Let the random variable representing the distance be denoted as \(X\), where \(X \sim N(35.0, 11.6^2)\).

We need to find \(P(30 < X < 40)\).

First, standardize the variable using the formula for the z-score:

\(z = \frac{x - ext{mean}}{ ext{standard deviation}}\)

For \(x = 40\):

\(z = \frac{40 - 35.0}{11.6} = 0.431\)

For \(x = 30\):

\(z = \frac{30 - 35.0}{11.6} = -0.431\)

Using the standard normal distribution table, find \(\, \Phi(0.431)\) and \(\, \Phi(-0.431)\).

\(P(30 < X < 40) = \Phi(0.431) - (1 - \Phi(0.431))\)

\(= 0.334\)