(i) Given that 79% of people have visits lasting less than 10 minutes, we find the z-score corresponding to 0.79, which is 0.807. Using the formula for the z-score:

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

where \(X = 10\), \(\mu = 8.2\), and \(z = 0.807\). Solving for \(\sigma\):

\(0.807 = \frac{10 - 8.2}{\sigma}\)

\(\sigma = \frac{1.8}{0.807} \approx 2.23\)

(ii) We need to find \(P(|X - 8.2| > 1)\), which is equivalent to \(P(X > 9.2) + P(X < 7.2)\). Standardizing these values:

\(P\left(\left| z \right| > \frac{1}{2.23}\right)\)

\(P(|z| > 0.4484) = 2 \times (1 - \Phi(0.4484))\)

\(= 2 \times (1 - 0.6729) = 0.654\)

(iii) Let \(p = 0.21\) be the probability of a visit lasting longer than 10 minutes. We want \(P(X > 2)\) for a binomial distribution with \(n = 6\) and \(p = 0.21\):

\(P(X > 2) = 1 - P(X \leq 2)\)

\(= 1 - \{ (0.79)^6 + 6 \times (0.21) \times (0.79)^5 + 15 \times (0.21)^2 \times (0.79)^4 \}\)

\(= 1 - \{ 0.2621 + 0.2371 + 0.3888 \}\)

\(= 0.112\)