Since the lifespan is normally distributed, we first find the z-score from the standard normal probability table that is associated with the 4th percentile (0.04 probability). That z-score is -1.75.
Let x be the lifespan associated with the 4th percentile.
The z-score = (x - mean) / standard deviation, where mean given as 13 and std dev given as 1.3.
-1.75 = (x - 13)/1.3. Solve for x.