(i) We know that 225 out of 900 cartons contain more than 1002 millilitres. Therefore, the probability \(P(X > 1002) = \frac{225}{900} = 0.25\).

Using the standard normal distribution, \(P(Z > z) = 0.25\) corresponds to \(z = 0.674\) (using standard normal tables or calculator).

Standardizing, we have:

\(\frac{1002 - \mu}{8} = 0.674\)

Solving for \(\mu\):

\(1002 - \mu = 0.674 \times 8\)

\(1002 - \mu = 5.392\)

\(\mu = 1002 - 5.392 = 996.608\)

Rounding to the nearest whole number, \(\mu = 997\).

(ii) The probability that a carton contains more than 1002 millilitres is \(p = 0.25\). We want the probability that exactly 2 out of 3 chosen cartons contain more than 1002 millilitres.

This is a binomial probability problem with \(n = 3\), \(k = 2\), and \(p = 0.25\).

\(P(2) = \binom{3}{2} \times (0.25)^2 \times (0.75)^1\)

\(P(2) = 3 \times 0.0625 \times 0.75\)

\(P(2) = 0.140625\)

Rounding to three significant figures, the probability is 0.140.

The problem states that the daily minimum temperature follows a normal distribution \(N(\mu, 40.0)\). We are given that the probability of the temperature being above 0°C is 0.8888.

Using the standard normal distribution table, a probability of 0.8888 corresponds to a \(z\)-score of approximately 1.22. However, since we are dealing with the probability above 0°C, we use \(z = -1.22\).

The \(z\)-score formula is:

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

where \(X = 0\) (the temperature), \(\mu\) is the mean, and \(\sigma = \sqrt{40}\).

Substitute the values into the formula:

\(-1.22 = \frac{0 - \mu}{\sqrt{40}}\)

Solving for \(\mu\):

\(-1.22 \times \sqrt{40} = -\mu\)

\(\mu = 1.22 \times \sqrt{40}\)

\(\mu \approx 7.72\)

(i) We know that 25% of infections clear up in less than 7 days, which corresponds to a z-score of approximately -0.674. Using the standardization formula:

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

Substitute the known values:

\(-0.674 = \frac{7 - \mu}{2.6}\)

Solving for \(\mu\):

\(-0.674 \times 2.6 = 7 - \mu\)

\(-1.7524 = 7 - \mu\)

\(\mu = 7 + 1.7524 = 8.75\)

(ii) We need to find \(P(X > 6.2)\) where \(X\) is normally distributed with mean 6.5 and standard deviation 2.6. Standardizing:

\(P(X > 6.2) = P\left( Z > \frac{6.2 - 6.5}{2.6} \right)\)

\(= P(Z > -0.1154)\)

Using the symmetry of the normal distribution, \(P(Z > -0.1154) = 1 - P(Z < -0.1154)\). Since \(P(Z < -0.1154) \approx 0.454\),

\(P(Z > -0.1154) = 1 - 0.454 = 0.546\)

(i) To find the standard deviation \(\sigma\), we use the standard normal distribution formula:

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

Given \(P(X > 5.5) = 0.0465\), we find the corresponding \(z\)-value from the standard normal distribution table, which is approximately 1.68. Thus,

\(z = \frac{5.5 - 4.5}{\sigma} = 1.68\)

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

\(\sigma = \frac{1}{1.68} = 0.595\)

(ii) To find the probability that \(X\) lies between 3.8 and 4.8, we calculate the \(z\)-values for 3.8 and 4.8:

\(z_1 = \frac{3.8 - 4.5}{0.595} = -1.176\)

\(z_2 = \frac{4.8 - 4.5}{0.595} = 0.504\)

The probability is given by:

\(P(3.8 < X < 4.8) = \Phi(0.504) - (1 - \Phi(1.176))\)

Using the standard normal distribution table, \(\Phi(0.504) \approx 0.6929\) and \(\Phi(1.176) \approx 0.8802\), so:

\(P(3.8 < X < 4.8) = 0.6929 - (1 - 0.8802) = 0.573\)

(a) Let the mean be \(\mu = 2s\). The standard normal variable \(Z\) is given by:

\(Z = \frac{X - \mu}{s} = \frac{5.2 - 2s}{s}\)

Given \(P(X > 5.2) = 0.9\), we have \(P(Z > z) = 0.9\), which implies \(P(Z < z) = 0.1\). From standard normal tables, \(z = -1.282\).

Thus, \(\frac{5.2 - 2s}{s} = -1.282\).

Solving for \(s\):

\(5.2 - 2s = -1.282s\)

\(5.2 = 0.718s\)

\(s = \frac{5.2}{0.718} \approx 7.24 \text{ or } 7.23\)

(b) The probability that a value lies between \(\mu - \sigma\) and \(\mu + \sigma\) is \(P(|Z| < 1)\).

From standard normal tables, \(P(|Z| < 1) = 0.6826\).

Thus, the expected number of observations is:

\(0.6826 \times 800 = 546.08\)

Therefore, approximately 546 observations (accept 547).