Every morning, Peter has to take a bus to school. On average, he reaches the bus stop at 6:30 am each day. His arrival time at the bus stop is normally distributed with a standard deviation of 10 minutes. The first bus will leave the bus stop at 6:40 am sharp and the second bus will leave at 6:50 am sharp. Regardless of the time the buses leave the bus stop, the bus journey will follow a normal distribution with mean 45 minutes and standard deviation 20 minutes. Peter will be late if he arrives at school after 7:30 am or if he misses the 6:50 am bus. Assuming that the time Peter reaches the bus stop and the time taken for the bus journey are independent, find the probability that

1. Define the random variable and its normal distribution.

Let $T$ be the time Peter reaches the bus stop, in minutes after 6:30 am.
Since his average arrival time is 6:30 am and the standard deviation is $10$ minutes, we have:
$T \sim \mathrm{N}(0, 10^2)$

2. Calculate the probability that he arrives after 6:50 am.

To miss the 6:50 am bus, Peter must arrive after 6:50 am, which is $20$ minutes after 6:30 am.
$P(T > 20) = 0.022750...$
$P(T > 20) \approx 0.0228 \text{ (to 3 s.f.)}$

1. Define the normal distributions for the arrival time at the bus stop and the journey duration.

Let $T$ be Peter's arrival time at the bus stop in minutes past 6:00 am.
$T \sim \text{N}(30, 10^2)$
Let $J$ be the duration of the bus journey in minutes.
$J \sim \text{N}(45, 20^2)$

2. Identify the mutually exclusive scenarios in which Peter is late.

Peter is late in three mutually exclusive scenarios:

1. He misses the 6:50 am bus: $T > 50$.
2. He takes the 6:40 am bus ($T \le 40$) and the journey takes more than 50 minutes ($6:40\text{ am} + 50\text{ mins} = 7:30\text{ am}$), so $J > 50$.
3. He takes the 6:50 am bus ($40 < T \le 50$) and the journey takes more than 40 minutes ($6:50\text{ am} + 40\text{ mins} = 7:30\text{ am}$), so $J > 40$.

3. Calculate the required normal probabilities for $T$ and $J$.

From the standard normal distribution:
$\text{P}(T > 50) \approx 0.022750$
$\text{P}(T \le 40) \approx 0.841345$
$\text{P}(40 < T \le 50) \approx 0.135905$
$\text{P}(J > 50) \approx 0.401294$
$\text{P}(J > 40) \approx 0.598706$

4. Calculate the total probability of being late by summing the probabilities of each scenario.

Since $T$ and $J$ are independent, we multiply their probabilities for each scenario and sum them:
$\text{P}(\text{Late}) = \text{P}(T > 50) + \text{P}(T \le 40)\text{P}(J > 50) + \text{P}(40 < T \le 50)\text{P}(J > 40)$
$\text{P}(\text{Late}) \approx 0.022750 + (0.841345 \times 0.401294) + (0.135905 \times 0.598706)$
$\text{P}(\text{Late}) \approx 0.022750 + 0.337626 + 0.081367$
$\text{P}(\text{Late}) \approx 0.441743 \approx 0.442 \text{ (shown)}$

1. Define the random variable and state its distribution using the probability from part (ii).

Let $W$ be the number of days Peter is late for school in a 5-day week.
Given from part (ii) that the probability of being late on any given day is $0.442$,
$W \sim \mathrm{B}(5, 0.442)$

2. Calculate the probability that Peter is late on at least 2 days.

$\mathrm{P}(W \ge 2) = 1 - \mathrm{P}(W \le 1)$
$= 1 - 0.26858$
$= 0.73142 \approx 0.731$ (to 3 s.f.)

1. Define the random variable for the number of late days in a week and find its mean and variance.

Let $X$ be the number of days Peter is late in a 5-day week.
$X \sim \mathrm{B}(5, 0.442)$
$\mathrm{E}(X) = np = 5(0.442) = 2.21$
$\mathrm{Var}(X) = np(1-p) = 5(0.442)(1-0.442) = 1.23318$

2. Apply the Central Limit Theorem to state the distribution of the sample mean.

Let $\bar{X}$ be the average number of late days per week over a sample of $n=40$ weeks.
Since the sample size $n = 40$ is large ($n \ge 30$), the Central Limit Theorem applies, and $\bar{X}$ follows an approximately normal distribution:
$\bar{X} \sim \mathrm{N}\left(\mathrm{E}(X), \frac{\mathrm{Var}(X)}{n}\right)$
$\bar{X} \sim \mathrm{N}\left(2.21, \frac{1.23318}{40}\right)$
$\bar{X} \sim \mathrm{N}(2.21, 0.0308295)$

3. Calculate the probability that the average number of late days per week is at least 2.

We require $\mathrm{P}(\bar{X} \ge 2)$.
Using a graphing calculator,
$\mathrm{P}(\bar{X} \ge 2) \approx 0.88417 \approx 0.884$ (to 3 s.f.)