In a small town, there are two clinics, Lee's Clinic and Hope Clinic. At Lee's Clinic, the waiting time of a patient follows a normal distribution with mean 25 minutes and standard deviation 8 minutes. The waiting time for patient at Hope Clinic follows a normal distribution with mean 37 minutes and standard deviation 4 minutes.

1. Define the random variables and find the distribution of the sample mean for Lee's Clinic.

Let $L$ and $H$ be the waiting times (in minutes) at Lee's Clinic and Hope Clinic respectively.
$L \sim \mathrm{N}(25, 8^2)$ and $H \sim \mathrm{N}(37, 4^2)$
Let $\bar{L}$ be the sample mean waiting time of $5$ patients at Lee's Clinic.
$\bar{L} \sim \mathrm{N}\left(25, \frac{8^2}{5}\right) \equiv \mathrm{N}(25, 12.8)$

2. Formulate the linear combination for the difference and determine its mean and variance.

Let $D = \bar{L} - H$.
$\mathrm{E}(D) = \mathrm{E}(\bar{L}) - \mathrm{E}(H) = 25 - 37 = -12$
$\mathrm{Var}(D) = \mathrm{Var}(\bar{L}) + \mathrm{Var}(H) = 12.8 + 16 = 28.8$
Therefore, $D \sim \mathrm{N}(-12, 28.8)$.

3. Set up the probability expression for a difference of at least 5 minutes.

The probability that the difference is at least $5$ minutes is given by $\mathrm{P}(|\bar{L} - H| \ge 5)$.
$\mathrm{P}(|D| \ge 5) = \mathrm{P}(D \le -5) + \mathrm{P}(D \ge 5)$

4. Evaluate the probability using a graphing calculator.

Using a graphing calculator for $D \sim \mathrm{N}(-12, 28.8)$:
$\mathrm{P}(D \le -5) \approx 0.90393$
$\mathrm{P}(D \ge 5) \approx 0.000768$
Total Probability $= 0.90393 + 0.000768 = 0.90470 \approx 0.905$ (to 3 s.f.)
(Alternatively, $1 - \mathrm{P}(-5 < D < 5) = 1 - 0.095299 \approx 0.905$)

1. State the assumption required for combining the distributions of two different random variables

To calculate the variance of the difference $\bar{L} - H$, we must assume that the random variables are independent.

Assumption: The waiting time of a patient at Lee's Clinic is independent of the waiting time of a patient at Hope Clinic.

1. Identify the probability of the given condition and define the binomial distribution.

Let $W_L$ be the waiting time of a patient at Lee's Clinic. We are given $W_L \sim \mathrm{N}(25, 8^2)$.
Since the normal distribution is symmetric about its mean, the probability that a patient waits more than 25 minutes is $\mathrm{P}(W_L > 25) = 0.5$.
Let $X$ be the number of patients with waiting times of more than 25 minutes out of $n$ patients.
Then $X \sim \mathrm{B}(n, 0.5)$.

2. State the suitable normal approximation.

Since $n > 40$, $np = 0.5n > 5$ and $n(1-p) = 0.5n > 5$, $X$ can be approximated by a normal distribution:
$X \sim \mathrm{N}(np, np(1-p))$
$X \sim \mathrm{N}(0.5n, 0.25n)$ approximately.

3. Formulate the probability inequality using continuity correction and standardise.

We are given $\mathrm{P}(X \le 40) < 0.95$.
Applying continuity correction for the approximation:
$\mathrm{P}(X < 40.5) < 0.95$
Standardising the variable:
$\mathrm{P}\left(Z < \frac{40.5 - 0.5n}{\sqrt{0.25n}}\right) < 0.95$

4. Find the critical value from the standard normal distribution and simplify the inequality.

From the standard normal distribution, $\mathrm{P}(Z < 1.64485) \approx 0.95$.
Thus, we require:
$\frac{40.5 - 0.5n}{0.5\sqrt{n}} < 1.64485$
Since $0.5\sqrt{n} > 0$, we can multiply both sides by it:
$40.5 - 0.5n < 0.82243\sqrt{n}$
$0.5n + 0.82243\sqrt{n} - 40.5 > 0$

5. Solve the quadratic inequality to find the least integer $n$.

Let $y = \sqrt{n}$. The inequality becomes:
$0.5y^2 + 0.82243y - 40.5 > 0$
Solving the quadratic equation $0.5y^2 + 0.82243y - 40.5 = 0$:
$y = \frac{-0.82243 + \sqrt{0.82243^2 - 4(0.5)(-40.5)}}{2(0.5)}$
$y \approx 8.215 \text{ or } y \approx -9.860 \text{ (rejected as } y > 0)$
So we need $y > 8.215$, which means:
$\sqrt{n} > 8.215$
$n > 67.48$
Since $n$ must be an integer, the least value of $n$ is 68.

1. Define the distribution for the number of influenza patients in a single sample.

Let $X$ be the number of patients treated for influenza in a sample of 20 patients.
Given that the probability of a patient being treated for influenza is $\frac{1}{5} = 0.2$,
$X \sim \mathrm{B}(20, 0.2)$

$\mathrm{E}(X) = np = 20(0.2) = 4$
$\mathrm{Var}(X) = np(1-p) = 20(0.2)(0.8) = 3.2$

2. State the approximate distribution of the sample mean using the Central Limit Theorem.

Let $\bar{X}$ be the average number of influenza patients per sample for $n = 60$ samples.
Since the number of samples $n = 60$ is large ($n \ge 30$), by the Central Limit Theorem, $\bar{X}$ is approximately normally distributed.

$\mathrm{E}(\bar{X}) = \mathrm{E}(X) = 4$
$\mathrm{Var}(\bar{X}) = \frac{\mathrm{Var}(X)}{n} = \frac{3.2}{60} = \frac{4}{75}$

Therefore, $\bar{X} \sim \mathrm{N}\left(4, \frac{4}{75}\right)$ approximately.

3. Calculate the probability that the sample mean is greater than 3.5.

We want to find $\mathrm{P}(\bar{X} > 3.5)$.

Using a graphing calculator,
$\mathrm{P}(\bar{X} > 3.5) \approx 0.98480 \approx 0.985$ (to 3 s.f.)

1. State the inappropriateness of the normal model and provide the reasoning based on the mixture of two distributions with different means.

A normal model is not likely to be appropriate.

The waiting times of patients at Lee's Clinic and Hope Clinic follow normal distributions with significantly different means (25 minutes and 37 minutes) and different standard deviations. The combined group will consist of a mixture of these two distinct distributions, which will likely result in a bimodal (two-peaked) or skewed distribution rather than the symmetric, unimodal, bell-shaped curve required for a normal model.