The finishing times in a 10km race with a large number of runners follow a normal distribution. After 40 minutes, 10% of the runners have completed the race. After one hour, 35% of the runners have yet to complete the race. The first 20% of runners who finish the race receive a medal.

1. Set up the probability statements and obtain standard normal equations using the inverse normal function.

Let $T$ be the running time in minutes. $T \sim \mathrm{N}(\mu, \sigma^2)$.

Given 10% complete the race under 40 minutes:
$\mathrm{P}(T \leqslant 40) = 0.10 \implies \frac{40-\mu}{\sigma} \approx -1.28155$

Given 35% take more than 60 minutes:
$\mathrm{P}(T > 60) = 0.35 \implies \mathrm{P}(T \leqslant 60) = 0.65 \implies \frac{60-\mu}{\sigma} \approx 0.38532$

2. Form simultaneous linear equations and solve for $\sigma$.

$40 - \mu = -1.28155\sigma \quad \text{--- (1)}$
$60 - \mu = 0.38532\sigma \quad \text{--- (2)}$

Subtracting equation (1) from (2):
$20 = (0.38532 - (-1.28155))\sigma$
$20 = 1.66687\sigma$
$\sigma = \frac{20}{1.66687} \approx 11.9985$

3. Solve for $\mu$ and state the answers correct to 1 decimal place.

Substitute $\sigma$ into equation (2):
$60 - \mu = 0.38532(11.9985)$
$60 - \mu \approx 4.6232$
$\mu = 60 - 4.6232 \approx 55.3768$

Rounding to 1 decimal place:
$\mu \approx 55.4 \text{ minutes}$
$\sigma \approx 12.0 \text{ minutes}$ (Shown)

1. Set up the probability equation for the top 20% of runners who finish first.

Let $T$ be the time taken by a runner to finish the race.
The first 20% of runners are those with the shortest finishing times. We need to find the maximum time $t$ such that:
$\mathrm{P}(T < t) = 0.2$

2. Solve for the maximum time $t$ using the inverse normal distribution.

Using the unrounded parameters from part (i), $\mu \approx 55.377$ and $\sigma \approx 11.999$, we use the inverse normal function on the graphing calculator:
$t \approx 45.279$
$t \approx 45.3 \text{ minutes (to 3 s.f.)}$

1. Define the distribution for the number of runners receiving a medal

Let $X$ be the number of runners in the sample of 12 who receive a medal.
Since the first 20% of runners receive a medal, the probability of any randomly selected runner receiving a medal is $p = 0.2$.
$X \sim \mathrm{B}(12, 0.2)$

2. Calculate the probability that more than four runners receive a medal

$\mathrm{P}(X > 4) = 1 - \mathrm{P}(X \le 4)$
Using a graphing calculator:
$\mathrm{P}(X > 4) = 1 - 0.92744$
$\mathrm{P}(X > 4) = 0.072556 \approx 0.0726 \text{ (to 3 s.f.)}$

1. Translate the scenario into conditional probability events for the sample of 12 runners.

Let $A$ be the event that none of the 12 runners receives a medal.
Let $B$ be the event that the slowest of the 12 runners completes the race in under 60 minutes.

For the slowest runner to take under 60 minutes, all 12 runners must take under 60 minutes.
We require $\mathrm{P}(B \mid A) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(A)}$.

2. Determine the required probabilities for a single runner using the exact percentages given in the problem stem.

Let $T$ be the completion time of a single runner, and $t_m$ be the maximum time to receive a medal.
From the given information:
$\mathrm{P}(T < t_m) = 0.20$ (the first 20% receive medals)
$\mathrm{P}(T > 60) = 0.35 \implies \mathrm{P}(T \le 60) = 0.65$

For a single runner to not receive a medal, $T > t_m$:
$\mathrm{P}(T > t_m) = 1 - 0.20 = 0.80$

For a single runner to not receive a medal AND complete in under 60 minutes, $t_m \le T < 60$:
$\mathrm{P}(t_m \le T < 60) = \mathrm{P}(T < 60) - \mathrm{P}(T \le t_m) = 0.65 - 0.20 = 0.45$

3. Calculate the joint and marginal probabilities for the sample of 12 independent runners.

Since the 12 runners are randomly selected and independent:
$\mathrm{P}(A) = [\mathrm{P}(T > t_m)]^{12} = 0.80^{12}$
$\mathrm{P}(A \cap B) = [\mathrm{P}(t_m \le T < 60)]^{12} = 0.45^{12}$

4. Evaluate the final conditional probability.

$\begin{aligned} \mathrm{P}(B \mid A) &= \frac{0.45^{12}}{0.80^{12}} \\ &= \left(\frac{0.45}{0.80}\right)^{12} \\ &= (0.5625)^{12} \\ &\approx 0.000993 \text{ (to 3 s.f.)} \end{aligned}$

(Note: Using the rounded normal distribution parameters $\mu = 55.4$ and $\sigma = 12.0$ from part (i) yields a slightly inaccurate $0.000974$, which typically loses an accuracy mark. Relying on exact given percentages is the intended method.)