Let the random variable \(X\) represent the number of overweight people in the sample of 400. Given that 2 out of 5 people are overweight, the probability \(p\) of a person being overweight is \(p = \frac{2}{5} = 0.4\).

The expected number of overweight people \(\mu\) is given by \(\mu = np = 400 \times 0.4 = 160\).

The variance \(\sigma^2\) is given by \(\sigma^2 = np(1-p) = 400 \times 0.4 \times 0.6 = 96\).

We use a normal approximation to the binomial distribution. The standard deviation \(\sigma\) is \(\sqrt{96} \approx 9.798\).

We apply continuity correction for the normal approximation: \(P(X < 165) \approx P(X \leq 164.5)\).

Standardizing, we find:

\(Z = \frac{164.5 - 160}{\sqrt{96}} = \frac{4.5}{9.798} \approx 0.4593\)

Using standard normal distribution tables, \(\Phi(0.4593) \approx 0.677\).

Thus, the probability that fewer than 165 people in the sample are overweight is approximately 0.677.