(i) To find the probability that a pencil has a length greater than 10.9 cm, we standardize the normal distribution:

\(P(x > 10.9) = P\left(z > \frac{10.9 - 11}{0.095}\right)\)

\(= P(z > -1.0526)\)

Using the standard normal distribution table, \(P(z > -1.0526) = 0.8538\), which rounds to 0.854.

(ii) To find the probability that at least two pencils in a sample of 6 have lengths less than 10.9 cm, we first find the probability of a single pencil being less than 10.9 cm:

\(P(x < 10.9) = 1 - P(x > 10.9) = 1 - 0.8538 = 0.1462\)

We use the binomial distribution for 6 trials with probability \(p = 0.1462\):

\(P(\text{at least 2} < 10.9) = 1 - P(0, 1)\)

\(= 1 - \left(0.8538^6 + 6 \cdot 0.1462 \cdot 0.8538^5\right)\)

\(= 1 - (0.8538^6 + 6 \cdot 0.1462 \cdot 0.8538^5)\)

\(= 0.215\)