During the Great Mega Sale, the waiting times, in minutes, for men and women, at their respective fitting rooms, have means 4.2 and 8.5, and standard deviations 1.6 and 2.2 respectively.

1. Define the random variable for the sample mean and state its distribution using the Central Limit Theorem.

Let $W$ be the waiting time of a woman, with mean $\mu_W = 8.5$ and variance $\sigma_W^2 = 2.2^2$.
Let $\bar{W}$ be the sample mean waiting time of 50 randomly chosen women.
Since the sample size $n = 50$ is large ($n \ge 30$), by the Central Limit Theorem, $\bar{W}$ is approximately normally distributed:
$\bar{W} \sim \text{N}\left(8.5, \frac{2.2^2}{50}\right)$
$\bar{W} \sim \text{N}(8.5, 0.0968)$

2. Formulate the inequality based on the given condition.

The problem states that the probability of the average waiting time exceeding $k$ minutes is at most $10\%$. Hence,
$\text{P}(\bar{W} > k) \le 0.1$

3. Find the critical threshold for the probability using a graphing calculator.

First, find the threshold value $w$ such that $\text{P}(\bar{W} > w) = 0.1$.
Using the inverse normal function on a graphing calculator:
$w \approx 8.8987$
To satisfy $\text{P}(\bar{W} > k) \le 0.1$, since the probability decreases as $k$ increases, we must have:
$k \ge 8.8987$

4. Determine the least integer value for $k$.

Since $k$ must be an integer and $k \ge 8.8987$, the possible integer values are $9, 10, 11, \dots$
The least integer value is:
$k = 9$

1. Define the linear combination for the difference in waiting times and calculate its expected value.

Let $M$ and $W$ be the waiting times of a man and a woman respectively.
Let $D$ be the difference between the combined waiting time of 6 men and three times the waiting time of 1 woman:
$D = (M_1 + M_2 + M_3 + M_4 + M_5 + M_6) - 3W$.
$\text{E}(D) = 6\text{E}(M) - 3\text{E}(W) = 6(4.2) - 3(8.5) = -0.3$

2. Calculate the variance of the linear combination.

$\text{Var}(D) = 6\text{Var}(M) + 3^2\text{Var}(W)$
$\text{Var}(D) = 6(1.6^2) + 9(2.2^2) = 15.36 + 43.56 = 58.92$

3. Determine the distribution of the linear combination and calculate the required probability using a graphing calculator.

Since the waiting times are normally distributed and assumed to be independent, $D \sim \text{N}(-0.3, 58.92)$.
We require the probability that the difference is strictly more than 4 minutes, which is $\text{P}(|D| > 4)$.
$\text{P}(|D| > 4) = \text{P}(D < -4) + \text{P}(D > 4)$
Using the graphing calculator:
$\text{P}(|D| > 4) \approx 0.31489 + 0.28767 = 0.60256 \approx 0.603 \text{ (to 3 s.f.)}$

4. State the assumption required for the variance calculation to be valid.

Assumption: The waiting times of all the randomly chosen men and the randomly chosen woman are independent of one another.

1. Find the probability that a single randomly chosen woman waits for more than 8.5 minutes.

Let $W$ be the waiting time for a woman. Based on the assumption from the previous part, $W$ follows a normal distribution with mean $\mu = 8.5$.
Due to the symmetry of the normal distribution curve about its mean:
$P(W > 8.5) = 0.5$

2. Model the scenario using a Binomial distribution and calculate the probability of at least 3 women waiting more than 8.5 minutes.

Let $X$ be the number of women, out of 4, who wait for more than 8.5 minutes.
Since the women are chosen randomly, their waiting times are independent, and we can model this as $X \sim \text{B}(4, 0.5)$.
We need to find $P(X \ge 3)$:
$\begin{aligned} P(X \ge 3) &= P(X = 3) + P(X = 4) \\ &= \binom{4}{3}(0.5)^3(0.5)^1 + \binom{4}{4}(0.5)^4(0.5)^0 \\ &= 4(0.0625) + 1(0.0625) \\ &= 5(0.0625) \\ &= 0.3125 \end{aligned}$