We are given that the weight of the tea packets follows a normal distribution with mean \(\mu = 260\) g and standard deviation \(\sigma\) g. A packet is considered underweight if it weighs less than 250 g, and 1% of packets are underweight.

We need to find the value of \(\sigma\) such that the probability of a packet being underweight is 0.01. This corresponds to the left tail of the normal distribution.

Using the standard normal distribution table, the z-score corresponding to the 1st percentile is approximately \(z = -2.326\).

We use the standardization formula:

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

Substituting the known values:

\(-2.326 = \frac{250 - 260}{\sigma}\)

\(-2.326 = \frac{-10}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{-10}{-2.326}\)

\(\sigma \approx 4.30\)

(i) To find the probability that a randomly chosen person sleeps for less than 8 hours, we standardize the variable:

\(P(X < 8) = P\left( Z < \frac{8 - 7.15}{0.88} \right)\)

\(= P(Z < 0.9659)\)

Using the standard normal distribution table, \(\Phi(0.9659) = 0.833\).

Thus, the probability is 0.833.

(ii) To find the value of \(q\) such that \(P(X < q) = 0.75\), we find the z-value corresponding to 0.75 in the standard normal distribution table, which is approximately 0.674.

Using the standardization formula:

\(\frac{q - 7.15}{0.88} = 0.674\)

Solving for \(q\):

\(q = 0.674 \times 0.88 + 7.15\)

\(q = 7.74\)

Given that 25% of the sheep weigh more than 67.5 kg, this corresponds to the 75th percentile of the normal distribution. The z-score for the 75th percentile is approximately 0.674.

Using the standardization formula:

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

Substitute the known values:

\(0.674 = \frac{67.5 - \mu}{4.92}\)

Solving for \(\mu\):

\(67.5 - \mu = 0.674 \times 4.92\)

\(67.5 - \mu = 3.31608\)

\(\mu = 67.5 - 3.31608\)

\(\mu = 64.18392\)

Rounding to one decimal place, \(\mu = 64.2\).

(i) To find the value of t, we use the standard normal distribution. We know that 90% of calls take longer than t, so 10% take less time. This corresponds to a z-score of -1.282 (since the left tail is 0.10).

Using the standardization formula:

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

\(-1.282 = \frac{t - 6.5}{1.76}\)

Solving for t gives:

\(t = 6.5 + (-1.282 \times 1.76) = 4.24\) minutes

(ii) We need to find the probability that more than 7 out of 9 calls are within 1 standard deviation of the mean. The probability of a call being within 1 standard deviation is given by:

\(P(\text{within 1 sd of mean}) = 2\Phi(1) - 1 = 0.6826\)

We use the binomial distribution with \(n = 9\) and \(p = 0.6826\). We need \(P(X > 7) = P(X = 8) + P(X = 9)\).

\(P(X = 8) = \binom{9}{8} (0.6826)^8 (0.3174)^1\)

\(P(X = 9) = \binom{9}{9} (0.6826)^9\)

Calculating these gives:

\(P(X = 8) = 9 \times 0.6826^8 \times 0.3174\)

\(P(X = 9) = 0.6826^9\)

Adding these probabilities gives \(P(X > 7) = 0.167\).

Given that the lengths are normally distributed, we use the standard normal distribution to find \(\mu\) and \(\sigma\).

For 4% longer than 12 cm, the z-score is:

\(z = \frac{12 - \mu}{\sigma}\)

From standard normal tables, \(z = 1.751\) for 4% in the upper tail.

Thus, \(1.751 = \frac{12 - \mu}{\sigma}\).

For 32% longer than 9 cm, the z-score is:

\(z = \frac{9 - \mu}{\sigma}\)

From standard normal tables, \(z = 0.468\) for 32% in the upper tail.

Thus, \(0.468 = \frac{9 - \mu}{\sigma}\).

We have two equations:

1. \(1.751 = \frac{12 - \mu}{\sigma}\)
2. \(0.468 = \frac{9 - \mu}{\sigma}\)

Subtract the second equation from the first:

\(1.751 - 0.468 = \frac{12 - \mu}{\sigma} - \frac{9 - \mu}{\sigma}\)

\(1.283 = \frac{3}{\sigma}\)

\(\sigma = \frac{3}{1.283} = 2.34\)

Substitute \(\sigma = 2.34\) into the first equation:

\(1.751 = \frac{12 - \mu}{2.34}\)

\(12 - \mu = 1.751 \times 2.34\)

\(12 - \mu = 4.1\)

\(\mu = 12 - 4.1 = 7.91\)