The probability that a person is neither a student, office worker, nor shop assistant is given by:

\(1 - (0.36 + 0.22 + 0.29) = 0.13\)

For a binomial distribution with parameters \(n = 300\) and \(p = 0.13\), the mean and variance are:

\(\text{mean} = 300 \times 0.13 = 39\)

\(\text{variance} = 300 \times 0.13 \times 0.87 = 33.93\)

We approximate the binomial distribution with a normal distribution \(N(39, 33.93)\).

Using continuity correction, we find:

\(P(31 \, \text{to} \, 49) = P(30.5 < x < 49.5)\)

Standardizing, we have:

\(P\left( \frac{30.5 - 39}{\sqrt{33.93}} < z < \frac{49.5 - 39}{\sqrt{33.93}} \right)\)

\(= P(-1.4592 < z < 1.8026)\)

Using the standard normal distribution table:

\(= \Phi(1.8026) + \Phi(1.4592) - 1\)

\(= 0.9643 + 0.9278 - 1\)

\(= 0.892\)