Given that \(Y \sim N(\mu, \sigma^2)\) and \(2\sigma = 3\mu\), we need to find \(P(Y > 4\mu)\).

First, express \(\sigma\) in terms of \(\mu\):

\(2\sigma = 3\mu \Rightarrow \sigma = \frac{3\mu}{2}\)

Standardize the variable:

\(P(Y > 4\mu) = P\left( z > \frac{4\mu - \mu}{\sigma} \right) = P\left( z > \frac{3\mu}{\frac{3\mu}{2}} \right) = P(z > 2)\)

Using the standard normal distribution table, \(P(z > 2) = 1 - P(z \leq 2) = 1 - 0.9772 = 0.0228\).

(i) To find the number of sheep weighing between 70 kg and 72.5 kg, we first standardize the weights using the formula for the z-score:

\(z_1 = \frac{70 - 66.4}{5.6} = 0.6429\)

\(z_2 = \frac{72.5 - 66.4}{5.6} = 1.089\)

Next, we find the area under the standard normal curve between these z-scores:

\(\Phi(1.089) - \Phi(0.643) = 0.8620 - 0.7399 = 0.1221\)

Multiply this probability by the total number of sheep:

\(0.1221 \times 250 = 30.5\)

Therefore, approximately 30 or 31 sheep have a weight between 70 kg and 72.5 kg.

(ii) To find the value of \(y\), we use the symmetry of the normal distribution. The mean is 66.4 kg, and the proportion of sheep weighing less than 59.2 kg is equal to the proportion weighing more than \(y\) kg. Calculate the difference from the mean:

\(66.4 - 59.2 = 7.2\)

Thus, \(y = 66.4 + 7.2 = 73.6\) kg.

Let the random variable representing petrol consumption be denoted by \(X\), where \(X \sim N(24, 4.7^2)\).

We need to find \(P(21.6 < X < 28.7)\).

Standardize the variable using the formula:

\(z = \frac{x - \mu}{\sigma}\)

For \(x = 21.6\):

\(z_1 = \frac{21.6 - 24}{4.7} = -0.5106\)

For \(x = 28.7\):

\(z_2 = \frac{28.7 - 24}{4.7} = 1\)

Thus, we need to find \(P(-0.5106 < z < 1)\).

This is equivalent to \(\Phi(1) - \Phi(-0.5106)\).

Using standard normal distribution tables:

\(\Phi(1) = 0.8413\)

\(\Phi(-0.5106) = 1 - \Phi(0.5106) = 1 - 0.6953 = 0.3047\)

Therefore, \(P(-0.5106 < z < 1) = 0.8413 - 0.3047 = 0.537\).

First, we find the probability that a single flower pot has a base diameter between 13.6 cm and 14.8 cm. We standardize the values using the normal distribution:

\(P(13.6 < X < 14.8) = P\left( \frac{13.6 - 14}{0.52} < z < \frac{14.8 - 14}{0.52} \right)\)

\(= P(-0.7692 < z < 1.538)\)

Using the standard normal distribution table, we find:

\(= \Phi(1.538) - [1 - \Phi(0.7692)]\)

\(= 0.9380 - [1 - 0.7791]\)

\(= 0.7171\)

Now, we use the binomial distribution to find the probability that exactly 8 out of 10 flower pots have diameters in this range:

\(P(8) = (0.7171)^8 (0.2829)^2 \binom{10}{8}\)

\(= 0.252\)

To find the probability that \(X < -2.4\), we first standardize the variable using the formula for the standard normal distribution:

\(P(X < -2.4) = P\left( Z < \frac{-2.4 - 1.5}{3.2} \right)\)

Calculating the standardized value:

\(\frac{-2.4 - 1.5}{3.2} = \frac{-3.9}{3.2} = -1.219\)

Now, we find \(P(Z < -1.219)\) using the standard normal distribution table or a calculator:

\(P(Z < -1.219) = 1 - P(Z < 1.219)\)

From the standard normal distribution table, \(P(Z < 1.219) \approx 0.8886\).

Therefore, \(P(Z < -1.219) = 1 - 0.8886 = 0.1114\).

Rounding to three decimal places, the probability is \(0.111\).