Given that 69% of the distribution is less than 28, we find the corresponding z-score from the standard normal distribution table, which is approximately 0.496. Thus, we have:

\(28 - \mu = 0.496\sigma\)

Similarly, for 90% less than 35, the z-score is approximately 1.282. Thus, we have:

\(35 - \mu = 1.282\sigma\)

We now have two equations:

1. \(28 - \mu = 0.496\sigma\)
2. \(35 - \mu = 1.282\sigma\)

Subtract the first equation from the second:

\((35 - \mu) - (28 - \mu) = 1.282\sigma - 0.496\sigma\)

\(7 = 0.786\sigma\)

Solving for \(\sigma\):

\(\sigma = \frac{7}{0.786} \approx 8.91\)

Substitute \(\sigma = 8.91\) back into the first equation:

\(28 - \mu = 0.496 \times 8.91\)

\(28 - \mu = 4.42\)

\(\mu = 28 - 4.42 = 23.6\)

Thus, the mean \(\mu = 23.6\) and the standard deviation \(\sigma = 8.91\).

Let the height of the sunflowers be a normally distributed random variable \(X\) with mean \(\mu = 115\) cm and standard deviation \(\sigma\).

We know that 80% of the heights are greater than 103 cm, which means 20% are less than 103 cm. This corresponds to a cumulative probability of 0.20.

Using the standard normal distribution table, we find that the z-score corresponding to a cumulative probability of 0.20 is approximately \(z = -0.842\).

We use the z-score formula:

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

Substituting the known values:

\(-0.842 = \frac{103 - 115}{\sigma}\)

\(-0.842 = \frac{-12}{\sigma}\)

Solving for \(\sigma\):

\(\sigma = \frac{-12}{-0.842}\)

\(\sigma \approx 14.3\)

To find the least distance that must be thrown to qualify for a certificate, we need to find the 90th percentile of the normal distribution, as the top 10% of pupils qualify.

The mean \\(\mu \\\) is 35.0 m and the standard deviation \\(\sigma \\\) is 11.6 m. We need to find the value of \\(x \\\) such that the cumulative probability is 0.90.

Using the standard normal distribution table, the z-score corresponding to the 90th percentile is approximately \\(z = 1.282 \\\).

We use the z-score formula:

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

Substitute the known values:

\\(1.282 = \frac{x - 35.0}{11.6} \\\)

Solve for \\(x \\\):

\\(x - 35.0 = 1.282 \times 11.6 \\\)

\\(x - 35.0 = 14.8592 \\\)

\\(x = 35.0 + 14.8592 \\\)

\\(x \approx 49.9 \\\)

Therefore, the least distance that must be thrown to qualify for a certificate is 49.9 m.

(i) Since \(P(X > 3.6) = 0.5\), the mean \(\mu\) is 3.6 because the probability of being greater than the mean in a normal distribution is 0.5.

For \(P(X > 2.8) = 0.6554\), we standardize: \(\frac{2.8 - \mu}{\sigma} = -0.4\).

Substitute \(\mu = 3.6\): \(\frac{2.8 - 3.6}{\sigma} = -0.4\).

Solve for \(\sigma\): \(\sigma = \frac{-0.8}{-0.4} = 2\).

(ii) Let \(p = 0.6554\) be the probability that an observation is greater than 2.8. We use the binomial distribution with \(n = 4\) and \(p = 0.6554\).

The probability that at least two observations are greater than 2.8 is:

\(P(X \geq 2) = \binom{4}{2} (0.6554)^2 (0.3446)^2 + \binom{4}{3} (0.6554)^3 (0.3446) + \binom{4}{4} (0.6554)^4\).

Calculate each term:

\(\binom{4}{2} (0.6554)^2 (0.3446)^2 = 0.3061\)

\(\binom{4}{3} (0.6554)^3 (0.3446) = 0.3881\)

\(\binom{4}{4} (0.6554)^4 = 0.1845\)

Sum these probabilities: \(0.3061 + 0.3881 + 0.1845 = 0.879\).

(a) To find the probability that a randomly chosen male leopard weighs between 46 and 62 kg, we standardize the values using the formula:

\(P(46 < X < 62) = P\left( \frac{46 - 55}{6} < Z < \frac{62 - 55}{6} \right)\)

This simplifies to:

\(P(-1.5 < Z < 1.1667)\)

Using the standard normal distribution table, we find:

\(\Phi(1.1667) - \Phi(-1.5) = 0.8784 + (0.9332 - 1) = 0.812\)

(b) Given that 25% of female leopards weigh less than 36 kg, we find the z-value corresponding to 0.25, which is approximately -0.674. Using the standardization formula:

\(\frac{36 - 42}{\sigma} = -0.674\)

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

\(\sigma = \frac{36 - 42}{-0.674} = 8.9\)

(c) To find the probability that both leopards weigh less than 46 kg, we calculate:

For male leopards: \(P(X < 46) = 1 - \Phi\left(\frac{46 - 55}{6}\right) = 0.0668\)

For female leopards: \(P(Y < 46) = \Phi\left(\frac{46 - 42}{8.9}\right) = 0.6732\)

The probability that both events occur is:

\(P(\text{both}) = 0.0668 \times 0.6732 = 0.0450\)