Let the random variable \(X\) represent the number of damaged tapes in a sample of 1600 tapes. Given that 1 in 5 tapes are damaged, the probability of a tape being damaged is \(p = 0.2\).

The expected number of damaged tapes is \(\mu = np = 1600 \times 0.2 = 320\).

The variance is \(\sigma^2 = np(1-p) = 1600 \times 0.2 \times 0.8 = 256\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(320, 256)\).

We need to find \(P(X \geq 290)\). Using continuity correction, this is \(P(X > 289.5)\).

Standardize: \(Z = \frac{289.5 - 320}{\sqrt{256}} = \frac{-30.5}{16} = -1.906\).

Thus, \(P(X \geq 290) = 1 - \Phi(-1.906) = \Phi(1.906)\).

Using standard normal distribution tables, \(\Phi(1.906) = 0.972\).