IB Mathematics SL 4.7: Concept of discrete random variables AI SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
(i) \(P(C \ge 2) = P(C=2) + P(C=3) = 0.28 + 0.17 = \mathbf{0.45}\).
(ii) \(E(C) = \sum c \times P(C=c) = (0 \times 0.21) + (1 \times 0.34) + (2 \times 0.28) + (3 \times 0.17)\).
\(E(C) = 0 + 0.34 + 0.56 + 0.51 = \mathbf{1.41}\).
(b)
\(E(T) = 115\).
Value = \(1.41 \times 115 = \mathbf{162}\) (162.15) seconds (or 2m 42s).
(c)
\(T \sim N(115, 28^2)\). Need \(P(T > 180)\) (3 minutes).
Using GDC Normal CDF: Lower=180, Upper=1E99, \(\mu=115, \sigma=28\).
\(P \approx \mathbf{0.0101}\) (0.01013…).
(d)
Total time for two customers \(T_2 = T_1 + T_2\).
Mean \(= 115 + 115 = 230\). Variance \(= 28^2 + 28^2 = 2 \times 28^2\). SD \(= \sqrt{2 \times 28^2} \approx 39.6\).
Need \(P(T_2 > 180)\).
Using GDC: \(P \approx \mathbf{0.897}\) (0.8966…).
(e)
(i) Probability of waiting > 3 mins depends on how many people in queue.
– If 0 in queue: Wait is 0 (Prob 0).
– If 1 in queue: Wait is 1 service time. \(P(T > 180) \approx 0.0101\) (from c).
– If 2 in queue: Wait is 2 service times. \(P(T_2 > 180) \approx 0.897\) (from d).
– If 3 in queue: Wait is 3 service times. \(P(T_3 > 180) = 1\) (Assumption given).
Total Probability = \((0.21 \times 0) + (0.34 \times 0.0101) + (0.28 \times 0.897) + (0.17 \times 1)\).
\(P \approx 0 + 0.0034 + 0.251 + 0.17 = \mathbf{0.425}\) (0.4245…).
(ii) Since \(0.425 > 0.4\), she will employ more staff.