Let the random variable \(X\) represent the number of people who enjoy watching Historical Drama. \(X\) follows a binomial distribution with parameters \(n = 160\) and \(p = 0.1\).

We approximate this binomial distribution with a normal distribution. First, calculate the mean \(np\) and variance \(npq\):

\(np = 160 \times 0.1 = 16\)

\(npq = 160 \times 0.1 \times 0.9 = 14.4\)

Using the normal approximation, we have \(X \sim N(16, 14.4)\).

We need to find \(P(X > 17)\). Using continuity correction, this is \(P(X \geq 17.5)\).

Standardize the variable:

\(P(X \geq 17.5) = P\left( Z \geq \frac{17.5 - 16}{\sqrt{14.4}} \right) = P(Z \geq 0.3953)\)

Using standard normal distribution tables, \(P(Z \geq 0.3953) = 1 - P(Z < 0.3953) = 1 - 0.6536 = 0.346\).

The problem can be modeled using a binomial distribution with parameters:

- Number of trials, \(n = 365\) (days in a year) - Probability of success, \(p = 0.15\) (probability of eating potato)

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

\(\mu = np = 365 \times 0.15 = 54.75\)

\(\sigma^2 = np(1-p) = 365 \times 0.15 \times 0.85 = 46.5375\)

We approximate the binomial distribution with a normal distribution \(N(\mu, \sigma^2)\).

To find the probability that Annabel eats potato on more than 44 days, we standardize the variable:

\(P(X > 44) = P\left( Z > \frac{44.5 - 54.75}{\sqrt{46.5375}} \right)\)

\(= P(Z > -1.5025)\)

Using the standard normal distribution table, \(P(Z > -1.5025) = 0.933\).

The problem involves a binomial distribution with parameters:

- Number of trials, \(n = 100\)

- Probability of success (asleep), \(p = 0.2\)

We approximate the binomial distribution with a normal distribution:

- Mean, \(\mu = np = 100 \times 0.2 = 20\)

- Variance, \(\sigma^2 = np(1-p) = 100 \times 0.2 \times 0.8 = 16\)

- Standard deviation, \(\sigma = \sqrt{16} = 4\)

We want to find \(P(X \leq 30)\). Using continuity correction, we find \(P(X \leq 30.5)\).

Standardizing, we calculate the z-score:

\(z = \frac{30.5 - 20}{4} = 2.625\)

Using the standard normal distribution table, we find:

\(P(Z < 2.625) = 0.996\)

The probability of exactly 7 passengers carrying backpacks in a minibus is given by the binomial probability:

\(P(7) = \binom{12}{7} (0.65)^7 (0.35)^5 = 0.2039\)

For 250 minibuses, the mean and variance of the number of minibuses with exactly 7 passengers carrying backpacks are:

Mean: \(250 \times 0.2039 = 50.9798\)

Variance: \(250 \times 0.2039 \times (1 - 0.2039) = 40.5851\)

We approximate the distribution of the number of minibuses with exactly 7 passengers carrying backpacks by a normal distribution with the above mean and variance.

We need to find \(P(X > 54)\), where \(X\) is the number of minibuses with exactly 7 passengers carrying backpacks.

Using continuity correction, we find:

\(P(X > 54) = P\left( \frac{54.5 - 50.9798}{\sqrt{40.5851}} \right) = P(Z > 0.5526)\)

Using the standard normal distribution table, \(P(Z > 0.5526) = 1 - \Phi(0.5526) = 1 - 0.7098 = 0.2902\)

Thus, the probability is approximately \(0.290\).

(i) The probability of waiting between 5 and 10 minutes is given by the difference between the probability of waiting less than 10 minutes and the probability of waiting less than 5 minutes:

\(P(5 < X < 10) = P(X < 10) - P(X < 5) = 0.88 - 0.16 = 0.72\)

(ii) Let \(X\) be the number of people who wait between 5 and 10 minutes. \(X\) follows a binomial distribution \(X \sim B(180, 0.72)\).

Using a normal approximation, \(X\) can be approximated by a normal distribution \(X \sim N(np, npq)\) where:

\(np = 180 \times 0.72 = 129.6\)

\(npq = 180 \times 0.72 \times 0.28 = 36.288\)

Thus, \(X \sim N(129.6, 36.288)\).

We need to find \(P(X > 115)\). Using continuity correction, this becomes \(P(X > 115.5)\).

Standardizing, we have:

\(P\left( z > \frac{115.5 - 129.6}{\sqrt{36.288}} \right) = P(z > -2.341)\)

Using standard normal distribution tables, \(P(z > -2.341) = 0.990\).