The problem involves a binomial distribution where the probability of being an office worker is given as 0.22. The probability of not being an office worker is therefore 1 - 0.22 = 0.78.

We need to find the probability that between 4 and 6 people inclusive are office workers out of 8 people. This is calculated using the binomial probability formula:

\(P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}\)

where \(n = 8\), \(p = 0.22\), and \(r\) is the number of office workers.

Calculate for \(r = 4, 5, 6\):

\(P(4) = \binom{8}{4} (0.22)^4 (0.78)^4\)

\(P(5) = \binom{8}{5} (0.22)^5 (0.78)^3\)

\(P(6) = \binom{8}{6} (0.22)^6 (0.78)^2\)

Sum these probabilities:

\(P(4, 5, 6) = P(4) + P(5) + P(6)\)

\(= \binom{8}{4} (0.22)^4 (0.78)^4 + \binom{8}{5} (0.22)^5 (0.78)^3 + \binom{8}{6} (0.22)^6 (0.78)^2\)

\(= 0.0763\)