Let’s go through each part step by step. --- ## **Background: Non-Homogeneous Poisson Process (NHPP)** - **Intensity function:** \(\lambda(t) = 1 + t\), for \(t \ge 0\). - **Cumulative rate function:** \(\Lambda(t) = \int_0^t \lambda(s)\, ds = t + \frac{1}{2} t^2\). - **Start time:** 0pm (i.e., \(t = 0\) at 0pm). --- ## **1. Probability that the 1st passenger arrives between 1pm and 2pm** This means the first arrival time \(T_1\) is in \([1,2]\) hours after 0pm. - **CDF of the 1st arrival:** For NHPP, the probability that no event occurs by time \(t\) is \(P(N(t) = 0) = e^{-\Lambda(t)}\). - So, the probability the first arrival is after time \(t\) is \(P(T_1 > t) = e^{-\Lambda(t)}\). Thus, \[ P(1 < T_1 < 2) = P(T_1 > 1) - P(T_1 > 2) \] Calculate the cumulative rate: - \(\Lambda(1) = 1 + \frac{1}{2} = 1.5\) - \(\Lambda(2) = 2 + \frac{1}{2}\cdot 4 = 2 + 2 = 4\) So, \[ P(T_1 > 1) = e^{-\Lambda(1)} = e^{-1.5} \] \[ P(T_1 > 2) = e^{-\Lambda(2)} = e^{-4} \] Therefore, \[ \boxed{ P(1 < T_1 < 2) = e^{-1.5} - e^{-4} \approx 0.2231 - 0.0183 = 0.2048 } \] --- ## **2. Probability that the 2nd passenger arrives between 1pm and 2pm** Let \(T_2\) be the time of the 2nd arrival. For NHPP, \(T_k\) (the time of the \(k\)th arrival) has the property that \(\Lambda(T_k)\) is distributed as a Gamma random variable with shape \(k\) and rate \(1\): \[ \Lambda(T_k) \sim \text{Gamma}(k, 1) \] So, \[ P(1 < T_2 < 2) = P(T_2 > 1) - P(T_2 > 2) \] But for \(T_2\), the CDF is: \[ P(T_2 > t) = P(\text{at most one arrival by time } t) = P(N(t) \le 1) = e^{-\Lambda(t)} [1 + \Lambda(t)] \] So, \[ P(T_2 > 1) = e^{-\Lambda(1)} [1 + \Lambda(1)] = e^{-1.5} \times (1 + 1.5) = e^{-1.5} \times 2.5 \] \[ P(T_2 > 2) = e^{-\Lambda(2)} [1 + \Lambda(2)] = e^{-4} \times (1 + 4) = e^{-4} \times 5 \] Thus, \[ P(1 < T_2 < 2) = e^{-1.5} \times 2.5 - e^{-4} \times 5 \] Plug in the numbers: - \(e^{-1.5} \approx 0.2231\) - \(e^{-4} \approx 0.0183\) So, \[ P(1 < T_2 < 2) = 0.2231 \times 2.5 - 0.0183 \times 5 = 0.5578 - 0.0915 = 0.4663 \] \[ \boxed{ P(1 < T_2 < 2) \approx 0.466 } \] --- ## **3. Expectation and SD of number of passengers taken by the first bus** - **Bus arrivals:** Homogeneous Poisson, rate = 2/hr. - **First bus arrival time:** Exponential(2). - **Passengers up to first bus:** This is the number of passenger arrivals up to random time \(S \sim \text{Exp}(2)\). Let’s find \(E[N(S)]\) and \(\operatorname{SD}(N(S))\). ### **A. Expected number** For a NHPP with cumulative rate \(\Lambda(t)\): - If \(S\) is independent of the process, then \[ E[N(S)] = E[\Lambda(S)] \] because the expected number up to a random time \(S\) is the expected value of \(\Lambda(S)\). Here, \[ \Lambda(S) = S + \frac{1}{2}S^2 \] So, \[ E[N(S)] = E[S] + \frac{1}{2} E[S^2] \] For \(S \sim \text{Exp}(2)\), - \(E[S] = 1/2\) - \(E[S^2] = \operatorname{Var}(S) + (E[S])^2 = 1/4 + 1/4 = 1/2\) Therefore, \[ E[N(S)] = \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \] \[ \boxed{E[N(S)] = 0.75} \] --- ### **B. Variance (and SD)** For a NHPP, conditioned on \(S\), \(N(S)\) is Poisson with mean \(\Lambda(S)\): By the law of total variance: \[ \operatorname{Var}(N(S)) = E[\operatorname{Var}(N(S) | S)] + \operatorname{Var}(E[N(S) | S]) \] But \(N(S) | S \sim \mathrm{Poisson}(\Lambda(S))\), so: - \(\operatorname{Var}(N(S) | S) = \Lambda(S)\) - \(E[N(S) | S] = \Lambda(S)\) Therefore, \[ \operatorname{Var}(N(S)) = E[\Lambda(S)] + \operatorname{Var}(\Lambda(S)) \] We already have \(E[\Lambda(S)] = 0.75\). Calculate \(\operatorname{Var}(\Lambda(S))\): \[ \Lambda(S) = S + \frac{1}{2}S^2 \] So, \[ E[\Lambda(S)^2] = E\left[ S^2 + S^3 + \frac{1}{4} S^4 \right] = E[S^2] + E[S^3] + \frac{1}{4}E[S^4] \] Recall for \(S \sim \mathrm{Exp}(\lambda)\): - \(E[S^n] = \frac{n!}{\lambda^n}\) With \(\lambda = 2\): - \(E[S^2] = 2! / 2^2 = 2/4 = 0.5\) - \(E[S^3] = 6 / 8 = 0.75\) - \(E[S^4] = 24 / 16 = 1.5\) So, \[ E[\Lambda(S)^2] = 0.5 + 0.75 + 0.25 \times 1.5 = 0.5 + 0.75 + 0.375 = 1.625 \] Now, \(\operatorname{Var}(\Lambda(S)) = E[\Lambda(S)^2] - (E[\Lambda(S)])^2\): - \(E[\Lambda(S)] = 0.75\) - So \((E[\Lambda(S)])^2 = 0.5625\) Thus, \[ \operatorname{Var}(\Lambda(S)) = 1.625 - 0.5625 = 1.0625 \] Now, \[ \operatorname{Var}(N(S)) = E[\Lambda(S)] + \operatorname{Var}(\Lambda(S)) = 0.75 + 1.0625 = 1.8125 \] \[ \operatorname{SD}(N(S)) = \sqrt{1.8125} \approx 1.346 \] --- ### **Summary Table** | Quantity | Value | |----------|-------| | \(E[N(S)]\) | 0.75 | | \(\operatorname{SD}(N(S))\) | 1.35 (rounded) | --- ## **Final Answers** **1. Probability 1st passenger arrives between 1pm and 2pm:** \[ \boxed{0.205} \] **2. Probability 2nd passenger arrives between 1pm and 2pm:** \[ \boxed{0.466} \] **3. Expectation and SD of number of passengers taken by 1st bus:** \[ \boxed{ \text{Expected: } 0.75;\quad \text{SD: } 1.35 } \] --- **If you want more intuition or step-by-step explanations, just ask!**