Let the mean be \(\mu\) and the standard deviation be \(s\). We know that \(\mu = 4s\).

The probability that a value is greater than 5 is 0.15, which corresponds to a z-score of approximately 1.036 or 1.037.

Using the z-score formula:

\(z = \frac{5 - \mu}{s}\)

Substitute \(\mu = 4s\) into the equation:

\(1.036 = \frac{5 - 4s}{s}\)

Solve for \(s\):

\(1.036s = 5 - 4s\)

\(5.036s = 5\)

\(s = \frac{5}{5.036} \approx 0.993\)

Substitute back to find \(\mu\):

\(\mu = 4s = 4 \times 0.993 = 3.97\)

(i) To find the standard deviation \(\sigma\), we use the z-score formula:

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

Given \(X = 73\), \(\mu = 75\), and \(z = -1.036\) (from the z-table for 15%), we have:

\(-1.036 = \frac{73 - 75}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{2}{1.036} \approx 1.93\)

(ii) The probability that a roll is longer than 77 m is 0.15. We need to find the probability that fewer than 3 out of 8 rolls are longer than 77 m. This is a binomial probability problem with \(n = 8\) and \(p = 0.15\).

\(P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)\)

\(= (0.85)^8 + \binom{8}{1}(0.15)(0.85)^7 + \binom{8}{2}(0.15)^2(0.85)^6\)

\(= 0.895\)

(i) To find the mean \(\mu\) and standard deviation \(\sigma\), we use the standard normal distribution properties. Given that 96% of lengths are less than 34.1 cm, we find the z-score corresponding to 0.96, which is approximately 1.751. Thus,

\(\frac{34.1 - \mu}{\sigma} = 1.751\)

Similarly, 70% of lengths are more than 26.7 cm, meaning 30% are less than 26.7 cm. The z-score for 0.30 is approximately -0.524. Thus,

\(\frac{26.7 - \mu}{\sigma} = -0.524\)

Solving these two equations simultaneously, we find:

\(\mu = 28.4\), \(\sigma = 3.25\).

(iii) We need to find \(t\) such that \(P(31.8 < X < t) = 0.5\). The distribution of \(X\) is \(N(32.9, 2.4^2)\).

First, find the z-score for 31.8:

\(z = \frac{31.8 - 32.9}{2.4} = -0.4583\)

The cumulative probability for \(z = -0.4583\) is approximately 0.3235. Therefore, \(P(X > 31.8) = 1 - 0.3235 = 0.6765\).

We need \(P(31.8 < X < t) = 0.5\), so \(P(X < t) = 0.5 + 0.6765 = 0.8235\).

Find the z-score for 0.8235, which is approximately 0.929. Thus,

\(\frac{t - 32.9}{2.4} = 0.929\)

Solving for \(t\), we get:

\(t = 35.1\).

(i) To find the probability that the symphony takes longer than 42 minutes, we standardize the normal variable:

\(P(>42) = P\left( z > \frac{42 - 41.1}{3.4} \right) = P(z > 0.2647)\)

Using the standard normal distribution table, \(P(z > 0.2647) = 1 - 0.6045 = 0.3955\).

The probability that the symphony takes longer than 42 minutes on exactly 1 of 3 occasions is given by the binomial distribution:

\(\text{Prob} = (0.3955)(0.6045)^2 \times \binom{3}{1} = 0.433 \text{ or } 0.434\)

(ii) For Beethoven’s Fifth Symphony, we use the given probabilities to find the mean \(\mu\) and standard deviation \(\sigma\):

\(-1.282 = \frac{26.5 - \mu}{\sigma}\)

\(1.645 = \frac{34.6 - \mu}{\sigma}\)

Solving these equations simultaneously gives \(\mu = 30.0\) and \(\sigma = 2.77\).

(iii) To find the probability that both symphonies are less than 34.6 minutes:

For the Sixth Symphony:

\(P(B6 < 34.6) = P\left( z < \frac{34.6 - 41.1}{3.4} \right) = P(z < -1.912)\)

\(P(z < -1.912) = 1 - 0.9720 = 0.0280\)

For the Fifth Symphony, \(P(B5 < 34.6) = 0.95\).

The probability that both are less than 34.6 minutes is:

\(P(\text{both} < 34.6) = 0.028 \times 0.95 = 0.0266\)

The weights of the apples are normally distributed with mean \(\mu = 182\) and standard deviation \(\sigma = 20\). We need to find \(w\) such that \(P(X > w) = 0.72\).

First, convert the problem to a standard normal distribution problem:

\(P\left(Z > \frac{w - 182}{20}\right) = 0.72\)

This implies:

\(P\left(Z < \frac{w - 182}{20}\right) = 0.28\)

Using the standard normal distribution table, find the z-value for which \(P(Z < z) = 0.28\). This gives \(z = -0.583\).

Now, equate and solve for \(w\):

\(\frac{w - 182}{20} = -0.583\)

\(w - 182 = -0.583 \times 20\)

\(w = 182 - 11.66\)

\(w = 170.34\)

Therefore, \(170 \leq w < 170.35\).