Continuous random variables and the normal distribution — Question 48
Original QCE Vault practice · 5 marks
Q48 · Practice questionTechnology-activeComplex unfamiliar5 marks
QUESTION 48 (5 marks)
A continuous random variable X has the triangular density shown, given by f(x)=x for 0≤x≤1 and f(x)=2−x for 1<x≤2. Determine P(X>1.5∣X>0.5). 
WORKED SOLUTION
Practice marking scheme
5 marksANSWER71≈0.143. Worked solution
P(X>1.5)=∫1.52(2−x)dx=0.125. Also P(X>0.5)=1−∫00.5xdx=1−0.125=0.875. Therefore the conditional probability is 0.125/0.875=1/7≈0.143. Determines P(X>1.5)=0.125. [1 mark]Determines P(X≤0.5)=0.125. [1 mark]Determines P(X>0.5)=0.875. [1 mark]Uses the conditional probability ratio.
[1 mark]Obtains 1/7≈0.143. [1 mark]Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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