Let the random variable \(X\) represent the weight of a large bag of pasta. \(X\) is normally distributed with mean \(\mu = 1.5\) kg and standard deviation \(\sigma = 0.05\) kg.

We need to find \(P(1.42 < X < 1.52)\).

Standardize the variable using the formula:

\(Z = \frac{X - \mu}{\sigma}\)

For \(X = 1.42\):

\(Z = \frac{1.42 - 1.5}{0.05} = -1.6\)

For \(X = 1.52\):

\(Z = \frac{1.52 - 1.5}{0.05} = 0.4\)

Thus, \(P(1.42 < X < 1.52) = P(-1.6 < Z < 0.4)\).

Using standard normal distribution tables or a calculator, find:

\(\Phi(0.4) = 0.6554\)

\(\Phi(-1.6) = 0.0548\)

Therefore, \(P(-1.6 < Z < 0.4) = \Phi(0.4) - \Phi(-1.6) = 0.6554 - 0.0548 = 0.6006\).

Rounding to three decimal places, the probability is \(0.601\).