Let the mean be \(\mu\) and the standard deviation be \(\sigma\). We are given that \(\mu = 5\sigma\).

The probability \(P(Y > 20) = 0.0732\) corresponds to a standard normal variable \(Z\) such that \(P(Z > z) = 0.0732\). From standard normal distribution tables, \(z \approx 1.452\).

Using the transformation for a normal distribution, \(Z = \frac{Y - \mu}{\sigma}\), we have:

\(1.452 = \frac{20 - \mu}{\mu / 5}\)

Solving for \(\mu\):

\(1.452 = \frac{20 - \mu}{\mu / 5}\)

\(1.452 \times \frac{\mu}{5} = 20 - \mu\)

\(0.2904\mu = 20 - \mu\)

\(1.2904\mu = 20\)

\(\mu = \frac{20}{1.2904} \approx 15.5\)