Let the number of people with visits lasting less than 8.2 minutes be a binomial random variable with parameters:

\(n = 35\) (number of trials) and \(p = 0.5\) (probability of success for each trial).

The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = n \times p = 35 \times 0.5 = 17.5\)

\(\sigma^2 = n \times p \times (1-p) = 35 \times 0.5 \times 0.5 = 8.75\)

We approximate the binomial distribution with a normal distribution \(N(17.5, 8.75)\).

Using continuity correction, we find:

\(P(X < 16) = \Phi\left( \frac{15.5 - 17.5}{\sqrt{8.75}} \right)\)

\(= 1 - \Phi(0.676)\)

\(= 1 - 0.7505\)

\(= 0.2495 \approx 0.250\)

Alternatively, using the binomial distribution directly:

\(P(X < 16) = \sum_{x=0}^{15} \binom{35}{x} (0.5)^x (0.5)^{35-x}\)

\(= 0.250\)