Peter has an appointment at $11:00$. He wants to arrive between $10:40$ and $11:00$. Suppose that, if he leaves from home $t$ minutes after $10:00$, then the time of his arrival has normal distribution with mean $t + 30$ and variance $5$ (time is measured in minutes).
$(a)$ How to choose $t$ s.t. the probability of arriving between $10:40$ and $11:00$ is maximum? Justify your statement by proving that the probability is smaller for any other $t$.
$(b)$ With this optimal $t$, given that he arrives after $11:00$, what is the probability that he arrives after $11:10$?
To put in another way, in the case of being late, what is the probability that the delay is at least ten minutes?
My solution: using The normal law of probability distribution we can found: $$f(x)=\frac{1}{var \sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2var^2}}=$$ $$=\frac{1}{5 \sqrt{2\pi}}e^{-\frac{(x-t-30)^2}{50}}$$
(a) $P(10:40<x<11:00)=Φ(\frac{10}{5})-Φ(\frac{5}{5})=Φ(2)-Φ(1)=0.9772-0.8413=0.1359$ (b) I dont know
Part (a)
From your problem statement, the probability that Peter arrives between 10:40-11:00 if he leaves home $t$ minutes after 10:00 is $$p(t):=\int_{40}^{60}\,\frac{\text{e}^{-\frac{(x-t-30)^2}{2\cdot 5}}}{\sqrt{2\pi\cdot 5}}\,\text{d}x=\int_{40}^{60}\,\frac{\text{e}^{-\frac{(x-t-30)^2}{10}}}{\sqrt{10\pi}}\,\text{d}x\,,$$ whence $$p'(t)=\int_{40}^{60}\,\frac{(x-t-30)\,\text{e}^{-\frac{(x-t-30)^2}{10}}}{5\sqrt{10\pi}}\,\text{d}x\,.$$ Write $u:=x-t-30$. We have $$p'(t)=\int_{10-t}^{30-t}\,\frac{u\,\text{e}^{-\frac{u^2}{10}}}{5\sqrt{10\pi}}\,\text{d}u.$$
If $t\ge 30$, then clearly, $p'(t)<0$. If $20<t<30$, then $$p'(t)=\int_{10-t}^{t-30}\,\frac{u\,\text{e}^{-\frac{u^2}{10}}}{5\sqrt{10\pi}}\,\text{d}u\,.$$ If $10<t<20$, then $$p'(t)=\int_{t-10}^{t-30}\,\frac{u\,\text{e}^{-\frac{u^2}{10}}}{5\sqrt{10\pi}}\,\text{d}u>0\,.$$ If $t\leq 10$, then clearly, $p'(t)>0$. Therefore, the function $p:\mathbb{R}\to\mathbb{R}$ is strictly increasing on the interval $(-\infty,20)$, and strictly decreasing on the interval $t\in(20,\infty)$. Thus, $p(t)$ is maximized at $t=20$ (see also here).
The maximum value $p_\text{max}$ of $p(t)$ is given by $p(20)$. That is, $$p_\text{max}=\int_{40}^{60}\,\frac{\text{e}^{-\frac{(x-20-30)^2}{10}}}{\sqrt{10\pi}}\,\text{d}x=\int_{-10}^{+10}\,\frac{\text{e}^{-\frac{u^2}{10}}}{\sqrt{10\pi}}\,\text{d}u\,.$$ Consequently, if $v:=\dfrac{u}{\sqrt{10}}$, then $$p_\text{max}=\int_{-\sqrt{10}}^{+\sqrt{10}}\,\frac{\text{e}^{-v^2}}{\sqrt{\pi}}\,\text{d}v=\text{erf}(\sqrt{10})\approx 1-7.74422\times{10}^{-6}\,,$$ where $\text{erf}$ is the error function.
Part (b)
Denote by $X$ the time in minutes after 10:00 that Peter arrives. Using $t:=20$, we have $$\begin{align}\mathbb{P}[X>60]&=\int_{60}^{\infty}\,\frac{\text{e}^{-\frac{(x-20-30)^2}{10}}}{\sqrt{10\pi}}\,\text{d}x=\int_{\sqrt{10}}^\infty\,\frac{\text{e}^{-v^2}}{\sqrt{\pi}}\,\text{d}v\\&=\frac{\text{erfc}(\sqrt{10})}{2}=\frac{1-p_\text{max}}{2}\approx 3.87211\times 10^{-6}\,,\end{align}$$ where $\text{erfc}$ is the complementary error function. Moreover, $$\begin{align}\mathbb{P}[X>70]&=\int_{70}^{\infty}\,\frac{\text{e}^{-\frac{(x-20-30)^2}{10}}}{\sqrt{10\pi}}\,\text{d}x=\int_{2\sqrt{10}}^\infty\,\frac{\text{e}^{-v^2}}{\sqrt{\pi}}\,\text{d}v\\&=\frac{\text{erfc}(2\sqrt{10})}{2}\approx 1.87205\times 10^{-19}\,.\end{align}$$ That is, $$\mathbb{P}[X>70|X>60]=\frac{\mathbb{P}[X>70]}{\mathbb{P}[X>60]}=\frac{\text{erfc}(2\sqrt{10})}{\text{erfc}(\sqrt{10})}\approx 4.83470\times 10^{-14}\,.$$
In terms of the cumulative distribution function $\Phi$ of the standard normal distribution, we have $$\text{erf}(z)=2\,\Phi(\sqrt{2}\,z)-1$$ and $$\text{erfc}(z)=1-\text{erf}(z)=2\,\big(1-\Phi(\sqrt{2}z)\big)$$ for all $z\in\mathbb{R}$. Ergo, $$p_\text{max}=2\,\Phi(2\sqrt{5})-1\,,$$ $$\mathbb{P}[X>60]=1-\Phi(2\sqrt{5})\,,$$ $$\mathbb{P}[X>70]=1-\Phi(4\sqrt{5})\,,$$ and $$\mathbb{P}[X>70|X>60]=\frac{1-\Phi(2\sqrt{5})}{1-\Phi(4\sqrt{5})}\,.$$