Each day a train travels from Town A to Town B and it is scheduled to arrive at Town B at 0800 hrs. During the first year of operation it leaves Town A at 0700 hrs and is found to be late arriving at Town B on an average of once every 4 days. In the second year it leaves Town A at 0650 hrs and is late once every 20 days. Assuming the time of journey has a normal distribution, find, to the nearest minute, the standard deviation of the time of journey.

1. Set up the probability inequality for the train arriving late.

Let $X$ be the travel time in minutes. We are given $X \sim \mathrm{N}(80, 3^2)$.
Let $t$ be the allowed travel time. The train is late if $X > t$.
We require $\mathrm{P}(X > t) \le 0.02$.

2. Solve for the minimum allowed travel time $t$.

Standardising the variable:
$\mathrm{P}\left(Z > \frac{t - 80}{3}\right) \le 0.02$
Using the inverse normal distribution for an upper tail area of $0.02$ (or cumulative area of $0.98$):
$\frac{t - 80}{3} \ge 2.0537$
$t \ge 80 + 3(2.0537) = 86.161\text{ minutes}$

3. Calculate the latest departure time that satisfies the condition.

The minimum travel time is $86.161$ minutes (or $1$ hour and $26.161$ minutes).
The exact latest departure time is $1200\text{ hrs} - 1\text{ h } 26.161\text{ m} = 1033.839\text{ hrs}$ (which is 10:33 and 50 seconds).
Rounding to the nearest minute purely by numerical value gives $1034\text{ hrs}$. However, a travel time of $86$ minutes gives $\mathrm{P}(X > 86) \approx 0.0228 > 0.02$.
To strictly ensure the probability does not exceed $0.02$, we must round down the clock time to depart earlier:
Latest departure time $= 1033\text{ hrs}$.

1. State the distribution of the sample mean journey time over 20 days.

From the earlier context, let the journey time be $T$. We are given $T \sim \text{N}(80, 9)$.
For a sample of $n = 20$ days, the sample mean $\bar{T}$ follows a normal distribution with the same mean but a reduced variance:
$\bar{T} \sim \text{N}\left(80, \frac{9}{20}\right)$
$\bar{T} \sim \text{N}(80, 0.45)$

2. Determine the required range for $\bar{T}$ based on the arrival times.

The scheduled arrival time is 1200 hrs. Since the departure is fixed at 1040 hrs, the journey must take exactly $1200 - 1040 = 80$ minutes to arrive exactly on schedule.
For the average arrival time to be within 2 minutes of 1200 hrs, the average journey time $\bar{T}$ must be within $80 \pm 2$ minutes.
Therefore, we need to find $\text{P}(78 < \bar{T} < 82)$.

3. Calculate the probability using the normal distribution.

Using a graphing calculator to find the area under the normal curve for $\bar{T} \sim \text{N}(80, 0.45)$:
$\text{P}(78 < \bar{T} < 82) \approx 0.997135$
$\text{P}(78 < \bar{T} < 82) = 0.997 \text{ (to 3 s.f.)}$