IB Mathematics SL 4.8 Binomial distribution. Mean and variance-AI SL Paper 2 - Exam Style Questions - New Syllabus
▶️ Answer/Explanation
(a)
(i) Scheduled $18:00$, Mean Arrival $18:20$. Mean delay $\mu = \mathbf{20}$ minutes.
(ii) Standard deviation $\sigma = 10$. Variance $\sigma^2 = 10^2 = \mathbf{100}$.
(b)
Using Normal CDF on a GDC with Lower bound $= 15$, Upper bound $= 25$, $\mu = 20, \sigma = 10$:
$P(15 \le X \le 25) \approx \mathbf{0.383}$ (to 3 sig figs).
(c)
Probability of a fine: $P(X > 30)$. Using Normal CDF (Lower $= 30$, Upper $= \infty$): $P \approx 0.158655$.
Expected days $= 0.158655 \times 365 \approx 57.909$.
$57.9$ days (or $58$ days).
(d)
Janine needs $P(\text{Arrival Delay} + 10 \le \text{Departure Delay}) \ge 0.99$.
Let $k$ be the delay such that $P(X \le k) = 0.99$.
Using Inverse Normal (Area $= 0.99, \mu = 20, \sigma = 10$): $k \approx 43.26$ minutes.
Total time needed from $18:00$ is $k + 10 = 43.26 + 10 = 53.26$ minutes.
Time is $18:00 + 53.26$ mins $\approx 18:53:15$.
To ensure at least $99\%$, the train must depart at or after this. To the nearest minute: $18:54$.
(e)
(i) “Delayed” means $X > 0$. $P(X > 0)$ with $\mu = 20, \sigma = 10$:
$P = \mathbf{0.977}$ (0.97725…).
(ii) Let $Y \sim B(7, 0.97725)$.
$P(Y = 5) = \binom{7}{5}(0.97725)^5(1 – 0.97725)^2 \approx \mathbf{0.00969}$.
(iii) For at least $5$ consecutive days, we sum the probabilities of the following patterns (where $L$ is Late and $N$ is Not Late):
$5$ consecutive: $LLLLLNN$, $NLLLLLN$, $NNLLLLL$ (3 cases)
$6$ consecutive: $LLLLLLN$, $NLLLLLL$ (2 cases)
$7$ consecutive: $LLLLLLL$ (1 case)
Let $p = 0.97725$ and $q = 0.02275$.
$P = (p^7) + 2(p^6 q) + 3(p^5 q^2) \approx \mathbf{0.892}$.