在本题中，你应该写明你所使用的任何分布的参数。 为了规范付费会员对其设施的使用，某健身房规定了一项指南，建议每位付费会员每次使用的最长时间为 90 分钟。然而，该健身房的记录显示，其付费会员的使用时间 $X$（分钟）服从分布 $\text{N}(95, \sigma^2)$，其中 $\sigma$ 是标准差。只有 30% 的付费会员遵守该指南。

1. Formulate the probability equation based on the context.

Since only 30% of the paying members adhere to the guideline of a maximum usage time of 90 minutes,
$P(X \le 90) = 0.3$

2. Standardise the variable and find the corresponding $z$-value.

$P\left(Z \le \frac{90 - 95}{\sigma}\right) = 0.3$
Using the inverse normal function on a calculator,
$\frac{-5}{\sigma} \approx -0.52440$

3. Solve for $\sigma$.

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

1. Determine the maximum time John has to work out before the gym closes.

Time available from $9.15\text{ pm}$ to $11.00\text{ pm}$ is $1$ hour $45$ minutes, which is $105$ minutes.
Therefore, he cannot finish his workout if his usage time $X > 105$.

2. Find the probability that his workout time exceeds the available time.

Using the unrounded standard deviation $\sigma \approx 9.5347$ from part (a), we have $X \sim \text{N}(95, 9.5347^2)$.
$\text{P}(X > 105) \approx 0.14713 \approx 0.147 \text{ (to 3 s.f.)}$

1. State the probability of a member adhering to the guideline and define the condition for the required event.

From the problem context, the probability that a paying member adheres to the guideline is $p = 0.3$.
The probability of not adhering is $1 - 0.3 = 0.7$.

For the 9th member to be the 3rd and last one to adhere out of 10 members, the following independent events must occur in sequence:

1. Exactly 2 out of the first 8 members adhere.
2. The 9th member adheres.
3. The 10th member does not adhere (since the 9th is the last one).

2. Calculate the combined probability using the binomial probability formula for the first 8 members.

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

$\begin{aligned} P(\text{required event}) &= P(Y=2) \times P(\text{9th adheres}) \times P(\text{10th does not adhere}) \\ &= \left[ \binom{8}{2} (0.3)^2 (0.7)^6 \right] \times 0.3 \times 0.7 \\ &= 28 \times (0.3)^3 \times (0.7)^7 \\ &\approx 0.0622598 \\ &\approx 0.0623 \text{ (to 3 s.f.)} \end{aligned}$

1. Determine the relative positions of the peaks of both distributions.

$W \sim \text{N}(85, 80)$ and $X \sim \text{N}(95, 9.53^2)$.

The mean of $W$ is $85$ and the mean of $X$ is $95$.
Therefore, the peak of the curve for $W$ occurs at $x = 85$, which is to the left of the peak for $X$ at $x = 95$.

2. Determine the relative heights and spreads of both distributions to complete the sketch.

The variance of $W$ is $80$, so its standard deviation is $\sigma_W = \sqrt{80} \approx 8.94$.
The standard deviation of $X$ is $\sigma_X = 9.53$ (from part a).

Since $\sigma_W < \sigma_X$, the curve for $W$ has a smaller spread and a correspondingly higher peak than the curve for $X$.
Both curves should be drawn as bell-shapes on the same axis, asymptotic to the horizontal axis, with their respective peaks at $85$ and $95$, and clearly labelled $W$ and $X$.

1. Define the appropriate linear combination of variables and find its expected value.

Let $Y = X_1 + X_2 - 2W$.
$\mathrm{E}(Y) = \mathrm{E}(X_1) + \mathrm{E}(X_2) - 2\mathrm{E}(W)$
$\mathrm{E}(Y) = 95 + 95 - 2(85) = 20$

2. Find the variance of the linear combination.

$\mathrm{Var}(Y) = \mathrm{Var}(X_1) + \mathrm{Var}(X_2) + (-2)^2 \mathrm{Var}(W)$
Using $\sigma \approx 9.5347$ from part (a):
$\mathrm{Var}(Y) = 9.5347^2 + 9.5347^2 + 4(80)$
$\mathrm{Var}(Y) = 181.821 + 320 = 501.821$

3. State the distribution of the linear combination.

Since $X$ and $W$ are independent normal random variables, $Y$ is also normally distributed:
$Y \sim \mathrm{N}(20, 501.821)$

4. Calculate the probability that the difference is at least 15 minutes.

We require $\mathrm{P}(|Y| \ge 15)$, which means the difference is either $\le -15$ or $\ge 15$.
$\mathrm{P}(|Y| \ge 15) = \mathrm{P}(Y \le -15) + \mathrm{P}(Y \ge 15)$
Using a graphing calculator:
$\mathrm{P}(Y \le -15) \approx 0.059097$
$\mathrm{P}(Y \ge 15) \approx 0.588307$
$\mathrm{P}(|Y| \ge 15) \approx 0.647404 = 0.647$ (to 3 s.f.)

1. State the assumption required for combining the variances of the normal distributions.

For the calculation of $\mathrm{Var}(X_1 + X_2 - 2W) = \mathrm{Var}(X_1) + \mathrm{Var}(X_2) + 4\mathrm{Var}(W)$ to be valid, the random variables must be independent. Thus, we assume that the usage times of the randomly chosen paying members and the trial member are independent of one another.