(i) Given that 94% of the letters are within 12 g of the mean, we use the z-score for 47% (since 94% is the total for both tails, 47% for one tail) which is approximately 1.882. The equation is:

\(1.882 = \frac{32 - 20}{\sigma}\)

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

\(\sigma = \frac{12}{1.882} \approx 6.38\)

(ii) To find the probability that a letter weighs more than 13 g, we standardize:

\(P(x > 13) = P\left( z > \frac{13 - 20}{6.38} \right)\)

\(= P(z > -1.0978)\)

Using the standard normal distribution table, \(P(z > -1.0978) = 0.864\).

(iii) The probability that a letter weighs more than 32 g is 0.03 (since 6% are more than 12 g above the mean). We use the binomial distribution for at least 2 out of 7:

\(P(\text{at least 2}) = 1 - P(0, 1)\)

\(= 1 - (0.97)^7 - (0.03)(0.97)^6 \times \binom{7}{1}\)

\(= 1 - (0.97)^7 - 7 \times (0.03)(0.97)^6\)

\(= 0.0171\)