Problem source

Fifty times a year, 1024 tourists disembark from a cruise liner at a port. From there they must travel to the city centre either by bus or by taxi. Tourists  are equally likely to be directed to the bus station or to the taxi rank. Each bus of the bus company holds 32 passengers, and the company currently runs 15 buses. The company makes a profit of $\pounds$1 for each passenger carried. It carries as many passengers as it can, with any excess being (eventually) transported by taxi. Show that  the largest annual licence fee, in pounds, that  the company should consider paying to be allowed to run an extra bus is approximately
\[
1600 \Phi(2) -  \frac{800}{\sqrt{2\pi}}\big(1- \e^{-2}\big)\,,
\]
where  
$\displaystyle \Phi(x) =\dfrac1{\sqrt{2\pi}} \int_{-\infty}^x \e^{-\frac12t^2}\d t\,$.
\textit{You should not consider continuity corrections.}

Solution source

The the number of people being directed towards the buses (each cruise) is $X \sim B(1024, \tfrac12) \approx N(512, 256) \approx 16Z + 512$. Therefore without an extra bus, the expected profit is $\mathbb{E}[\min(X, 15 \times 32)]$. With the extra bus, the extra profit is $\mathbb{E}[\min(X, 16 \times 32)]$, therefore the expected extra profit is: $\mathbb{E}[\min(X, 16 \times 32)]-\mathbb{E}[\min(X, 15 \times 32)] = \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] $

\begin{align*}
\text{Expected extra profit} &=  \mathbb{E}[\min(X, 16 \times 32)-\min(X, 15 \times 32)] \\
&=  \mathbb{E}[\min(16Z+512, 16 \times 32)-\min(16Z+512, 15 \times 32)] \\
&=  16\mathbb{E}[\min(Z+32, 32)-\min(Z+32, 30)] \\
&=16\int_{-\infty}^{\infty} \left (\min(Z+32, 32)-\min(Z+32, 30) \right)p_Z(z) \d z \\
&= 16 \left ( \int_{-2}^{0} (z+32-30) p_Z(z) \d z + \int_0^\infty (32-30)p_Z(z) \d z \right) \\
&= 16 \left ( \int_{-2}^{0} (z+2) p_Z(z) \d z + \int_0^\infty 2p_Z(z) \d z \right) \\
&= 16 \left ( \int_{-2}^{0} zp_Z(z) \d z + 2\int_{-2}^\infty p_Z(z) \d z \right) \\
&= 16 \left ( \int_{-2}^{0} z \frac{1}{\sqrt{2\pi}} e^{-\frac12 z^2} \d z + 2(1-\Phi(2)) \right) \\
&= 32(1-\Phi(2)) + \frac{16}{\sqrt{2\pi}} \left [ -e^{-\frac12z^2} \right]_{-2}^0 \\
&= 32(1-\Phi(2)) - \frac{16}{\sqrt{2\pi}} \left ( 1-e^{-2}\right)
\end{align*}

Across $50$ different runs, this profit is  \[ 1600(1-\Phi(2)) - \frac{800}{\sqrt{2\pi}} \left ( 1-e^{-2}\right) \]