IB Mathematics SL 4.7 Concept of discrete random variables AI SL Paper 2- Exam Style Questions - New Syllabus
▶️ Answer/Explanation
Markscheme (with methods)
Set-up
Sum of probabilities: \(k+k^2+a+k^3=1\). A1
Expected value: \(E(X)=1\cdot k+2\cdot k^2+3\cdot a+4\cdot k^3=2.6\). A1
Expected value: \(E(X)=1\cdot k+2\cdot k^2+3\cdot a+4\cdot k^3=2.6\). A1
Method 1 (intersection of functions)
From the two equations, \[ a=1-(k+k^2+k^3),\qquad a=\frac{2.6-(k+2k^2+4k^3)}{3}. \] Plot or table the two functions of \(k\) and find their intersection for \(0<k<1\) (since probabilities are positive and sum to 1). M1
Using technology gives \(k\approx 0.185928\), hence \[ a=1-(k+k^2+k^3)\approx 1-(0.185928+0.034579+0.006439)=0.773073\ldots \] Therefore \(\boxed{a\approx 0.773}\). M1 A1
Using technology gives \(k\approx 0.185928\), hence \[ a=1-(k+k^2+k^3)\approx 1-(0.185928+0.034579+0.006439)=0.773073\ldots \] Therefore \(\boxed{a\approx 0.773}\). M1 A1
Method 2 (algebraic elimination to a cubic)
Eliminate \(a\) using \(a=1-(k+k^2+k^3)\) in \(E(X)=2.6\): \[ k+2k^2+3\bigl[1-(k+k^2+k^3)\bigr]+4k^3=2.6. \] Simplify: \[ k+2k^2+3-3k-3k^2-3k^3+4k^3=2.6 \] \[ \Rightarrow\; k^3 – k^2 – 2k + 0.4=0. \] Solve the cubic (e.g. GDC / numerical solver) to get the admissible root \(k\approx 0.185928\) (other roots are \(>1\) or negative). M1 M1
Then \[ a=1-(k+k^2+k^3)\approx 0.773073\ldots\ \Rightarrow\ \boxed{a\approx 0.773}. \] A1
Then \[ a=1-(k+k^2+k^3)\approx 0.773073\ldots\ \Rightarrow\ \boxed{a\approx 0.773}. \] A1
Final answer: \(\boxed{a=0.773\text{ (to 3 s.f.)}}\).