Let the random variable \(X\) represent the number of seeds that germinate. \(X\) follows a binomial distribution with parameters \(n = 250\) and \(p = 0.86\).

We approximate the binomial distribution with a normal distribution since \(n\) is large and \(p\) is not too close to 0 or 1.

The mean \(\mu\) of the distribution is given by:

\(\mu = np = 250 \times 0.86 = 215\)

The variance \(\sigma^2\) is given by:

\(\sigma^2 = np(1-p) = 250 \times 0.86 \times 0.14 = 30.1\)

Thus, the standard deviation \(\sigma\) is:

\(\sigma = \sqrt{30.1}\)

We use a continuity correction for the normal approximation. We want \(P(X > 210)\), which is approximated by \(P(X > 210.5)\).

Standardizing, we find:

\(P(X > 210.5) = 1 - \Phi\left( \frac{210.5 - 215}{\sqrt{30.1}} \right)\)

\(= 1 - \Phi(-0.820)\)

\(= \Phi(0.820)\)

Using standard normal distribution tables, \(\Phi(0.820) = 0.794\).