Let the random variable \(X\) represent the number of cars made by Ford in a sample of 50 cars. \(X\) follows a binomial distribution with parameters \(n = 50\) and \(p = 0.28\).

To use a normal approximation, we calculate the mean \(\mu\) and variance \(\sigma^2\) of the distribution:

\(\mu = 50 \times 0.28 = 14\)

\(\sigma^2 = 50 \times 0.28 \times 0.72 = 10.08\)

We apply a continuity correction to find \(P(X > 18)\), which is approximated by \(P(X > 18.5)\).

Standardize the variable:

\(Z = \frac{18.5 - 14}{\sqrt{10.08}} = 1.417\)

Using the standard normal distribution table, find \(\Phi(1.417) \approx 0.9218\) or \(0.9217\).

Thus, \(P(X > 18) = 1 - \Phi(1.417) = 1 - 0.9218 = 0.0782\) or \(1 - 0.9217 = 0.0783\).

Let the random variable \(X\) represent the number of students with blue eyes in a sample of 80 students. \(X\) follows a binomial distribution with parameters \(n = 80\) and \(p = 0.32\).

To approximate, we use a normal distribution. The mean \(\mu\) and variance \(\sigma^2\) are given by:

\(\mu = np = 80 \times 0.32 = 25.6\)

\(\sigma^2 = np(1-p) = 80 \times 0.32 \times 0.68 = 17.408\)

Using the normal approximation, we apply continuity correction for \(P(X < 20)\):

\(P(X < 20) = P\left(Z < \frac{19.5 - 25.6}{\sqrt{17.408}}\right) = P(Z < -1.462)\)

Using standard normal distribution tables or a calculator, find:

\(P(Z < -1.462) = 1 - \Phi(1.462) = 1 - 0.9282 = 0.0718\)

Thus, the probability that fewer than 20 students have blue eyes is approximately \(0.0718 \leq p < 0.0719\).

Let the random variable \(X\) represent the number of damaged tapes in a sample of 1600 tapes. Given that 1 in 5 tapes are damaged, the probability of a tape being damaged is \(p = 0.2\).

The expected number of damaged tapes is \(\mu = np = 1600 \times 0.2 = 320\).

The variance is \(\sigma^2 = np(1-p) = 1600 \times 0.2 \times 0.8 = 256\).

We approximate the binomial distribution with a normal distribution: \(X \sim N(320, 256)\).

We need to find \(P(X \geq 290)\). Using continuity correction, this is \(P(X > 289.5)\).

Standardize: \(Z = \frac{289.5 - 320}{\sqrt{256}} = \frac{-30.5}{16} = -1.906\).

Thus, \(P(X \geq 290) = 1 - \Phi(-1.906) = \Phi(1.906)\).

Using standard normal distribution tables, \(\Phi(1.906) = 0.972\).

(i) The number of eggs laid by the hens follows a binomial distribution with parameters \(n = 30\) and \(p = 0.7\). We need to find \(P(X = 24)\).

Using the binomial probability formula:

\(P(X = 24) = \binom{30}{24} (0.7)^{24} (0.3)^{6}\)

\(= (0.7)^{24} \times (0.3)^{6} \times \binom{30}{24}\)

\(= 0.0829\)

(ii) To approximate the probability that fewer than 20 eggs are laid, we use a normal approximation to the binomial distribution.

The mean \(\mu\) and variance \(\sigma^2\) of the binomial distribution are given by:

\(\mu = np = 30 \times 0.7 = 21\)

\(\sigma^2 = np(1-p) = 30 \times 0.7 \times 0.3 = 6.3\)

Using the normal approximation, we standardize the variable:

\(P(X < 20) \approx P\left(Z < \frac{19.5 - 21}{\sqrt{6.3}}\right)\)

\(= P(Z < -0.5976)\)

Using standard normal distribution tables:

\(P(Z < -0.5976) = 1 - 0.7251 = 0.275\)

(i) Let the probability of choosing a chocolate biscuit be \(p = \frac{1}{4}\) and a cream biscuit be \(1 - p = \frac{3}{4}\). The probability of having 5 chocolate and 5 cream biscuits in a box of 10 is given by:

\(P(\text{equal}) = \left(0.25\right)^5 \times \left(0.75\right)^5 \times \binom{10}{5} = 0.0584\)

(ii) The probability that exactly 1 out of 8 boxes contains equal numbers of cream and chocolate biscuits is:

\((0.0584)^1 \times (0.9416)^7 \times \binom{8}{1} = 0.307\)

(iii) For a large box of 120 biscuits, use the normal approximation. Let \(X\) be the number of chocolate biscuits. Then \(\mu = 120 \times 0.25 = 30\) and \(\sigma^2 = 30 \times 0.75 = 22.5\).

Using continuity correction, find \(P(X < 35)\):

\(P(X < 35) = \Phi\left(\frac{34.5 - 30}{\sqrt{22.5}}\right) = \Phi(0.949)\)

From standard normal tables, \(\Phi(0.949) = 0.829\).