Let the time taken be represented by the random variable \(X\), where \(X \sim N(85, 6.8^2)\).

We need to find \(P(79 < X < 91)\).

Standardize the variable using the formula \(Z = \frac{X - \mu}{\sigma}\), where \(\mu = 85\) and \(\sigma = 6.8\).

Calculate the z-scores:

For \(X = 79\), \(Z = \frac{79 - 85}{6.8} = -0.8824\).

For \(X = 91\), \(Z = \frac{91 - 85}{6.8} = 0.8824\).

Thus, \(P(79 < X < 91) = P(-0.8824 < Z < 0.8824)\).

Using the standard normal distribution table, find:

\(\Phi(0.8824) = 0.8111\) and \(\Phi(-0.8824) = 1 - 0.8111 = 0.1889\).

Therefore, \(P(-0.8824 < Z < 0.8824) = 0.8111 - 0.1889 = 0.622\).