Given that X ~ N(20, 49), we have a normal distribution with mean bc = 20 and variance 3c2 = 49, so the standard deviation 3c = 7.

\(We need to find k such that P(X > k) = 0.25. This implies that P(X 3c k) = 0.75.\)

Using the standard normal distribution table, we find the z-score corresponding to 0.75 is approximately 0.674.

Standardizing, we have:

\(3cbr>z = 3cfrac{k - 20}{7}\)

Substituting the z-score, we get:

\(0.674 = 3cfrac{k - 20}{7}\)

Solving for k:

\(k - 20 = 0.674 3c 7\)

\(k = 0.674 3c 7 + 20\)

\(k = 4.718 + 20\)

\(k = 24.7\)