Q275 · Practice questionTechnology-freeComplex unfamiliar6 marks
QUESTION 275 (6 marks)
A two-stage population follows
(An+1Jn+1)=(fs2f0)(AnJn),f>0,0<s≤1.
The initial vector is (40,20)T and the vector after one step is (32,20)T.
a)
Determine f and s and interpret their roles.
[3 marks]
b)
Determine the vector after three steps from the start. A student instead uses (32,20)T as the initial vector and applies the matrix three times. Explain the timing error.
(a) f=0.4, s=0.5; births per juvenile are 0.4, births per adult are 0.8, juvenile survival is 0.5. (b) n3=(24.32,14.4)T; the student calculates step 4.
Worked solution
(a) The youngest-class equation gives 32=f(40)+2f(20)=80f. The adult equation gives 20=40s. These imply the stated rates; the zero lower-right entry means adults do not remain in that stage for another step.
(b) Using L=(0.40.50.80), n2=(28.8,16)T and n3=(24.32,14.4)T. The measured vector is already n1, so applying three more steps gives n4, one step too late.
Equivalent justified methods accepted; respect any requested proof method.
Exact values unless specified. Sketches assessed by mathematical features, not artistic quality.
Part a: Form the youngest-class equation and solve f.
[1 mark]
Part a: Form the survival equation and solve s.
[1 mark]
Part a: Interpret f, 2f and s as the corresponding rates.
[1 mark]
Part b: Calculate the step-2 vector.
[1 mark]
Part b: Calculate the step-3 vector (24.32,14.4)T.
[1 mark]
Part b: Explain the off-by-one timing error.
[1 mark]
Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.