Given that the heights are normally distributed, we have:

\(P(X < 10) = \frac{48}{500} = 0.096\)

Using the standard normal distribution table, \(z = -1.305\) for \(P(Z < z) = 0.096\).

Similarly, \(P(X > 24) = \frac{76}{500} = 0.152\)

Using the standard normal distribution table, \(z = 1.028\) for \(P(Z > z) = 0.152\).

We form the equations:

\(10 - \mu = -1.305\sigma\)

\(24 - \mu = 1.028\sigma\)

Subtracting these equations gives:

\(14 = 2.333\sigma\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{14}{2.333} = 6.00\)

Substituting \(\sigma = 6.00\) back into one of the equations:

\(10 - \mu = -1.305 \times 6.00\)

\(10 - \mu = -7.83\)

\(\mu = 17.83\)

Thus, the mean \(\mu\) is approximately 17.8 and the standard deviation \(\sigma\) is 6.00.