) - \frac{1^3}{3}) \right)$ $= \frac{3}{32} \left( (18 - 9) - (2 - \frac{1}{3}) \right)$ $= \frac{3}{32} \left( 9 - \frac{5}{3} \right)$ $= \frac{3}{32} \left( \frac{27-5}{3} \right) = \frac{3}{32} \left( \frac{22}{3} \right)$ $= \frac{22}{32} = \frac{11}{16}$ [A1 for correct final answer]

First, we find the expectation, $E(T)$. $E(T) = \int_{0}^{\infty} t f(t) \, dt = \int_{0}^{\infty} t \cdot \frac{1}{4}e^{-\frac{t}{4}} \, dt$ This requires integration by parts: $\int u \frac{dv}{dt} dt = uv - \int v \frac{du}{dt} dt$. Let $u=t \implies \frac{du}{dt}=1$. Let $\frac{dv}{dt}=\frac{1}{4}e^{-\frac{t}{4}} \implies v = -e^{-\frac{t}{4}}$. [M1 for setting up correct integral and identifying integration by parts] $E(T) = \left[ -t e^{-\frac{t}{4}} \right]_{0}^{\infty} - \int_{0}^{\infty} (-e^{-\frac{t}{4}}) \, dt$ $= \left[ -t e^{-\frac{t}{4}} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-\frac{t}{4}} \, dt$ As $t \to \infty$, $t e^{-\frac{t}{4}} \to 0$. At $t=0$, the term is 0. So the first part is 0. $E(T) = \left[ -4e^{-\frac{t}{4}} \right]_{0}^{\infty} = (0) - (-4e^0) = 4$. The mean lifetime is 4 years. [A1 for E(T)]

Next, we find $E(T^2)$ to calculate the variance. $E(T^2) = \int_{0}^{\infty} t^2 f(t) \, dt = \int_{0}^{\infty} \frac{1}{4}t^2 e^{-\frac{t}{4}} \, dt$ Again, use integration by parts. Let $u=t^2 \implies \frac{du}{dt}=2t$. Let $\frac{dv}{dt}=\frac{1}{4}e^{-\frac{t}{4}} \implies v = -e^{-\frac{t}{4}}$. [M1 for setting up E(T^2) integral] $E(T^2) = \left[ -t^2 e^{-\frac{t}{4}} \right]_{0}^{\infty} - \int_{0}^{\infty} (-e^{-\frac{t}{4}})(2t) \, dt$ The first term is 0. $E(T^2) = 2 \int_{0}^{\infty} t e^{-\frac{t}{4}} \, dt = 8 \int_{0}^{\infty} \frac{1}{4} t e^{-\frac{t}{4}} \, dt$ We recognise the integral as $E(T)$, which we found to be 4. $E(T^2) = 8 \times E(T) = 8 \times 4 = 32$. [A1 for E(T^2)]

Finally, calculate the variance. $Var(T) = E(T^2) - [E(T)]^2 = 32 - 4^2 = 32 - 16 = 16$. [M1 A1 for Var(T)] The variance of the lifetime is 16 years$^2$.