The problem states that the daily minimum temperature follows a normal distribution \(N(\mu, 40.0)\). We are given that the probability of the temperature being above 0°C is 0.8888.

Using the standard normal distribution table, a probability of 0.8888 corresponds to a \(z\)-score of approximately 1.22. However, since we are dealing with the probability above 0°C, we use \(z = -1.22\).

The \(z\)-score formula is:

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

where \(X = 0\) (the temperature), \(\mu\) is the mean, and \(\sigma = \sqrt{40}\).

Substitute the values into the formula:

\(-1.22 = \frac{0 - \mu}{\sqrt{40}}\)

Solving for \(\mu\):

\(-1.22 \times \sqrt{40} = -\mu\)

\(\mu = 1.22 \times \sqrt{40}\)

\(\mu \approx 7.72\)