Given that the mean \(\mu = 1.62\) m and the probability \(P(X > 1.8) = 0.15\), we need to find \(\sigma\).

First, find the z-score corresponding to a probability of 0.15 in the standard normal distribution table. This gives \(z = 1.037\).

Using the standardization formula:

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

Substitute the known values:

\(1.037 = \frac{1.8 - 1.62}{\sigma}\)

Solve for \(\sigma\):

\(\sigma = \frac{0.18}{1.037} = 0.174\)

(i) To find the mean \(m\), we use the standard normal distribution. Given that 95% of times are longer than 0.9 hours, the corresponding z-value is \(z = -1.645\) (since 5% is in the left tail).

Using the standardization formula:

\(z = \frac{0.9 - m}{0.35}\)

Substitute \(z = -1.645\):

\(-1.645 = \frac{0.9 - m}{0.35}\)

Solving for \(m\):

\(0.9 - m = -1.645 \times 0.35\)

\(0.9 - m = -0.57575\)

\(m = 0.9 + 0.57575\)

\(m = 1.47575\)

Rounding to 3 significant figures, \(m = 1.48\).

(ii) We need to find the probability that none of the 4 cars takes more than 2 hours to fit. First, find the probability that a single car takes less than 2 hours.

Standardize 2 hours:

\(z = \frac{2 - 1.48}{0.35}\)

\(z = \frac{0.52}{0.35}\)

\(z = 1.4857\)

Using the standard normal distribution table, \(P(z < 1.4857) \approx 0.933\).

The probability that none of the 4 cars takes more than 2 hours is:

\((0.933)^4 \approx 0.758\)

(i) To find \(\sigma\), we use the standard normal distribution. Given that 13% of seeds take longer than 136 hours, we find the corresponding \(z\)-value for 0.87 (since 1 - 0.13 = 0.87) in the standard normal distribution table, which is approximately 1.127.

Using the formula for standardization:

\(z = \frac{136 - 125}{\sigma}\)

Substitute \(z = 1.127\):

\(1.127 = \frac{136 - 125}{\sigma}\)

\(\sigma = \frac{11}{1.127} \approx 9.76\)

(ii) To find the expected number of seeds taking between 131 and 141 hours, we standardize these values:

\(P(131 < x < 141) = P\left( \frac{131 - 125}{9.76} < z < \frac{141 - 125}{9.76} \right)\)

\(= P(0.6147 < z < 1.639)\)

Using the standard normal distribution table:

\(\Phi(1.639) - \Phi(0.6147) = 0.9493 - 0.7307 = 0.2186\)

The expected number of seeds is:

\(0.2186 \times 170 \approx 37.2\)

Thus, the expected number of seeds is 37 or 38.

(i) To find the number of days the sales exceed 3900 litres, we standardize the value:

\(P(x > 3900) = P\left( z > \frac{3900 - 4520}{560} \right) = P(z > -1.107) = \Phi(1.107)\)

Using the standard normal distribution table, \(\Phi(1.107) = 0.8657\).

The expected number of days is \(365 \times 0.8657 = 315.98\), so approximately 315 or 316 days.

(ii) Given \(P(X > 8000) = 0.122\), we find the z-score:

\(z = 1.165\)

Using the standardization formula:

\(1.165 = \frac{8000 - m}{560}\)

Solving for \(m\):

\(m = 8000 - 1.165 \times 560 = 7350\)

(iii) The probability that sales exceed 8000 litres on fewer than 2 of 6 days is calculated using the binomial distribution:

\(P(0, 1) = (0.878)^6 + \binom{6}{1}(0.122)^1(0.878)^5\)

Calculating gives \(0.840\), which can be accepted as 0.84.

Since \(P(X < 54.1) = 0.5\), the mean \(\mu = 54.1\) because the median of a normal distribution is the mean.

For \(P(X > 50.9) = 0.8665\), we find the corresponding \(z\)-score. The probability \(P(X > 50.9) = 0.8665\) implies \(P(X < 50.9) = 1 - 0.8665 = 0.1335\).

Using the standard normal distribution table, \(P(Z < z) = 0.1335\) corresponds to \(z = -1.11\).

Standardizing, we have:

\(-1.11 = \frac{50.9 - 54.1}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{50.9 - 54.1}{-1.11} = 2.88\)