Given that the probability \(P(X > 195) = 0.128\), we need to find the mean \(\mu\) of the normal distribution.

First, find the z-score corresponding to the probability 0.128. From standard normal distribution tables, \(P(Z > 1.136) = 0.128\), so \(z = 1.136\).

Using the standardization formula:

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

Substitute the known values:

\(1.136 = \frac{195 - \mu}{22}\)

Solving for \(\mu\):

\(1.136 \times 22 = 195 - \mu\)

\(24.992 = 195 - \mu\)

\(\mu = 195 - 24.992\)

\(\mu = 170.008\)

Rounding to the nearest whole number, \(\mu = 170\).