If $X_1$, $X_2$, $X_3$ are mutually independent exponential($\lambda$) random variables, what is the $96$th percentile of $3\min\{X_1,X_2,X_3\}$?
The answer I got is $\frac{-1.44}{\lambda}$ and I'm not sure if that makes sense or not... Could somebody please verify or discredit this answer?
"exponential($\lambda$)" is ambiguous since sometimes it means the density is proportional to $e^{-\lambda x}$ and sometimes it means it's proportional to $e^{-x/\lambda}$.
Assuming the former is meant, we have $\Pr(X_1>x) = e^{-\lambda x}$ and $$ \Pr(\min>x)=\Pr(X_1>x\ \&\ X_2>x\ \&\ X_3>x) = \Big(\Pr(X_1>x)\Big)^3 = \Big(e^{-\lambda x}\Big)^3 = e^{-3\lambda x} $$ So $$ e^{-3\lambda x} = 0.04 \iff x = \frac{\log0.04}{-3\lambda} = \frac{\log 25}{3\lambda}. $$ (Note $1/0.04=25$. The natural logarithm of $25$ is somewhat more than $3$, so the bottom line is somewhat more than $1/\lambda$.)
PS: I neglected the $3$, so let's deal with that. $$ \Pr(3\min>x) = \Pr(\min>x/3), $$ so just put $x/3$ in place of $x$ in what was done above: $$ \Pr(\min>x/3) = \Big(e^{-\lambda x/3}\Big)^3 = e^{-\lambda x}. $$ So we end up with $\dfrac{\log 25}{\lambda}$.