Given that the weights are normally distributed with mean \(\mu = 0.75\) kg and standard deviation \(\sigma\) kg, and 68% of the weights are less than 0.9 kg, we can use the standard normal distribution to find \(\sigma\).

The probability statement is:

\(P(X < 0.9) = P\left( Z < \frac{0.9 - 0.75}{\sigma} \right) = 0.68\)

From standard normal distribution tables, \(P(Z < 0.468) \approx 0.68\).

Thus, \(\frac{0.9 - 0.75}{\sigma} = 0.468\).

Solving for \(\sigma\):

\(\sigma = \frac{0.15}{0.468} \approx 0.3205\)

Therefore, the value of \(\sigma\) is approximately between 0.3205 and 0.3215.

(b) We use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\). Given:

\(z_1 = \frac{3 - \mu}{\sigma} = -1.329\)

\(z_2 = \frac{8 - \mu}{\sigma} = 0.878\)

Solving these equations simultaneously, we find:

\(\sigma = 2.27, \mu = 6.01\)

(c) To find the number of leaves more than 1 standard deviation from the mean, we calculate:

\(P(Z < -1) + P(Z > 1) = 2 \times \Phi(1) - \Phi(-1)\)

\(= 2 - 2 \times 0.8413 = 0.3174\)

Number of leaves: \(2000 \times 0.3174 = 634.8\), so approximately 634 or 635 leaves.

(i) Given that 5% of the fish are longer than 50 cm, we use the z-score for 95% which is 1.645. The equation is:

\(1.645 = \frac{50 - 38}{\sigma}\)

Solving for \(\sigma\), we get:

\(\sigma = \frac{50 - 38}{1.645} = 7.29\)

(ii) To find the proportion of fish shorter than 30 cm, we standardize:

\(z = \frac{30 - 38}{7.29} = -1.097\)

Using the standard normal distribution table, \(P(z < -1.097) = 1 - \Phi(1.097)\)

\(= 1 - 0.8637 = 0.136\)

(iii) The probability that a fish is longer than 50 cm is 0.05. For 9 fish, the probability that none are longer than 50 cm is \((0.95)^9\).

The probability that at least one is longer than 50 cm is:

\(1 - (0.95)^9 = 0.370\)

Given that the tyre pressures follow a normal distribution with mean \(\mu = 1.9\) bars and standard deviation \(\sigma = 0.15\) bars, we need to find the value of \(b\) such that 80% of the data lies within \(1.9 - b\) and \(1.9 + b\).

Since 80% of the data is within these limits, we have \(P(1.9 - b < X < 1.9 + b) = 0.8\).

This corresponds to a cumulative probability of 0.8, which means the z-scores are \(\pm 1.282\) (or approximately \(\pm 1.28\)).

Using the z-score formula: \(z = \frac{b}{0.15}\), we have:

\(\pm 1.282 = \frac{b}{0.15}\)

Solving for \(b\), we get:

\(b = 1.282 \times 0.15 = 0.1923\)

Thus, the safety limits are:

\(1.9 - 0.1923 = 1.7077 \approx 1.71\)

\(1.9 + 0.1923 = 2.0923 \approx 2.09\)

Therefore, the limits are between 1.71 and 2.09 bars.

(i) We know that the probability of the lunch break lasting more than 52 minutes is \(\frac{1}{4} = 0.25\). This corresponds to a standard normal value \(z\) such that \(P(Z > z) = 0.25\). From standard normal distribution tables, \(z \approx 0.674\).

Using the formula for the standard normal variable, \(z = \frac{52 - \mu}{5}\), we have:

\(\frac{52 - \mu}{5} = 0.674\)

Solving for \(\mu\):

\(52 - \mu = 5 \times 0.674\)

\(52 - \mu = 3.37\)

\(\mu = 52 - 3.37 = 48.63\)

Rounding to one decimal place, \(\mu = 48.6\).

(ii) We need to find the probability that the lunch break lasts between 40 and 46 minutes. First, standardize these values:

\(z_1 = \frac{40 - 48.63}{5} = -1.726\)

\(z_2 = \frac{46 - 48.63}{5} = -0.526\)

Using the standard normal distribution table, find the probabilities:

\(P(Z < -1.726) \approx 0.0422\)

\(P(Z < -0.526) \approx 0.2995\)

The probability that the lunch break lasts between 40 and 46 minutes is:

\(P(-1.726 < Z < -0.526) = 0.2995 - 0.0422 = 0.2573\)

The probability that this happens on every one of the next four days is:

\((0.2573)^4 = 0.00438\)