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

For lengths less than 6 cm: \(Z = \frac{6 - \mu}{\sigma} = -1.253\)

For lengths more than 12 cm: \(Z = \frac{12 - \mu}{\sigma} = 0.648\)

Solving these equations simultaneously gives:

\(\mu = 9.9\)

\(\sigma = 3.15 \text{ or } 3.16\)

(ii) We need \(P(Z < -1 \text{ or } Z > 1)\).

This is equivalent to \(1 - \Phi(1) + \Phi(-1)\).

\(= 2 - 2 \times 0.8413\)

\(= 0.3174\)

In a sample of 1000, the expected number is \(0.3174 \times 1000 = 317\).

(i) Given that the mean μ is three times the standard deviation σ, we have μ = 3σ. The probability P(X < 25) = 0.648 corresponds to a standard normal variable z = 0.38. Using the standardization formula:

\(z = \frac{25 - \mu}{\sigma} = 0.38\)

Substitute \(\mu = 3\sigma\) into the equation:

\(\frac{25 - 3\sigma}{\sigma} = 0.38\)

Solving for \(\sigma\):

\(25 - 3\sigma = 0.38\sigma\)

\(25 = 3.38\sigma\)

\(\sigma = \frac{25}{3.38} \approx 7.40\)

Then, \(\mu = 3\sigma = 3 \times 7.40 = 22.2\).

(ii) The probability that a single value of X is greater than 25 is 1 - 0.648 = 0.352. We need the probability that exactly 4 out of 6 values are greater than 25. This follows a binomial distribution:

\(P(4) = \binom{6}{4} \times (0.352)^4 \times (0.648)^2\)

\(P(4) = 15 \times (0.0152) \times (0.419904)\)

\(P(4) \approx 0.0967\)

(ii) To find the proportion of winter days with a minimum temperature below zero, we standardize the variable:

\(z = \frac{0 - \mu}{2\mu} = -0.5\)

The probability \(P(z < -0.5) = 1 - 0.6915 = 0.309\) or 30.9%.

(iii) To find how many of the 70 days have a temperature more than three times the mean:

\(z = \frac{3\mu - \mu}{2\mu} = 1\)

The probability \(P(z > 1) = 1 - 0.8413 = 0.1587\).

Thus, the expected number of days is \(70 \times 0.1587 = 11.1\), so approximately 11 or 12 days.

(iv) Given the probability of the temperature being above 6 °C is 0.0735, we find \(\mu\):

\(z = 1.45\)

\(1.45 = \frac{6 - \mu}{2\mu}\)

Solving for \(\mu\), we get \(\mu = 1.54\).

(i) Given that 94% of the letters are within 12 g of the mean, we use the z-score for 47% (since 94% is the total for both tails, 47% for one tail) which is approximately 1.882. The equation is:

\(1.882 = \frac{32 - 20}{\sigma}\)

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

\(\sigma = \frac{12}{1.882} \approx 6.38\)

(ii) To find the probability that a letter weighs more than 13 g, we standardize:

\(P(x > 13) = P\left( z > \frac{13 - 20}{6.38} \right)\)

\(= P(z > -1.0978)\)

Using the standard normal distribution table, \(P(z > -1.0978) = 0.864\).

(iii) The probability that a letter weighs more than 32 g is 0.03 (since 6% are more than 12 g above the mean). We use the binomial distribution for at least 2 out of 7:

\(P(\text{at least 2}) = 1 - P(0, 1)\)

\(= 1 - (0.97)^7 - (0.03)(0.97)^6 \times \binom{7}{1}\)

\(= 1 - (0.97)^7 - 7 \times (0.03)(0.97)^6\)

\(= 0.0171\)

(i) Given that 10% of the straws are shorter than 20 cm, we have \(P(x < 20) = 0.1\). Using the standard normal distribution, this corresponds to a \(z\)-value of approximately \(-1.282\).

Standardizing, \(z = \frac{20 - \mu}{0.8}\).

Thus, \(-1.282 = \frac{20 - \mu}{0.8}\).

Solving for \(\mu\), we get \(\mu = 20 + 1.282 \times 0.8 = 21.0256 \approx 21.0 \text{ cm}\).

(ii) We need to find \(P(21.5 < x < 22.5)\).

Standardizing, \(P\left( \frac{21.5 - 21.03}{0.8} < z < \frac{22.5 - 21.03}{0.8} \right)\).

This simplifies to \(P(0.5875 < z < 1.8375)\).

Using the standard normal distribution table, \(\Phi(1.8375) - \Phi(0.5875) = 0.9670 - 0.7217 = 0.2453\).

Now, we use the binomial distribution for 4 trials with probability \(p = 0.2453\).

We need \(P(X < 2) = P(0) + P(1)\).

\(P(0) = (0.7547)^4\).

\(P(1) = 4 \times (0.2453) \times (0.7547)^3\).

Calculating these gives \(P(0) + P(1) = 0.7547^4 + 4 \times 0.2453 \times 0.7547^3 = 0.746\).