(b) To complete the probability distribution table, we need to find \(P(X = 1)\) and \(P(X = 2)\).

Using the binomial distribution formula, \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n = 4\) and \(p = \frac{1}{4}\).

\(P(X = 1) = \binom{4}{1} \left(\frac{1}{4}\right)^1 \left(\frac{3}{4}\right)^3 = \frac{27}{64}\).

\(P(X = 2) = \binom{4}{2} \left(\frac{1}{4}\right)^2 \left(\frac{3}{4}\right)^2 = \frac{27}{128}\).

The completed table is:

(d) Let \(Y\) be the number of times Eli obtains at least two 2s in 96 throws. \(Y\) follows a binomial distribution with \(n = 96\) and \(p = P(X \geq 2)\).

\(P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1)) = 1 - \left(\frac{81}{256} + \frac{27}{64}\right) = \frac{67}{256}\).

Using a normal approximation for \(Y\), \(Y \sim N(\mu, \sigma^2)\) where \(\mu = np = 96 \times \frac{67}{256} = 25.125\) and \(\sigma^2 = np(1-p) = 96 \times \frac{67}{256} \times \frac{189}{256} = 18.549\).

We need \(P(Y < 20)\). Using continuity correction, \(P(Y < 20) \approx P(Z < \frac{19.5 - 25.125}{\sqrt{18.549}})\).

\(Z \approx \frac{19.5 - 25.125}{\sqrt{18.549}} = -1.306\).

\(P(Z < -1.306) = 1 - P(Z < 1.306) = 1 - 0.9042 = 0.0958\).

Thus, \(0.0957 \leq p \leq 0.0958\).