Given that 88% of shoppers spend more than t minutes, this implies that 12% spend less than t minutes. We need to find the z-score corresponding to the 12th percentile of a standard normal distribution.

The z-score for the 12th percentile is approximately -1.175.

Using the standardization formula:

\(z = \frac{t - \mu}{\sigma}\)

Substitute \(z = -1.175\), \(\mu = 96\), and \(\sigma = 18\):

\(-1.175 = \frac{t - 96}{18}\)

Solving for \(t\):

\(t - 96 = -1.175 \times 18\)

\(t - 96 = -21.15\)

\(t = 96 - 21.15\)

\(t = 74.85\)

Therefore, the value of \(t\) is approximately between 74.85 and 74.9.