Given that 87.5% of the batteries last longer than 70 hours, we find the corresponding z-score for the 12.5th percentile (since 100% - 87.5% = 12.5%).

The z-score for the 12.5th percentile is approximately \(z = -1.151\).

Using the standardization formula:

\(z = \frac{X - \mu}{s}\)

Substitute the known values:

\(-1.151 = \frac{70 - 120}{s}\)

Solve for \(s\):

\(-1.151s = 70 - 120\)

\(-1.151s = -50\)

\(s = \frac{-50}{-1.151}\)

\(s \approx 43.4 \text{ or } 43.5\)