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Paper 2 · Specimen 2026 · Continuous Random Variables

A continuous random variable XX has probability density function f(x)=0.1x+kf(x) = 0.1x + k for 4≤x≤64 \leq x \leq 6, f(x)=0.3f(x) = 0.3 for 6≤x≤86 \leq x \leq 8, and f(x)=0f(x) = 0 otherwise. Find the value of the constant kk.

Model answer

-0.3

Also accepted: -0,3, k=-0.3

Explanation

The density must integrate to 1: ∫46(0.1x+k) dx+∫680.3 dx=1\int_4^6 (0.1x+k)\,dx + \int_6^8 0.3\,dx = 1. The first integral is 0.05(62−42)+2k=1+2k0.05(6^2-4^2) + 2k = 1 + 2k and the second is 0.60.6, so 1+2k+0.6=11 + 2k + 0.6 = 1 and k=−0.3k = -0.3.

Derived from ZIMSEC Statistics 6046/2 Paper 2, Specimen Paper, Q1

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