A continuous random variable X has probability density function f(x) = dfracx12 for 0 ≤ x < 3 and f(x) = -dfrac120(x-8) for 3 ≤ x ≤ 8 (zero otherwise). What is the median of X, correct to 2 decimal places?

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Paper 2 · November 2019 · Continuous Random Variables

A continuous random variable $X$ has probability density function $f(x) = \dfrac{x}{12}$ for $0 \le x < 3$ and $f(x) = -\dfrac{1}{20}(x-8)$ for $3 \le x \le 8$ (zero otherwise). What is the median of $X$ , correct to 2 decimal places?

Model answer

3.53

Also accepted: 8-2√5, 8 - 2 sqrt(5)

Explanation

Since

\int_0^3 \frac{x}{12}dx = 0.375 < 0.5

, the median

m

lies in

[3,8]

. Solving

0.375 + \int_3^m -\frac{1}{20}(x-8)dx = 0.5

gives

(m-8)^2 = 20

, so

m = 8 - 2\sqrt{5} \approx 3.53

(taking the root

\le 8

Derived from ZIMSEC Statistics Paper 2, November 2019, Q1

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