Let \(z_1\) and \(z_2\) be the standard normal variables corresponding to the given percentages.

For 20% of students having times less than 14.5 minutes, \(z_1 = \frac{14.5 - \mu}{\sigma} = -0.842\).

For 67% of students having times greater than 18.5 minutes, 33% have times less than 18.5 minutes, so \(z_2 = \frac{18.5 - \mu}{\sigma} = -0.44\).

We have the equations:

1. \(\frac{14.5 - \mu}{\sigma} = -0.842\)
2. \(\frac{18.5 - \mu}{\sigma} = -0.44\)

Solving these equations simultaneously, we find:

\(\mu = 22.9\)

\(\sigma = 9.95\)

(i) To find the probability that Josie has to wait longer than 6 minutes, we standardize the normal variable:

\(P(T > 6) = P\left( Z > \frac{6 - 5.3}{2.1} \right) = P(Z > 0.333)\)

Using the standard normal distribution table, \(P(Z > 0.333) = 1 - \Phi(0.333) = 1 - 0.6304 = 0.3696\). Therefore, the probability is approximately 0.370 or 0.369.

(ii) For 5% of days, Josie waits longer than \(x\) minutes, so \(P(T > x) = 0.05\). This corresponds to a \(z\)-value of 1.645. Standardizing gives:

\(1.645 = \frac{x - 5.3}{2.1}\)

Solving for \(x\), we get \(x = 1.645 \times 2.1 + 5.3 = 8.7545\). Thus, \(x \approx 8.75\) or \(8.755\).

(iii) The probability that Josie waits longer than \(x\) minutes on fewer than 3 days in 10 days is a binomial probability problem with \(n = 10\) and \(p = 0.05\). We calculate:

\(P(0, 1, 2) = (0.95)^{10} + \binom{10}{1}(0.05)(0.95)^9 + \binom{10}{2}(0.05)^2(0.95)^8\)

\(= 0.988\) (0.9885 to 4 significant figures).

(iv) The probability that Josie misses the bus is \(P(T < 0)\). Standardizing gives:

\(P\left( Z < \frac{0 - 5.3}{2.1} \right) = P(Z < -2.524)\)

Using the standard normal distribution table, \(P(Z < -2.524) = 1 - \Phi(2.524) = 1 - 0.9942 = 0.0058\).

(i) To find the probability that a boy weighs more than 65 kg, we standardize the value:

\(P(X > 65) = P\left( Z > \frac{65 - 61.4}{12.3} \right) = P(Z > 0.2927)\)

Using the standard normal distribution table, \(P(Z > 0.2927) = 1 - 0.6153 = 0.385\).

(ii) We know \(P(X < 65) = 0.6153\), so \(P(65 < X < k) = 0.25\).

Thus, \(P(X < k) = 0.25 + 0.6153 = 0.8653\).

Find the z-value for 0.8653: \(z = 1.105\).

Standardize: \(1.105 = \frac{k - 61.4}{12.3}\).

Solve for \(k\): \(k = 1.105 \times 12.3 + 61.4 = 75.0\).

(iii) For Brigville, use the z-values for the given probabilities:

\(2.326 = \frac{97.2 - \mu}{\sigma}\) and \(-0.44 = \frac{55.2 - \mu}{\sigma}\).

Solve these simultaneous equations to find \(\mu\) and \(\sigma\):

From \(2.326\sigma = 97.2 - \mu\) and \(-0.44\sigma = 55.2 - \mu\),

Eliminate \(\mu\) to find \(\sigma = 15.2\).

Substitute back to find \(\mu = 61.9\).

(i) To find the proportion of large pineapples, calculate \(P(X > 570)\):

Standardize: \(z = \frac{570 - 500}{91.5} = 0.7650\)

\(P(Z > 0.7650) = 1 - P(Z < 0.7650) = 1 - 0.7779 = 0.222\)

To find the proportion of small pineapples, calculate \(P(X < 390)\):

Standardize: \(z = \frac{390 - 500}{91.5} = -1.202\)

\(P(Z < -1.202) = 0.1147\)

The proportion of medium pineapples is \(1 - (0.222 + 0.115) = 0.663\).

(ii) To find the weight exceeded by the heaviest 5% of pineapples, find \(x\) such that \(P(X > x) = 0.05\).

Find the critical value: \(z = 1.645\)

Standardize: \(1.645 = \frac{x - 500}{91.5}\)

\(x = 1.645 \times 91.5 + 500 = 651\)

(iii) To find \(k\) such that \(P(k < X < 610) = 0.3\):

\(P(X > 610) = 0.1147\) (from symmetry)

\(0.3 + 0.1147 = 0.4147 \Rightarrow \Phi(x) = 0.5853\)

Find \(z\) such that \(\Phi(z) = 0.5853\): \(z = 0.215\)

Standardize: \(0.215 = \frac{k - 500}{91.5}\)

\(k = 0.215 \times 91.5 + 500 = 520\)

Given that 87.5% of the batteries last longer than 70 hours, we find the corresponding z-score for the 12.5th percentile (since 100% - 87.5% = 12.5%).

The z-score for the 12.5th percentile is approximately \(z = -1.151\).

Using the standardization formula:

\(z = \frac{X - \mu}{s}\)

Substitute the known values:

\(-1.151 = \frac{70 - 120}{s}\)

Solve for \(s\):

\(-1.151s = 70 - 120\)

\(-1.151s = -50\)

\(s = \frac{-50}{-1.151}\)

\(s \approx 43.4 \text{ or } 43.5\)