Let the random variable \(X\) represent the number of students with blue eyes. Given that one in five students has blue eyes, the probability \(p\) is \(0.2\).

The expected number of students with blue eyes is \(\mu = 120 \times 0.2 = 24\).

The variance is \(\sigma^2 = 120 \times 0.2 \times 0.8 = 19.2\).

We approximate \(X\) by a normal distribution \(N(24, 19.2)\).

To find \(P(X < 33)\), we use a continuity correction: \(P(X < 33) \approx P(Z < \frac{32.5 - 24}{\sqrt{19.2}})\).

Calculate the standard score: \(Z = \frac{32.5 - 24}{\sqrt{19.2}} = 1.9398\).

Using standard normal distribution tables, \(P(Z < 1.9398) = 0.974\).

Let the random variable \(X\) represent the number of people whose birthday falls on a Monday. Since each day of the week is equally likely, the probability \(p\) that a person's birthday is on a Monday is \(\frac{1}{7}\).

We have \(n = 350\) people, so the expected number of people with a birthday on Monday is \(np = 350 \times \frac{1}{7} = 50\).

The variance is \(npq = 350 \times \frac{1}{7} \times \frac{6}{7} = 42.857\).

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

\(P(X \geq 47) = P\left( z > \frac{46.5 - 50}{\sqrt{42.857}} \right)\)

Calculating the z-score:

\(z = \frac{46.5 - 50}{\sqrt{42.857}} = -0.5346\)

Thus, \(P(z > -0.5346) = 0.704\).

Let the probability that a value is less than 5 be denoted by \(p\). Since the probability that a value is greater than 5 is 0.15, we have \(p = 0.85\).

The mean number of values less than 5 in 200 trials is \(\mu = 200 \times 0.85 = 170\).

The variance is \(\text{var} = 200 \times 0.85 \times 0.15 = 25.5\).

We need to find \(P(\text{at least 160})\), which is equivalent to \(P(X \geq 160)\).

Using continuity correction, this becomes \(P(X > 159.5)\).

Standardizing, we have:

\(P\left( z > \frac{159.5 - 170}{\sqrt{25.5}} \right) = P(z > -2.079)\)

Using standard normal distribution tables, \(P(z > -2.079) = 0.981\).

The probability that Ana is on time on any given day is:

\(p = 1 - 0.05 - 0.75 = 0.2\)

We are looking for the probability that she is on time on fewer than 20 out of 96 days. This is a binomial distribution problem with parameters \(n = 96\) and \(p = 0.2\).

We approximate the binomial distribution with a normal distribution:

\(\mu = np = 96 \times 0.2 = 19.2\)

\(\sigma^2 = np(1-p) = 96 \times 0.2 \times 0.8 = 15.36\)

Using continuity correction, we find:

\(P(X < 20) = P\left( Z < \frac{19.5 - 19.2}{\sqrt{15.36}} \right) = P(Z < 0.07654)\)

Using standard normal distribution tables or a calculator, we find:

\(P(Z < 0.07654) = 0.531\)

(b) We know that 72% of Eastern bluebirds have a length greater than 17.1 cm, which implies \(P(Z > \frac{17.1 - 18.4}{\sigma}) = 0.72\). This corresponds to a standard normal variable \(Z\) such that \(Z = -0.583\) (since \(P(Z > -0.583) = 0.72\)).

Thus, \(\frac{17.1 - 18.4}{\sigma} = -0.583\).

Solving for \(\sigma\), we get \(\sigma = \frac{17.1 - 18.4}{-0.583} = 2.23\).

(c) For a sample of 120 bluebirds, the mean number with length greater than 17.1 cm is \(120 \times 0.72 = 86.4\).

The variance is \(120 \times 0.72 \times 0.28 = 24.192\).

Using a normal approximation, \(P(X < 80) = P(Z < \frac{79.5 - 86.4}{\sqrt{24.192}})\) (using continuity correction).

\(Z = \frac{79.5 - 86.4}{\sqrt{24.192}} = -1.4029\).

Thus, \(P(Z < -1.4029) = 1 - \Phi(1.403) = 1 - 0.9196 = 0.0804\).