Given that \(X \sim N(\mu, \sigma^2)\), we have two probabilities:

1. \(P(X > 30.0) = 0.1480\)

2. \(P(X > 20.9) = 0.6228\)

Using the standard normal distribution, we convert these to \(z\)-scores:

For \(X = 30.0\):

\(z = \frac{30 - \mu}{\sigma}\)

From the standard normal table, \(P(Z > 1.045) = 0.1480\), so \(z = 1.045\).

Thus, \(\frac{30 - \mu}{\sigma} = 1.045\).

For \(X = 20.9\):

\(z = \frac{20.9 - \mu}{\sigma}\)

From the standard normal table, \(P(Z > -0.313) = 0.6228\), so \(z = -0.313\).

Thus, \(\frac{20.9 - \mu}{\sigma} = -0.313\).

We now have two equations:

1. \(30 - \mu = 1.045 \sigma\)

2. \(20.9 - \mu = -0.313 \sigma\)

Solving these simultaneously:

Subtract the second equation from the first:

\((30 - \mu) - (20.9 - \mu) = 1.045 \sigma + 0.313 \sigma\)

\(9.1 = 1.358 \sigma\)

\(\sigma = \frac{9.1}{1.358} = 6.70\)

Substitute \(\sigma = 6.70\) into the first equation:

\(30 - \mu = 1.045 \times 6.70\)

\(30 - \mu = 7.0015\)

\(\mu = 30 - 7.0015 = 23.0\)

Thus, \(\mu = 23.0\) and \(\sigma = 6.70\).