Let the probability that a value is less than 5 be denoted by \(p\). Since the probability that a value is greater than 5 is 0.15, we have \(p = 0.85\).

The mean number of values less than 5 in 200 trials is \(\mu = 200 \times 0.85 = 170\).

The variance is \(\text{var} = 200 \times 0.85 \times 0.15 = 25.5\).

We need to find \(P(\text{at least 160})\), which is equivalent to \(P(X \geq 160)\).

Using continuity correction, this becomes \(P(X > 159.5)\).

Standardizing, we have:

\(P\left( z > \frac{159.5 - 170}{\sqrt{25.5}} \right) = P(z > -2.079)\)

Using standard normal distribution tables, \(P(z > -2.079) = 0.981\).