每天早上，彼得都要坐公交车去上学。他平均每天在早上 6:30 到达公交站。他到达公交站的时间服从正态分布，标准差为 10 分钟。第一班公交车将于早上 6:40 准时出发，第二班公交车将于早上 6:50 准时出发。无论公交车何时离开公交站，公交车程都服从均值为 45 分钟、标准差为 20 分钟的正态分布。如果彼得在早上 7:30 之后到达学校，或者错过了早上 6:50 的公交车，他就迟到了。 假设彼得到达公交站的时间与公交车程所花费的时间是相互独立的，求以下概率：

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.)