Given that the tyre pressures follow a normal distribution with mean \(\mu = 1.9\) bars and standard deviation \(\sigma = 0.15\) bars, we need to find the value of \(b\) such that 80% of the data lies within \(1.9 - b\) and \(1.9 + b\).

Since 80% of the data is within these limits, we have \(P(1.9 - b < X < 1.9 + b) = 0.8\).

This corresponds to a cumulative probability of 0.8, which means the z-scores are \(\pm 1.282\) (or approximately \(\pm 1.28\)).

Using the z-score formula: \(z = \frac{b}{0.15}\), we have:

\(\pm 1.282 = \frac{b}{0.15}\)

Solving for \(b\), we get:

\(b = 1.282 \times 0.15 = 0.1923\)

Thus, the safety limits are:

\(1.9 - 0.1923 = 1.7077 \approx 1.71\)

\(1.9 + 0.1923 = 2.0923 \approx 2.09\)

Therefore, the limits are between 1.71 and 2.09 bars.