The speeds of an e-scooter ($X$ km/h) and a pedestrian ($Y$ km/h) measured on a particular stretch of footpath are normally distributed with mean and variance as follows:

1. State the value of $\mu$ using the symmetry of the normal distribution.

Since $\Pr(Y < 5.2) = \Pr(Y \ge 7.0)$, by the symmetry of the normal distribution, the mean $\mu$ is exactly in the middle of $5.2$ and $7.0$.
$\mu = \frac{5.2 + 7.0}{2} = 6.1$

2. Set up an equation involving the standard normal variable $Z$ to find $\sigma$.

Given $\Pr(Y < 5.2) = 0.379$, we standardize the variable:
$\Pr\left(Z < \frac{5.2 - 6.1}{\sigma}\right) = 0.379$
$\frac{-0.9}{\sigma} \approx -0.30808$

3. Solve for $\sigma$.

$\sigma = \frac{-0.9}{-0.30808} \approx 2.9213$
$\sigma \approx 2.92 \text{ (to 3 s.f.)}$

1. Translate the 'within a of the mean' condition into a probability equation

$X \sim \mathrm{N}(12.3,\,9.9)$ and we require
$\Pr(12.3 - a < X < 12.3 + a) = 0.5.$
By the symmetry of the normal curve, the two tails together carry the remaining $0.5$, so each tail is $0.25$:
$\Pr(X < 12.3 + a) = 0.75.$

2. Standardise and solve for a

Using the inverse normal (GC) with mean $12.3$ and standard deviation $\sqrt{9.9}$:
$12.3 + a = \mathrm{invNorm}(0.75,\,12.3,\,\sqrt{9.9}) = 14.422\ldots$
$a = 14.422\ldots - 12.3 = 2.122\ldots$
Equivalently, $\dfrac{a}{\sqrt{9.9}} = 0.6745$, so $a = 0.6745 \times 3.1464 = 2.12$.
$a \approx 2.12\text{ km/h}.$

1. Find the probability that a single randomly chosen e-scooter exceeds 10 km/h.

Let $X$ be the speed of an e-scooter in km/h. We are given $X \sim \mathrm{N}(12.3, 9.9)$.
The probability that an e-scooter exceeds 10 km/h is:
$p = \Pr(X > 10) \approx 0.767605$

2. Identify the conditions for the 50th e-scooter to be exactly the 35th one to exceed the limit.

For the 50th e-scooter to be the 35th one to exceed the speed limit, two independent events must occur sequentially:

1. Exactly 34 out of the first 49 e-scooters exceed the speed limit.
2. The 50th e-scooter exceeds the speed limit.

Let $W$ be the number of e-scooters out of the first 49 that exceed 10 km/h.
$W \sim \mathrm{B}(49, 0.767605)$

3. Calculate the final probability.

$\text{Required probability} = \Pr(W = 34) \times \Pr(X > 10)$
$= \left[ \binom{49}{34} (0.767605)^{34} (1 - 0.767605)^{15} \right] \times 0.767605$
$= 0.017114 \times 0.767605$
$\approx 0.013137$
$\approx 0.0131 \text{ (to 3 s.f.)}$

1. Determine the mean ($\mu$) and variance ($\sigma^2$) of the pedestrian speeds from the information given in part (i).

$\Pr(Y < 5.2) = \Pr(Y \ge 7.0) = 0.379$
By the symmetry of the normal distribution:
$\mu = \frac{5.2 + 7.0}{2} = 6.1$
Using the given probability to find $\sigma$:
$\Pr(Y < 5.2) = 0.379 \implies \Pr\left(Z < \frac{5.2 - 6.1}{\sigma}\right) = 0.379$
Using the inverse normal function on a graphing calculator:
$\frac{-0.9}{\sigma} \approx -0.308077 \implies \sigma \approx 2.92134$
$\sigma^2 \approx 8.53425$

2. State the distributions for the sample means $\bar{X}$ and $\bar{Y}$.

For $n=6$ e-scooters:
$\bar{X} \sim \mathrm{N}\left(12.3, \frac{9.9}{6}\right) \implies \bar{X} \sim \mathrm{N}(12.3, 1.65)$
For $n=15$ pedestrians:
$\bar{Y} \sim \mathrm{N}\left(6.1, \frac{8.53425}{15}\right) \implies \bar{Y} \sim \mathrm{N}(6.1, 0.56895)$

3. Formulate the linear combination $W = \bar{X} - 2\bar{Y}$ and calculate its expected value and variance.

Let $W = \bar{X} - 2\bar{Y}$.
$\mathrm{E}(W) = \mathrm{E}(\bar{X}) - 2\mathrm{E}(\bar{Y}) = 12.3 - 2(6.1) = 0.1$
$\mathrm{Var}(W) = \mathrm{Var}(\bar{X}) + 2^2\mathrm{Var}(\bar{Y}) = 1.65 + 4(0.56895) = 3.9258$
Thus, $W \sim \mathrm{N}(0.1, 3.9258)$.

4. Calculate the probability that the mean speed of the e-scooters is more than twice the mean speed of the pedestrians.

We need to find $\Pr(\bar{X} > 2\bar{Y})$, which is equivalent to $\Pr(W > 0)$.
Using a graphing calculator:
$\Pr(W > 0) \approx 0.52012$
$\Pr(W > 0) = 0.520 \text{ (to 3 s.f.)}$

1. State the distribution of the sample mean $\bar{X}$.

Since the population is normally distributed, $X \sim \mathrm{N}(12.3, 9.9)$.
For a sample of size $n$, the sample mean $\bar{X}$ is exactly normally distributed:
$\bar{X} \sim \mathrm{N}\left(12.3, \frac{9.9}{n}\right)$

2. Standardise the variable and evaluate the probability as $n$ becomes large.

$\begin{aligned} \mathrm{P}(\bar{X} > 10) &= \mathrm{P}\left(Z > \frac{10 - 12.3}{\sqrt{9.9/n}}\right) \\ &= \mathrm{P}\left(Z > -\frac{2.3}{\sqrt{9.9}} \sqrt{n}\right) \end{aligned}$
As $n \to \infty$, the term $-\frac{2.3}{\sqrt{9.9}} \sqrt{n} \to -\infty$.
Therefore, $\mathrm{P}(Z > -\infty) = 1$.