We need to find \(P(Y > 0)\). Since \(Y\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\), we can standardize \(Y\) to a standard normal variable \(Z\) using the formula:

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

Thus, \(P(Y > 0) = P\left( Z > \frac{0 - \mu}{\sigma} \right)\).

Given \(4\sigma = 3\mu\), we can express \(\mu\) in terms of \(\sigma\):

\(\mu = \frac{4\sigma}{3}\)

Substitute \(\mu\) into the standardization formula:

\(P\left( Z > \frac{0 - \frac{4\sigma}{3}}{\sigma} \right) = P\left( Z > \frac{-4\sigma/3}{\sigma} \right) = P\left( Z > -\frac{4}{3} \right)\)

Using the standard normal distribution table, \(P(Z > -4/3) = 0.909\).

To find the probability that \(X < 29.4\), we first standardize the variable using the formula for the standard normal distribution:

\(P(X < 29.4) = P\left(Z < \frac{29.4 - 31.4}{\sqrt{3.6}}\right)\)

Calculating the standardized value:

\(\frac{29.4 - 31.4}{\sqrt{3.6}} = -1.0541\)

Thus, \(P(X < 29.4) = P(Z < -1.0541)\).

Using the standard normal distribution table, we find:

\(P(Z < -1.0541) = 1 - 0.8540 = 0.146\)

Therefore, the probability that a randomly chosen value of \(X\) is less than 29.4 is 0.146.

First, we standardize the time of 20 minutes using the formula for the standard normal distribution:

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

where \(X = 20\), \(\mu = 14.6\), and \(\sigma = 5.2\).

\(Z = \frac{20 - 14.6}{5.2} = 1.03846\)

Next, we find the probability that \(Z > 1.03846\). Using standard normal distribution tables or a calculator, we find:

\(P(Z > 1.03846) = 1 - P(Z < 1.03846) = 1 - 0.8504 = 0.1496\)

This means approximately 14.96% of students take more than 20 minutes.

In a sample of 250 students, the expected number is:

\(250 \times 0.1496 = 37.4\)

Rounding to the nearest whole number, we expect about 37 or 38 students to take more than 20 minutes.

We start by standardizing the variable. We want to find \(P(X < 3\mu)\).

Standardizing, we have:

\(P(X < 3\mu) = P\left( z < \frac{3\mu - \mu}{4\mu/3} \right)\)

\(= P\left( z < \frac{9\sigma/4 - 3\sigma/4}{\sigma} \right)\)

\(= P\left( z < \frac{6}{4} \right)\)

Using the standard normal distribution table, \(P(z < 1.5) = 0.933\).

To find \(P(X > 0)\), we standardize the variable using the formula for the standard normal distribution:

\(P(X > 0) = P\left( z > \frac{0 - \mu}{\sigma} \right)\)

Substitute \(\mu = 1.5\sigma\):

\(P\left( z > \frac{0 - 1.5\sigma}{\sigma} \right) = P(z > -1.5)\)

Using the standard normal distribution table, \(P(z > -1.5) = 0.933\).