50th percentile of $Z = \exp(-0.05 T)$ where $T\sim \frac{2}{10\sqrt{2\pi}}e^{-t^2/200}$ I was asked to find 50th percentile $\xi_Z^{0.5}$ for random variable $Z = \exp(-0.05T)$ where $T$ has half-normal distribution, with p.d.f. of the form $\frac{2}{10\sqrt{2\pi}} e^{-t^2/200}$, where $t>0$.
From definitions we have $0.5 = P(Z\leq \xi_Z^{0.5})$, and if we plug in $\exp(-0.05\xi_T^{0.5})$ then we see that $$P(Z\leq \exp(-0.05\xi_T^{0.5}))=P(T>\xi_T^{0.5}) = 0.5$$ so that necessarily $\xi_Z^{0.5} = \exp(-0.05\xi_T^{0.5})$. Thus it suffices to find 50th percentile of $T$.
We have $$P(T\leq t) = \int_0^t \frac{2}{10\sqrt{2\pi}}e^{-t^2/200}\mathrm{d}t = \int_0^{0.1t} \frac{2}{\sqrt{2\pi}}e^{-t^2/2}\mathrm{d}t = 2(\Phi(0.1t)-0.5)$$ where $\Phi$ is c.d.f. of $N(0, 1)$. Then solving for $\xi_T^{0.5}$ we get $\Phi(0.1\xi_T^{0.5}) = 0.75$ and the 75th percentile of standard normal distribution is equal to $0.67449$ so that $\xi_T^{0.5} = 6.7449$ from which $\xi_Z^{0.5} = \exp(-0.05\cdot 6.7449) = 0.713734$, but according to the answers, this should be $0.7076$.
I want to make sure that my answer is correct.

I get your same result. For $q \in (0,1)$ we have $-\ln(q)>0$ so
$$\begin{aligned}P(Z\leq q)&=P(e^{-\lambda T}\leq q)=\\
&=P(-\lambda T\leq \ln(q))=\\
&=P\bigg(T\geq -\frac{\ln(q)}{\lambda}\bigg)=\\
&=1-\int_{[0,-\ln(q)/\lambda]}\frac{2}{\sqrt{2\pi \sigma^2}}e^{-\frac{t^2}{2\sigma^2}}dt=\\
&=1-\bigg(1+\textrm{erf}\bigg(\frac{-\ln(q)}{\lambda \sqrt{2\sigma^2}}\bigg)-1\bigg)=\\
&=1-\textrm{erf}\bigg(\frac{-\ln(q)}{\lambda \sqrt{2\sigma^2}}\bigg)
\end{aligned}$$
By manipulating the result and equating it to $0.5$
$$\begin{aligned}1-\textrm{erf}\bigg(\frac{-\ln(q)}{\lambda \sqrt{2\sigma^2}}\bigg)&=0.5\\
\textrm{erf}\bigg(\frac{-\ln(q)}{\lambda \sqrt{2\sigma^2}}\bigg)&=0.5\\
\frac{1}{2}\bigg(1+\textrm{erf}\bigg(\frac{-\ln(q)}{\lambda \sqrt{2\sigma^2}}\bigg)\bigg)&=0.75\\
\Phi\bigg(\frac{-\ln(q)}{\lambda\sigma} \bigg)&=0.75\\
q&=e^{-\lambda\sigma \Phi^{-1}(0.75)}\approx 0.7137
\end{aligned}$$
where $\sigma = 10,\,\lambda = 0.05$.
