(i) To find the value of c, we use the standard normal distribution. Given that 8% of carrots are shorter than c, we find the corresponding z-score for 0.08 in the standard normal distribution table, which is approximately -1.406.

Using the formula for standardization:

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

\(-1.406 = \frac{c - 14.2}{3.6}\)

Solving for c gives:

\(c = 14.2 + (-1.406 \times 3.6)\)

\(c = 9.14 \text{ cm}\)

(ii) To find the probability that at least 2 of the 7 carrots have lengths between 15 and 16 cm, we first find the probability that a single carrot is in this range.

Standardizing the values 15 and 16:

\(z_1 = \frac{15 - 14.2}{3.6} = 0.222\)

\(z_2 = \frac{16 - 14.2}{3.6} = 0.5\)

Using the standard normal distribution table, find:

\(P(0.222 < z < 0.5) = \Phi(0.5) - \Phi(0.222)\)

\(= 0.6915 - 0.5879 = 0.1036\)

Now, use the binomial distribution for 7 trials with probability 0.1036:

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

\(= 1 - (0.8964)^7 - 7 \times (0.8964)^6 \times 0.1036\)

\(= 1 - 0.8413\)

\(= 0.159\)