在一所初级学院中，体育（PE）部门正在分析两个年级 C1 和 C2 的女生完成 2.4 公里跑步测试所用的时间。 随机挑选的一名 C1 女生完成 2.4 公里跑步测试的时间（以分钟为单位）服从均值为 14.8 分钟、方差为 $2\ \text{minutes}^2$ 的正态分布。 体育部门对 C1 女生采用了新的训练策略。在进行新的训练策略后，随机挑选的一名 C1 女生完成 2.4 公里跑步测试的时间（以分钟为单位）现在据称服从分布 $N(\mu,\sigma^2$)。 如果女生能在不超过 15 分钟 40 秒内完成 2.4 公里跑步测试，她们将获得至少 B 级。女生 2.4 公里跑步测试的及格时间为 17 分钟 20 秒。

1. State the distribution of the sample mean running time, $\bar{X}$, for $k$ girls.

Let $\bar{X}$ be the sample mean running time. Then,
$\bar{X} \sim \text{N}\left(14.8, \frac{2}{k}\right)$

2. Convert 15 minutes 40 seconds to minutes and set up the probability inequality.

$15 \text{ min } 40 \text{ s} = 15 + \frac{40}{60} = \frac{47}{3} \text{ min}$
We require:
$\text{P}\left(\bar{X} \le \frac{47}{3}\right) > 0.99$

3. Standardise the inequality and solve for $k$ using the inverse normal distribution.

$\text{P}\left(Z \le \frac{\frac{47}{3} - 14.8}{\sqrt{\frac{2}{k}}}\right) > 0.99$
$\frac{13/15}{\sqrt{2/k}} > 2.3263$
$\sqrt{\frac{k}{2}} > \frac{2.3263}{13/15}$
$\sqrt{k} > 3.7964$
$k > 14.41$
Since $k$ must be an integer, the least value of $k$ is $15$.

1. Convert the given times into minutes and formulate the probability statements based on the context.

Let $X$ be the running time, in minutes, of a randomly chosen C1 girl after the new training strategy.
$X \sim \mathrm{N}(\mu, \sigma^2)$

$15\text{ minutes } 40\text{ seconds} = 15\frac{40}{60} = \frac{47}{3}\text{ minutes}$
$17\text{ minutes } 20\text{ seconds} = 17\frac{20}{60} = \frac{52}{3}\text{ minutes}$

Given the conditions for at least a B grade and failing:
$P\left(X \le \frac{47}{3}\right) = 0.70$
$P\left(X > \frac{52}{3}\right) = 0.05 \implies P\left(X \le \frac{52}{3}\right) = 0.95$

2. Standardise the variable $X$ to form linear equations in terms of $\mu$ and $\sigma$.

Standardising to $Z = \frac{X - \mu}{\sigma} \sim \mathrm{N}(0, 1)$:

$P\left(Z \le \frac{47/3 - \mu}{\sigma}\right) = 0.70$
From the inverse normal distribution, $\frac{47/3 - \mu}{\sigma} \approx 0.52440$
$\implies \frac{47}{3} - \mu = 0.52440\sigma \quad \text{--- (1)}$

$P\left(Z \le \frac{52/3 - \mu}{\sigma}\right) = 0.95$
From the inverse normal distribution, $\frac{52/3 - \mu}{\sigma} \approx 1.64485$
$\implies \frac{52}{3} - \mu = 1.64485\sigma \quad \text{--- (2)}$

3. Solve the simultaneous equations to find the values of $\mu$ and $\sigma$.

Subtracting equation (1) from equation (2):
$\frac{52}{3} - \frac{47}{3} = (1.64485 - 0.52440)\sigma$
$\frac{5}{3} = 1.12045\sigma$
$\sigma = \frac{5}{3(1.12045)} \approx 1.4875$

Substituting $\sigma \approx 1.4875$ back into equation (1):
$\mu = \frac{47}{3} - 0.52440(1.4875) \approx 14.8866$

Rounding to 3 significant figures, we get:
$\mu = 14.9$
$\sigma = 1.49$

1. Evaluate the spread of the distribution to find the probability of extreme low values.

Let $T$ be the running time in minutes. Given $T \sim \text{N}(15.1, 14)$, the standard deviation is $\sigma = \sqrt{14} \approx 3.742$ minutes.

