The probability of exactly 7 passengers carrying backpacks in a minibus is given by the binomial probability:

\(P(7) = \binom{12}{7} (0.65)^7 (0.35)^5 = 0.2039\)

For 250 minibuses, the mean and variance of the number of minibuses with exactly 7 passengers carrying backpacks are:

Mean: \(250 \times 0.2039 = 50.9798\)

Variance: \(250 \times 0.2039 \times (1 - 0.2039) = 40.5851\)

We approximate the distribution of the number of minibuses with exactly 7 passengers carrying backpacks by a normal distribution with the above mean and variance.

We need to find \(P(X > 54)\), where \(X\) is the number of minibuses with exactly 7 passengers carrying backpacks.

Using continuity correction, we find:

\(P(X > 54) = P\left( \frac{54.5 - 50.9798}{\sqrt{40.5851}} \right) = P(Z > 0.5526)\)

Using the standard normal distribution table, \(P(Z > 0.5526) = 1 - \Phi(0.5526) = 1 - 0.7098 = 0.2902\)

Thus, the probability is approximately \(0.290\).