Let the number of adults who own a cell phone be a binomial random variable with parameters \(n = 160\) and \(p = 0.75\).

Calculate the mean and variance:

\(np = 160 \times 0.75 = 120\)

\(npq = 160 \times 0.75 \times 0.25 = 30\)

Use the normal approximation to the binomial distribution:

\(P(X > 114) = P\left( z > \frac{114.5 - 120}{\sqrt{30}} \right)\)

\(= P(z > -1.004)\)

\(= \Phi(1.004) = 0.842\)

Thus, \(P(X > 114) = 1 - 0.842 = 0.158\).

For part (iii), the approximation is justified because both \(np\) and \(nq\) are greater than 5.