Let the number of plants per box be denoted by \(n\), and the probability of a plant producing yellow flowers be \(p\). The number of plants producing yellow flowers follows a binomial distribution \(B(n, p)\).

The mean of a binomial distribution is given by \(np = 11\) and the variance is \(np(1-p) = 4.95\).

From \(np = 11\), we have \(p = \frac{11}{n}\).

Substitute \(p\) into the variance equation:

\(np(1-p) = 4.95\)

\(n \left(\frac{11}{n}\right) \left(1 - \frac{11}{n}\right) = 4.95\)

\(11 \left(1 - \frac{11}{n}\right) = 4.95\)

\(11 - \frac{121}{n} = 4.95\)

\(11 - 4.95 = \frac{121}{n}\)

\(6.05 = \frac{121}{n}\)

\(n = \frac{121}{6.05} \approx 20\)

Thus, there are 20 plants per box.

For part (b), we need to find the probability that exactly 12 plants produce yellow flowers, i.e., \(P(X = 12)\).

Using \(n = 20\) and \(p = \frac{11}{20} = 0.55\), the probability mass function of a binomial distribution is:

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

\(P(X = 12) = \binom{20}{12} (0.55)^{12} (0.45)^{8}\)

\(P(X = 12) \approx 0.162\)