What is average life in radioactivity and what is its significance? By definition, average life of radioactive sample is the amount of time required for it to get decayed to 36.8% of its original amount.
But what is the significance of 36.8% and why has that value been chosen?
 A: When radioactive element A decays to produce element B, the (infinitesimal) number of decayed elements A, $dN$, that occurs in a small time interval, $dt$, is proportional to the initial population of A, $N$:
$$
-\frac{dN}{dt}\propto N
$$
Assuming the proportionality is a constant, then the above becomes
$$
-\frac{dN}{dt}=\lambda N
$$
which has a known solution:
$$
N(t)=N(0)e^{-\lambda t}=N(0)e^{-t/\tau}
$$
where $\tau$ is the mean lifetime. Now when 36.8% of the initial material remains then, $N(t)/N(0)=0.368$ and we get
$$
0.368=e^{-t/\tau}
$$
Taking the natural logarithm of this to eliminate the exponential, we find that
$$
\ln0.368=-1=-\frac{t}{\tau}
$$
Thus, the value 36.8% signifies the amount of material left over after one mean lifetime has passed.
A: I endorse Kyle's answer. Just two short comments.
The number 36.8% is literally
$$ 36.8 \approx 100 \exp(-1) =\frac{100}{2.71828\dots} $$
Moreover, it is right to call this quantity "average lifetime" or just "lifetime" because it is literally the average value of the time for which a nucleus (or something else) from the ensemble lives.
If the initial number is $N_0$, they decrease to
$$ N(t) = N_0 \cdot \exp (-t/t_0)$$
at time $t$ where $t_0$ is what we want to call the (average) lifetime. How many nuclei $dN\lt 0$ die (decay) in the short interval $(t,t+dt)$? Well, it's given by the derivative
$$ dN = dt\cdot \frac{dN(t)}{dt} = N_0\cdot dt\cdot \exp(-t/t_0)\cdot \left(-\frac{1}{t_0}\right)$$
To calculate the average "age at death" (a statistical expectation value), we must integrate
$$ \langle t \rangle = \int_0^\infty dt\cdot t\cdot P({\rm lifetime}=t) =\\
= -\int_0^\infty t\cdot  \frac{1}{N_0} \cdot dN/dt \cdot dt = \int_0^\infty dt\cdot t\cdot \exp(-t/t_0)\frac{1}{t_0} = t_0$$
where $P$ refers to the probability density that the lifetime was $t$ which can be calculated by integration by parts. So the average "age at death" for a large ensemble of nuclei will really be equal to the $t_0$ that appears in the exponent of $\exp(-t/t_0)$.
