We are given that the daily minimum temperature follows a normal distribution \(N(8, 24)\), where the mean \(\mu = 8\) and the variance \(\sigma^2 = 24\). The standard deviation \(\sigma = \sqrt{24}\).

To find the probability that the temperature is between 7°C and 12°C, we standardize these values to find the corresponding \(z\)-scores.

For 12°C:

\(z_1 = \frac{12 - 8}{\sqrt{24}} = \frac{4}{\sqrt{24}} = 0.816\)

Using the standard normal distribution table, \(\Phi(0.816) = 0.7926\).

For 7°C:

\(z_2 = \frac{7 - 8}{\sqrt{24}} = \frac{-1}{\sqrt{24}} = -0.204\)

Using the standard normal distribution table, \(\Phi(-0.204) = 1 - 0.5808 = 0.4192\).

The probability that the temperature is between 7°C and 12°C is:

\(P(7 < X < 12) = \Phi(0.816) - \Phi(-0.204) = 0.7926 - 0.4192 = 0.373\)