Given that the heights are normally distributed, we have:

\(P(X < 10) = \frac{48}{500} = 0.096\)

Using the standard normal distribution table, \(z = -1.305\) for \(P(Z < z) = 0.096\).

Similarly, \(P(X > 24) = \frac{76}{500} = 0.152\)

Using the standard normal distribution table, \(z = 1.028\) for \(P(Z > z) = 0.152\).

We form the equations:

\(10 - \mu = -1.305\sigma\)

\(24 - \mu = 1.028\sigma\)

Subtracting these equations gives:

\(14 = 2.333\sigma\)

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

\(\sigma = \frac{14}{2.333} = 6.00\)

Substituting \(\sigma = 6.00\) back into one of the equations:

\(10 - \mu = -1.305 \times 6.00\)

\(10 - \mu = -7.83\)

\(\mu = 17.83\)

Thus, the mean \(\mu\) is approximately 17.8 and the standard deviation \(\sigma\) is 6.00.

Given that 92% of athletes have PBs of more than t seconds, this implies that 8% have PBs less than t seconds. In a standard normal distribution, this corresponds to a z-score of approximately -1.406.

Using the standardization formula:

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

where \(\mu = 49.2\) and \(\sigma = 2.8\), we have:

\(\frac{t - 49.2}{2.8} = -1.406\)

Solving for \(t\):

\(t - 49.2 = -1.406 \times 2.8\)

\(t - 49.2 = -3.9368\)

\(t = 49.2 - 3.9368\)

\(t = 45.2632\)

Rounding to one decimal place, \(t = 45.3\).

Given that 96% of ferry crossings take longer than \(t\), this implies that 4% take less time than \(t\). We need to find the corresponding \(z\)-value for the 4th percentile of the standard normal distribution.

The \(z\)-value for 4% in the standard normal distribution is approximately \(z = -1.751\).

Using the standardization formula:

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

Substitute \(z = -1.751\), \(\mu = 85\), and \(\sigma = 6.8\):

\(-1.751 = \frac{t - 85}{6.8}\)

Solving for \(t\):

\(t - 85 = -1.751 \times 6.8\)

\(t - 85 = -11.9088\)

\(t = 85 - 11.9088\)

\(t = 73.0912\)

Rounding to one decimal place, \(t = 73.1\).

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

For 20% of pandas weighing more than 121 kg, the corresponding z-score is \(z = 0.842\). Using the standardization formula:

\(z = \frac{121 - \mu}{\sigma}\)

Substituting the values, we get:

\(0.842 = \frac{121 - \mu}{\sigma}\)

So, \(0.842\sigma = 121 - \mu\).

For 71.9% of pandas weighing more than 102 kg, the corresponding z-score is \(z = -0.58\). Using the standardization formula:

\(z = \frac{102 - \mu}{\sigma}\)

Substituting the values, we get:

\(-0.58 = \frac{102 - \mu}{\sigma}\)

So, \(-0.58\sigma = 102 - \mu\).

We now have two equations:

1. \(0.842\sigma = 121 - \mu\)
2. \(-0.58\sigma = 102 - \mu\)

Solving these simultaneously, we eliminate \(\mu\) and solve for \(\sigma\):

\(0.842\sigma + \mu = 121\)

\(-0.58\sigma + \mu = 102\)

Subtract the second equation from the first:

\((0.842 + 0.58)\sigma = 121 - 102\)

\(1.422\sigma = 19\)

\(\sigma = \frac{19}{1.422} \approx 13.4\)

Substitute \(\sigma = 13.4\) back into one of the original equations to find \(\mu\):

\(0.842 \times 13.4 + \mu = 121\)

\(11.2908 + \mu = 121\)

\(\mu = 121 - 11.2908\)

\(\mu \approx 110\)

Thus, the mean \(\mu\) is 110 and the standard deviation \(\sigma\) is 13.4.

(a) We know that \(P(X < 16) = 0.1\). Using the standard normal distribution, we find the z-value corresponding to 0.1, which is approximately \(-1.282\).

Using the standardization formula:

\(\frac{16 - 28}{\sigma} = -1.282\)

Solving for \(\sigma\):

\(\sigma = \frac{12}{1.282} \approx 9.36\)

(c) We need to find \(P(-1.3 < Z < 1.3)\).

Using the standard normal distribution table, \(\Phi(1.3) \approx 0.9032\).

Thus, \(P(-1.3 < Z < 1.3) = 2 \times 0.9032 - 1 = 0.8064\).

In 365 days, the expected number of days is:

\(0.8064 \times 365 \approx 294.336\)

Therefore, the expected number of days is 294 or 295.