Paper 2 · continuous random variables / probability density functions

A continuous random variable $X$ has probability density function $f(x) = \dfrac{x}{12}$ for $0 \le x < 3$ , $f(x) = k(x-8)$ for $3 \le x \le 8$ , and $f(x) = 0$ otherwise, where $k$ is a constant. Since the total area under $f$ must equal 1, what is the value of $k$ ?

A-0.05

B0.05

C-0.04

D-0.10

Explanation: For a valid pdf,

\int f(x)\,dx = 1

\int_0^3 \frac{x}{12}dx = \frac{9}{24} = \frac{3}{8}

\int_3^8 k(x-8)dx = k\left[\frac{(x-8)^2}{2}\right]_3^8 = k(0 - 12.5) = -12.5k

. So

\frac{3}{8} - 12.5k = 1

, giving

-12.5k = \frac{5}{8}

, so

k = -\frac{1}{20} = -0.05

Derived from ZIMSEC Statistics Paper 2, November 2019, Q1

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