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 \(0.601\).

(a) To find the probability that Raj runs for more than 43.2 minutes, we use the standard normal distribution. First, we calculate the z-score:

\(z = \frac{43.2 - 41.2}{3.6} = 0.5556\)

We need to find \(P(Z > 0.5556)\). Using the standard normal distribution table, \(\Phi(0.5556) = 0.7108\).

Therefore, \(P(Z > 0.5556) = 1 - 0.7108 = 0.2892\).

So, the probability that Raj runs for more than 43.2 minutes is approximately 0.289.

(b) To find the number of days Raj runs for less than 43.2 minutes, we calculate:

\(P(X < 43.2) = 1 - P(X > 43.2) = 1 - 0.2892 = 0.7108\).

Multiply this probability by the number of days in a year:

\(0.7108 \times 365 = 259.4\).

Therefore, the estimated number of days is approximately 259 or 260.

(a) To find the probability that a randomly chosen employee takes more than 28.6 minutes, we use the standard normal distribution. First, we calculate the z-score:

\(z = \frac{28.6 - 32.2}{9.6} = -0.375\)

We need \(P(Z > -0.375)\). Using the standard normal distribution table, \(\Phi(-0.375) = 0.3538\). Therefore, \(P(Z > -0.375) = 1 - 0.3538 = 0.6462\).

Thus, the probability is approximately 0.646.

(c) We need to find the probability that the time differs from the mean by less than 15.0 minutes, i.e., \(|X - 32.2| < 15\).

This is equivalent to \(17.2 < X < 47.2\). We calculate the z-scores:

\(z_1 = \frac{17.2 - 32.2}{9.6} = -1.5625\)

\(z_2 = \frac{47.2 - 32.2}{9.6} = 1.5625\)

We need \(P(-1.5625 < Z < 1.5625)\). Using the standard normal distribution table, \(\Phi(1.5625) = 0.9409\).

Thus, \(P(-1.5625 < Z < 1.5625) = 2 \times 0.9409 - 1 = 0.8818\).

Therefore, the probability is approximately 0.882.

(i) To find the number of days Karli spends more than 142 minutes on social media, we first calculate the probability that she spends more than 142 minutes on a given day. We use the standard normal distribution:

\(P(X > 142) = P\left( Z > \frac{142 - 125}{24} \right) = P(Z > 0.7083)\)

Using the standard normal distribution table, \(P(Z > 0.7083) = 1 - 0.7604 = 0.2396\).

Therefore, the expected number of days is \(0.2396 \times 365 = 87.454\).

Rounding to the nearest whole number, we get 87 or 88 days.

(ii) We need to find the probability that Karli spends more than 142 minutes on social media on fewer than 2 out of 10 days. This is a binomial probability problem with \(n = 10\) and \(p = 0.2396\).

\(P(0, 1) = 0.7604^{10} + \binom{10}{1} \times 0.2396^1 \times 0.7604^9\)

\(= 0.064628 + 0.20364\)

\(= 0.268\)

We need to find the probability that a rod's length is within 0.5 cm of the mean, i.e., between 24.7 cm and 25.7 cm.

Using the standardization formula, we calculate:

\(P\left( \frac{25.2 - (25.2 + 0.5)}{0.4} < z < \frac{25.2 - (25.2 - 0.5)}{0.4} \right)\)

This simplifies to:

\(P\left( \frac{-0.5}{0.4} < z < \frac{0.5}{0.4} \right) = P(-1.25 < z < 1.25)\)

Using the standard normal distribution table, we find:

\(P(-1.25 < z < 1.25) = 2 \Phi(1.25) - 1\)

\(\Phi(1.25) \approx 0.8944\)

Thus, \(2 \times 0.8944 - 1 = 0.7888\)

The expected number of rods is:

\(0.7888 \times 500 = 394.4\)

Rounding gives 394 or 395 rods.