Which sample,$A$ or $B$ shown in the figure,has a shorter mean-life?

(B) The activity of a radioactive sample is given by $I = I_0 e^{-\lambda t}$,where $\lambda$ is the decay constant. The mean-life $\tau$ is defined as $\tau = \frac{1}{\lambda}$.
From the given figure,at any time $t_0$,the activity of sample $B$ $(I_B)$ is less than the activity of sample $A$ $(I_A)$,i.e.,$I_B < I_A$.
Since both samples start with the same initial activity $I_0$,we have:
$I_A = I_0 e^{-\lambda_A t_0}$
$I_B = I_0 e^{-\lambda_B t_0}$
Given $I_B < I_A$,it follows that $e^{-\lambda_B t_0} < e^{-\lambda_A t_0}$.
Taking the natural logarithm on both sides:
$-\lambda_B t_0 < -\lambda_A t_0$
$\lambda_B > \lambda_A$
Since the mean-life $\tau$ is inversely proportional to the decay constant $\lambda$ $(\tau = \frac{1}{\lambda})$,a larger decay constant implies a shorter mean-life.
Therefore,$\tau_B < \tau_A$.
Thus,sample $B$ has a shorter mean-life.