Let the random variable representing the number of people with blood group A+ be denoted by \(X\). Given that 35% of the population are group A+, we have \(p = 0.35\).

The sample size is \(n = 150\). The mean \(\mu\) and variance \(\sigma^2\) of \(X\) are given by:

\(\mu = n \times p = 150 \times 0.35 = 52.5\)

\(\sigma^2 = n \times p \times (1-p) = 150 \times 0.35 \times 0.65 = 34.125\)

We use the normal approximation to the binomial distribution. Applying continuity correction, we find:

\(P(X > 60) = P\left( Z > \frac{60.5 - 52.5}{\sqrt{34.125}} \right)\)

Calculating the standard score:

\(Z = \frac{60.5 - 52.5}{\sqrt{34.125}} = 1.369\)

Using standard normal distribution tables, we find:

\(P(Z > 1.369) = 1 - \Phi(1.369) = 0.0854 \text{ or } 0.0855\)

Let the probability of landing on the blue side be \(p\). Then, the probability of landing on the red side is \(4p\), and the probability of landing on the green side is \(3p\).

Since the total probability must be 1, we have:

\(p + 4p + 3p = 1\)

\(8p = 1\)

\(p = \frac{1}{8}\)

Thus, the probability of landing on the blue side is \(\frac{1}{8}\).

For a binomial distribution with \(n = 136\) and \(p = \frac{1}{8}\), the mean \(\mu\) is:

\(\mu = np = 136 \times \frac{1}{8} = 17\)

The variance \(\sigma^2\) is:

\(\sigma^2 = np(1-p) = 136 \times \frac{1}{8} \times \frac{7}{8} = 14.875\)

Using a normal approximation, we standardize for \(P(X < 20)\) with continuity correction:

\(P(X < 20) = P\left( Z < \frac{19.5 - 17}{\sqrt{14.875}} \right)\)

\(= P(Z < 0.648)\)

Using standard normal distribution tables, \(\Phi(0.648) = 0.742\).

Therefore, the probability that the spinner lands on the blue side fewer than 20 times is 0.742.

Let the random variable \(X\) represent the number of seeds that germinate. \(X\) follows a binomial distribution with parameters \(n = 250\) and \(p = 0.86\).

We approximate the binomial distribution with a normal distribution since \(n\) is large and \(p\) is not too close to 0 or 1.

The mean \(\mu\) of the distribution is given by:

\(\mu = np = 250 \times 0.86 = 215\)

The variance \(\sigma^2\) is given by:

\(\sigma^2 = np(1-p) = 250 \times 0.86 \times 0.14 = 30.1\)

Thus, the standard deviation \(\sigma\) is:

\(\sigma = \sqrt{30.1}\)

We use a continuity correction for the normal approximation. We want \(P(X > 210)\), which is approximated by \(P(X > 210.5)\).

Standardizing, we find:

\(P(X > 210.5) = 1 - \Phi\left( \frac{210.5 - 215}{\sqrt{30.1}} \right)\)

\(= 1 - \Phi(-0.820)\)

\(= \Phi(0.820)\)

Using standard normal distribution tables, \(\Phi(0.820) = 0.794\).

(i) To find \(\mu\), we use the standard normal distribution. Given \(\text{P}(X > 20) = 0.04\), we find the corresponding \(z\)-value: \(z = \pm 1.751\).

Using the standardization formula: \(\frac{20 - \mu}{\mu/4} = 1.751\).

Solving for \(\mu\), we get \(\mu = 13.9\).

(ii) To find \(\text{P}(10 < X < 20)\), we first find \(\text{P}(X < 10)\).

Standardizing: \(\text{P}(z < \frac{10 - 13.91}{13.91/4}) = \text{P}(z < -1.124)\).

\(\text{P}(z < -1.124) = 1 - 0.8694 = 0.131\).

Thus, \(\text{P}(10 < X < 20) = 0.96 - 0.131 = 0.829 \text{ or } 0.830\).

(iii) For 250 observations, \(\mu = 250 \times 0.96 = 240\) and \(\sigma^2 = 250 \times 0.96 \times 0.04 = 9.6\).

We need \(\text{P}(\geq 235) = 1 - \Phi\left(\frac{234.5 - 240}{\sqrt{9.6}}\right)\).

\(\Phi(1.775) = 0.962\).

(i) Let \(X\) be the number of cloudy days in November. \(X\) follows a binomial distribution with parameters \(n = 30\) and \(p = 0.8\).

We approximate \(X\) using a normal distribution with mean \(\mu = np = 30 \times 0.8 = 24\) and variance \(\sigma^2 = npq = 30 \times 0.8 \times 0.2 = 4.8\).

Standard deviation \(\sigma = \sqrt{4.8}\).

To find the probability of fewer than 25 cloudy days, we use a continuity correction: \(P(X < 25) \approx P(Y < 24.5)\) where \(Y\) is normally distributed.

Calculate the z-score: \(z = \frac{24.5 - 24}{\sqrt{4.8}} = 0.228\).

Using standard normal distribution tables, \(P(Z < 0.228) = 0.590\).

(ii) The normal approximation is justified because both \(np = 24\) and \(nq = 6\) are greater than 5.