In this question you should state the parameters of any distributions that you use. To regulate the use of its facilities by its paying members, a gym provides a guideline of a maximum usage time of 90 minutes per visit per paying member. However, the gym's records show that the usage time, $X$ minutes, of its paying members follows the distribution $N(95, \sigma^{2})$, where $\sigma$ is the standard deviation. Only 30% of its paying members adhere to the guideline.

1. Formulate the probability equation based on the given context.

Let $X$ be the usage time in minutes.
$X \sim \mathrm{N}(95, \sigma^2)$
Since only 30% of the paying members adhere to the guideline of a maximum of 90 minutes:
$\mathrm{P}(X \le 90) = 0.3$

2. Standardise the normal variable and use the inverse normal function to find the z-score.

$\mathrm{P}\left(Z \le \frac{90 - 95}{\sigma}\right) = 0.3$
$\frac{-5}{\sigma} = -0.52440$

3. Solve for $\sigma$ and conclude the 'show that' requirement.

$\sigma = \frac{-5}{-0.52440}$
$\sigma = 9.5347\dots$
$\sigma = 9.53 \text{ (shown to 3 s.f.)}$

1. Find the maximum available time for John's workout.

The time duration from $9.15 \text{ pm}$ to closing time at $11.00 \text{ pm}$ is $1 \text{ hour } 45 \text{ minutes}$, which is equivalent to $105 \text{ minutes}$.

2. Calculate the probability that his intended usage time exceeds the available time.

John cannot finish his workout if his usage time $X$ exceeds $105$ minutes. We need to find $\mathrm{P}(X > 105)$.
Given $X \sim \mathrm{N}(95, \sigma^2)$ and using the unrounded $\sigma = 9.5347$ (from part (a)),
$\mathrm{P}(X > 105) \approx 0.14713$
$= 0.147 \text{ (to 3 s.f.)}$

1. Identify the conditions for the 9th member to be the third and last adherent out of 10 members.

Let $A$ be the event that a member adheres to the guideline, with $P(A) = 0.30$.
For the 9th member to be the third and last one to adhere out of 10:

1. Exactly 2 of the first 8 members adhere.
2. The 9th member adheres.
3. The 10th member does not adhere.

2. Calculate the combined probability using the binomial distribution.

Let $Y$ be the number of adherents in the first 8 members, so $Y \sim \text{B}(8, 0.3)$.

$\text{Required probability} = P(Y = 2) \times P(A) \times P(A')$
$= \left[ \binom{8}{2} (0.3)^2 (0.7)^6 \right] \times 0.3 \times 0.7$
$= 28 \times (0.3)^3 \times (0.7)^7$
$= 0.0622598...$
$\approx 0.0623 \text{ (to 3 s.f.)}$

1. Compare the means to determine the relative horizontal positions of the two normal curves.

From the problem, $X \sim \text{N}(95, 9.53^2)$ and $W \sim \text{N}(85, 80)$.
$\mu_W = 85 < \mu_X = 95$.
Therefore, the center of the curve for $W$ lies to the left of the center of the curve for $X$.

2. Compare the standard deviations to determine the relative widths and heights of the two curves.

$\text{Var}(W) = 80 \implies \sigma_W = \sqrt{80} \approx 8.94$
$\text{Var}(X) \approx 9.53^2 \approx 90.8 \implies \sigma_X \approx 9.53$
Since $\sigma_W < \sigma_X$, the spread of $W$ is smaller than $X$.
To maintain a total area of 1, the curve for $W$ must be narrower and have a higher peak than the curve for $X$.

1. Find the unrounded variance of $X$ from earlier parts.

From part (a), we know $\mathrm{P}(X < 90) = 0.3$.
$\frac{90 - 95}{\sigma} = \Phi^{-1}(0.3) \approx -0.52440$
$\sigma \approx 9.5347 \implies \mathrm{Var}(X) = \sigma^2 \approx 90.910$

2. Define the linear combination for the difference in times and compute its mean and variance.

Let $Y = X_1 + X_2 - 2W$, where $X_1$ and $X_2$ are the usage times of the two paying members and $W$ is the usage time of the trial member.
$\mathrm{E}(Y) = \mathrm{E}(X_1) + \mathrm{E}(X_2) - 2\mathrm{E}(W)$
$\mathrm{E}(Y) = 95 + 95 - 2(85) = 20$
$\mathrm{Var}(Y) = \mathrm{Var}(X_1) + \mathrm{Var}(X_2) + (-2)^2\mathrm{Var}(W)$
$\mathrm{Var}(Y) = 90.910 + 90.910 + 4(80) = 501.82$
$\therefore Y \sim \mathrm{N}(20, 501.82)$

3. Formulate the probability inequality for the absolute difference and evaluate it.

We require the probability that the absolute difference is at least 15 minutes, which is $\mathrm{P}(|Y| \ge 15)$.
$\mathrm{P}(|Y| \ge 15) = 1 - \mathrm{P}(-15 < Y < 15)$
Using a graphing calculator:
$\mathrm{P}(|Y| \ge 15) \approx 1 - 0.35258 = 0.64742$
$\mathrm{P}(|Y| \ge 15) \approx 0.647 \text{ (to 3 s.f.)}$

1. State the necessary condition for variance to be additive

For the calculation $\mathrm{Var}(X_1 + X_2 - 2W) = \mathrm{Var}(X_1) + \mathrm{Var}(X_2) + 2^2\mathrm{Var}(W)$ to be valid, the variables must be independent.