Given that the distribution is normal with mean \\(\mu = 15.8 \\\) and standard deviation \\(\sigma = 4.2 \\\), we need to find the value of \\(k \\\) such that \\(P(X < k) = 0.75 \\\).

First, find the z-score corresponding to the cumulative probability of 0.75. From standard normal distribution tables, \\(z = 0.674 \\\).

Use the z-score formula:

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

Substitute the known values:

\\(0.674 = \frac{k - 15.8}{4.2} \\\)

Solve for \\(k \\\):

\\(k - 15.8 = 0.674 \times 4.2 \\\)

\\(k - 15.8 = 2.8308 \\\)

\\(k = 15.8 + 2.8308 \\\)

\\(k = 18.6308 \\\)

Rounding to one decimal place, \\(k = 18.6 \\\).

(a) To find the probability that a tree is classified as short, we calculate:

\(P(X < 25) = P\left( Z < \frac{25 - 40}{12} \right) = P(Z < -1.25)\)

Using the standard normal distribution table, \(P(Z < -1.25) = 0.106\).

Thus, the probability is 0.106.

(b) Given that one third of the trees classified as tall or medium are tall, we have:

\(P(\text{tall}) = \frac{1}{3} \times (1 - 0.106) = \frac{0.8944}{3} = 0.298\).

Therefore, the probability that a randomly chosen tree is classified as tall is 0.298.

(c) To find the height above which trees are classified as tall, we use the probability found in part (b):

\(0.2981\) gives \(z = 0.53\).

Using the standard normal distribution formula:

\(\frac{h - 40}{12} = 0.53\)

Solving for \(h\):

\(h = 0.53 \times 12 + 40 = 46.4\).

Thus, trees are classified as tall if their height is above 46.4 m.

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

From the problem, 32 out of 200 snakes have lengths less than 45 cm. This corresponds to a probability of \(\frac{32}{200} = 0.16\). Using the standard normal distribution table, this probability corresponds to a z-score of approximately \(-0.994\).

Similarly, 17 out of 200 snakes have lengths more than 56 cm, corresponding to a probability of \(\frac{17}{200} = 0.085\). This corresponds to a z-score of approximately \(1.372\).

We have the following standardization equations:

\(\frac{45 - \mu}{\sigma} = -0.994\)

\(\frac{56 - \mu}{\sigma} = 1.372\)

Solving these two equations simultaneously, we first eliminate \(\mu\) by subtracting the first equation from the second:

\(\frac{56 - \mu}{\sigma} - \frac{45 - \mu}{\sigma} = 1.372 + 0.994\)

\(\frac{11}{\sigma} = 2.366\)

\(\sigma = \frac{11}{2.366} \approx 4.65\)

Substituting \(\sigma = 4.65\) back into one of the original equations to find \(\mu\):

\(\frac{45 - \mu}{4.65} = -0.994\)

\(45 - \mu = -0.994 \times 4.65\)

\(\mu = 45 + 4.6251 \approx 49.6\)

Thus, the estimates for the mean and standard deviation are \(\mu = 49.6\) and \(\sigma = 4.65\).

(a) We know that \(P(X > 87) = 0.22\). Using the standard normal distribution, we have:

\(P\left( Z > \frac{87 - 82}{\sigma} \right) = 0.22\)

This implies:

\(P\left( Z < \frac{5}{\sigma} \right) = 0.78\)

From standard normal distribution tables, \(\frac{5}{\sigma} \approx 0.772\).

Thus, \(\sigma = \frac{5}{0.772} \approx 6.48\).

(b) We need to find \(P(-4 < X - 82 < 4)\), which is equivalent to:

\(P\left( -\frac{4}{\sigma} < Z < \frac{4}{\sigma} \right)\)

Using \(\sigma = 6.48\), we have:

\(P(-0.6176 < Z < 0.6176)\)

From standard normal distribution tables, \(\Phi(0.6176) = 0.7317\).

Therefore, the probability is:

\(2 \times 0.7317 - 1 = 0.463\)

Given that the heights are normally distributed with mean \(\mu = 148\) cm and standard deviation \(\sigma = 8\) cm, we need to find the height \(h\) such that \(P(X < h) = 0.67\).

Using the standard normal distribution, we find the z-score corresponding to a probability of 0.67. From standard normal distribution tables, \(P(Z < 0.44) = 0.67\), so \(z = 0.44\).

The z-score formula is \(z = \frac{h - \mu}{\sigma}\).

Substitute the known values: \(0.44 = \frac{h - 148}{8}\).

Solve for \(h\):

\(h - 148 = 0.44 \times 8\)

\(h - 148 = 3.52\)

\(h = 148 + 3.52\)

\(h = 151.52\)

Rounding to the nearest whole number, \(h = 152\).