First, calculate the probability that a bag weighs more than 2.6 kg:

\(P(X > 2.6) = \frac{134}{5000} = 0.0268\)

The probability that a bag weighs less than 2.6 kg is:

\(P(X < 2.6) = 1 - 0.0268 = 0.9732\)

Using the standard normal distribution, find the z-score corresponding to this probability:

\(\frac{2.6 - 2.55}{\sigma} = 1.93\)

Solving for \(\sigma\):

\(\sigma = \frac{2.6 - 2.55}{1.93} = 0.0259\)

Given that \(X \sim N(\mu, \sigma^2)\), we have two probabilities:

1. \(P(X > 30.0) = 0.1480\)

2. \(P(X > 20.9) = 0.6228\)

Using the standard normal distribution, we convert these to \(z\)-scores:

For \(X = 30.0\):

\(z = \frac{30 - \mu}{\sigma}\)

From the standard normal table, \(P(Z > 1.045) = 0.1480\), so \(z = 1.045\).

Thus, \(\frac{30 - \mu}{\sigma} = 1.045\).

For \(X = 20.9\):

\(z = \frac{20.9 - \mu}{\sigma}\)

From the standard normal table, \(P(Z > -0.313) = 0.6228\), so \(z = -0.313\).

Thus, \(\frac{20.9 - \mu}{\sigma} = -0.313\).

We now have two equations:

1. \(30 - \mu = 1.045 \sigma\)

2. \(20.9 - \mu = -0.313 \sigma\)

Solving these simultaneously:

Subtract the second equation from the first:

\((30 - \mu) - (20.9 - \mu) = 1.045 \sigma + 0.313 \sigma\)

\(9.1 = 1.358 \sigma\)

\(\sigma = \frac{9.1}{1.358} = 6.70\)

Substitute \(\sigma = 6.70\) into the first equation:

\(30 - \mu = 1.045 \times 6.70\)

\(30 - \mu = 7.0015\)

\(\mu = 30 - 7.0015 = 23.0\)

Thus, \(\mu = 23.0\) and \(\sigma = 6.70\).

(i) To find the probability that a randomly chosen bar of soap weighs more than 128 grams, we standardize the variable:

\(P(X > 128) = P\left( z > \frac{128 - 125}{4.2} \right)\)

\(= P(z > 0.7143)\)

Using the standard normal distribution table, \(P(z > 0.7143) = 1 - 0.7623 = 0.238\).

(ii) We need to find the value of \(k\) such that \(P(k < X < 128) = 0.7465\).

\(P(X < 128) = 0.7465 + 0.2377 = 0.9842\)

Using the standard normal distribution table, \(z = -2.15\) corresponds to \(P(X < k)\).

\(-2.15 = \frac{k - 125}{4.2}\)

Solving for \(k\), we get \(k = 116\).

(iii) For five bars of soap, we find the probability that more than two weigh more than 128 grams:

\(P(X > 2) = P(3, 4, 5) = 1 - P(0, 1, 2)\)

Using the binomial distribution:

\(= \binom{5}{3} (0.2377)^3 (0.7623)^2 + \binom{5}{4} (0.2377)^4 (0.7623)^1 + \binom{5}{5} (0.2377)^5\)

\(= 0.07804 + 0.01216 + 0.0007588\)

\(= 0.0910\)

(i) The mean \(\mu\) is estimated as the median of the data, which is the middle value of the box-and-whisker plot. From the diagram, the median is 51 km h\(^{-1}\).

(ii) To estimate the standard deviation \(\sigma\), we use the interquartile range (IQR) and the properties of the normal distribution. The IQR is the difference between the upper quartile (63 km h\(^{-1}\)) and the median (51 km h\(^{-1}\)), which is 12 km h\(^{-1}\).

For a normal distribution, the IQR corresponds to approximately 1.34896 standard deviations (since \(z = \pm 0.674\) for the quartiles). Therefore, \(\sigma\) can be estimated by:

\(\sigma = \frac{63 - 51}{0.674} = \frac{12}{0.674} \approx 17.8\)

(i) To find the probability that a journey takes between 85 and 100 minutes, we standardize the values using the formula:

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

For \(x = 85\), \(z = \frac{85 - 100}{7} = -2.143\).

The probability \(P(85 < x < 100) = 0.5 - P(z < -2.143)\).

Using the standard normal distribution table, \(P(z < -2.143) = 1 - \Phi(2.143) = 0.0161\).

Thus, \(P(85 < x < 100) = 0.5 - 0.0161 = 0.484\).

(ii) For 'standard' journeys, which are the middle 34%, we find the z-scores corresponding to the cumulative probabilities of 0.33 and 0.67.

Using \(z = \Phi^{-1}(0.67) = 0.44\).

For the upper limit, \(0.44 = \frac{a - 100}{7}\), solving gives \(a = 103.1\) minutes.

For the lower limit, \(-0.44 = \frac{a - 100}{7}\), solving gives \(a = 96.9\) minutes.