Let the mean be \(\mu\) and the standard deviation be \(s\). We know that \(\mu = 4s\).

The probability that a value is greater than 5 is 0.15, which corresponds to a z-score of approximately 1.036 or 1.037.

Using the z-score formula:

\(z = \frac{5 - \mu}{s}\)

Substitute \(\mu = 4s\) into the equation:

\(1.036 = \frac{5 - 4s}{s}\)

Solve for \(s\):

\(1.036s = 5 - 4s\)

\(5.036s = 5\)

\(s = \frac{5}{5.036} \approx 0.993\)

Substitute back to find \(\mu\):

\(\mu = 4s = 4 \times 0.993 = 3.97\)