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 E(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 $E(X)$ , correct to 2 decimal places?

A3.53

B3.60

C3.67

D3.75

Explanation

E(X) = \int_0^3 x\cdot\frac{x}{12}dx + \int_3^8 x\cdot\left(-\frac{1}{20}\right)(x-8)dx = \left[\frac{x^3}{36}\right]_0^3 + \left(-\frac{1}{20}\right)\left[\frac{x^3}{3}-4x^2\right]_3^8 = 0.75 + \left(-\frac{1}{20}\right)\left(-\frac{175}{3}\right) = 0.75 + 2.9167 \approx 3.67

Derived from ZIMSEC Statistics Paper 2, November 2019, Q1

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