Considering the extreme lower end of the typical normal distribution spread ($3\sigma$ from the mean):
$\mu - 3\sigma = 15.1 - 3\sqrt{14} \approx 3.88 \text{ minutes}$

Alternatively, checking the probability of an exceptionally fast time (e.g., under 8 minutes):
$P(T < 8) = P\left(Z < \frac{8 - 15.1}{\sqrt{14}}\right) \approx P(Z < -1.897) \approx 0.0289$

2. Conclude that the model is incorrect based on the real-world physical context.

It is physically impossible for a randomly chosen JC girl to complete a 2.4 km run in 3.88 minutes. Similarly, having almost $3\%$ of girls finish in under 8 minutes is completely unrealistic for this context.

Therefore, the variance $\sigma^2 = 14$ is excessively large, making the given normal distribution model incorrect.

1. Define the linear combination $D$ for the difference and calculate its mean.

Let $X \sim \mathrm{N}(14.8, 2)$ be the running time in minutes of a randomly chosen C1 girl.
Let $Y \sim \mathrm{N}(15.1, 1.4)$ be the running time in minutes of a randomly chosen C2 girl.
Let $D = (X_1 + X_2 + X_3 + X_4) - 4Y$ be the required difference in time.
$\mathrm{E}(D) = 4\mathrm{E}(X) - 4\mathrm{E}(Y)$
$\mathrm{E}(D) = 4(14.8) - 4(15.1) = -1.2$

2. Calculate the variance of the linear combination $D$.

$\mathrm{Var}(D) = 4\mathrm{Var}(X) + 4^2\mathrm{Var}(Y)$
$\mathrm{Var}(D) = 4(2) + 16(1.4) = 8 + 22.4 = 30.4$
Therefore, $D \sim \mathrm{N}(-1.2, 30.4)$.

3. Find the probability that the magnitude of the difference is less than 1 minute.

We need the probability that the difference is less than a minute, which is $\mathrm{P}(|D| < 1)$ or equivalently $\mathrm{P}(-1 < D < 1)$.
Using the normal distribution function on a graphing calculator:
$\mathrm{P}(-1 < D < 1) \approx 0.140645$
$\mathrm{P}(-1 < D < 1) = 0.141 \text{ (to 3 s.f.)}$

1. State the assumption of independence in the context of the question.

Assumption 1: Whether any girl in the class fails the 2.4 km running test is independent of whether any other girl in the class fails.

2. State the assumption of constant probability in the context of the question.

Assumption 2: The probability of a girl failing the 2.4 km running test is constant for all girls in the class.

1. Calculate the probability that a C2 girl fails the 2.4 km running test.

Let $Y$ be the running time in minutes of a randomly chosen C2 girl.
From part (iv), $Y \sim N(15.1, 1.4)$.
The passing time is $17$ minutes $20$ seconds, which is $17\frac{1}{3}$ minutes.
The probability of a girl failing is $P(Y > 17\frac{1}{3})$.
$p = P(Y > 17\frac{1}{3}) \approx 0.0295461$

2. Set up the binomial model for the number of girls failing in a class of $n$ girls.

Let $X$ be the number of girls who fail the test in a class of $n$ girls.
$X \sim \text{B}(n, 0.0295461)$
We are given that the probability of less than 3 girls failing is $0.977$.
$P(X < 3) = 0.977$
$P(X \le 2) = 0.977$

3. Solve for the value of $n$ using the graphing calculator.

Using a graphing calculator to generate a table of values for $P(X \le 2)$ against $n$:
When $n = 20$, $P(X \le 2) \approx 0.980$
When $n = 21$, $P(X \le 2) \approx 0.977$
When $n = 22$, $P(X \le 2) \approx 0.974$

Comparing with the given probability of $0.977$, the value is $n = 21$.