IBDP Maths SL 4.12 Standardization of normal variables AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
\( \text{P}(W < 62) = 0.3 \), so \( \frac{62 – \mu}{\sigma} = \Phi^{-1}(0.3) \approx -0.524 \)
\( \text{P}(W > 98) = 0.25 \), so \( \text{P}(W < 98) = 0.75 \), and \( \frac{98 – \mu}{\sigma} = \Phi^{-1}(0.75) \approx 0.674 \)
\(\boxed{62 – \mu = -0.524\sigma, 98 – \mu = 0.674\sigma}\)
Solve: \( 62 – \mu = -0.524\sigma \), \( 98 – \mu = 0.674\sigma \)
Subtract: \( 98 – 62 = 0.674\sigma – (-0.524\sigma) \)
\( 36 = 1.198\sigma \), so \( \sigma \approx 30.05 \)
\( \mu = 98 – 0.674 \cdot 30.05 \approx 77.75 \)
\(\boxed{\mu \approx 77.7, \sigma \approx 30.0}\)
\( z = \frac{100 – 77.7}{30.0} \approx 0.743 \)
\( \text{P}(Z > 0.743) \approx 0.229 \)
\(\boxed{0.229}\)
\( X \sim \text{B}(10, 0.229) \)
\( \text{P}(X \geq 4) \approx 0.178 \)
\(\boxed{0.178}\)
\( \text{P}(X \geq 1) \approx 1 – (1 – 0.229)^{10} \approx 0.918 \)
\( \text{P}(1 \leq X \leq 3) \approx 0.741 \)
\( \text{P}(X < 4 | X \geq 1) = \frac{0.741}{0.918} \approx 0.808 \)
\(\boxed{0.808}\)
Average arrivals per 1800s (30 min) = \( \frac{1800}{36} \approx 50 \)
\( L \sim \text{Po}(50) \)
\( \text{P}(L > 60) = 1 – \text{P}(L \leq 60) \approx 0.0722 \)
\(\boxed{0.0722}\)
400 workers need 40 elevators (400 / 10).
\( \text{P}(L \geq 40) = 1 – \text{P}(L \leq 39) \approx 0.935 \)
\(\boxed{0.935}\)
Note: In Section B, accept answers that correctly round to 2 sf.
a.(i) let \(W\) be the weight of a worker and \(W \sim \text{N}(\mu, \sigma^2)\)
\(\text{P}\left(Z < \frac{62 – \mu}{\sigma}\right) = 0.3\) and \(\text{P}\left(Z < \frac{98 – \mu}{\sigma}\right) = 0.75\) (M1)
Note: Award M1 for a correctly shaded and labelled diagram.
\(\frac{62 – \mu}{\sigma} = \Phi^{-1}(0.3) \; ( = -0.524 \ldots )\) and
\(\frac{98 – \mu}{\sigma} = \Phi^{-1}(0.75) \; ( = 0.674 \ldots )\)
or linear equivalents A1A1
Note: Condone equations containing the GDC inverse normal command.
(ii) attempting to solve simultaneously (M1)
\(\mu = 77.7, \sigma = 30.0\) A1A1
[6 marks]
Note: In Section B, accept answers that correctly round to 2 sf.
\(\text{P}(W > 100) = 0.229\) A1
[1 mark]
Note: In Section B, accept answers that correctly round to 2 sf.
let \(X\) represent the number of workers over 100 kg in a lift of ten passengers
\(X \sim \text{B}(10, 0.229 \ldots)\) (M1)
\(\text{P}(X \geq 4) = 0.178\) A1
[2 marks]
Note: In Section B, accept answers that correctly round to 2 sf.
\(\text{P}(X < 4|X \geq 1) = \frac{\text{P}(1 \leq X \leq 3)}{\text{P}(X \geq 1)}\) M1(A1)
Note: Award the M1 for a clear indication of a conditional probability.
\( = 0.808\) A1
[3 marks]
Note: In Section B, accept answers that correctly round to 2 sf.
\(L \sim \text{Po}(50)\) (M1)
\(\text{P}(L > 60) = 1 – \text{P}(L \leq 60)\) (M1)
\( = 0.0722\) A1
[3 marks]
Note: In Section B, accept answers that correctly round to 2 sf.
400 workers require at least 40 elevators (A1)
\(\text{P}(L \geq 40) = 1 – \text{P}(L \leq 39)\) (M1)
\( = 0.935\) A1
[3 marks]
Total [18 marks]