First, standardize the values 1.82 and 1.92 using the formula for the z-score:

\(z_1 = \frac{1.92 - 1.9}{0.15} = 0.1333\)

\(z_2 = \frac{1.82 - 1.9}{0.15} = -0.5333\)

Next, find the probability that a single tyre has a pressure between 1.82 and 1.92 bars:

\(\text{area} = \Phi(0.1333) - \Phi(-0.5333)\)

Using standard normal distribution tables or a calculator:

\(\Phi(0.1333) = 0.5529\)

\(\Phi(-0.5333) = 1 - \Phi(0.5333) = 1 - 0.7029 = 0.2971\)

Thus, the probability for one tyre is:

\(0.5529 - 0.2971 = 0.256\)

Since the tyres are independent, the probability that all four tyres have pressures between 1.82 and 1.92 bars is:

\((0.256)^4 = 0.0043\)