平均而言，约翰在任何给定的 30 分钟时间段内，手机收到至少一条简讯（SMS）的概率为 0.95。一个上学日由 14 个这样的 30 分钟时间段组成。约翰收到至少一条简讯的时间段数量为随机变量 $X$。

1. State the assumption of constant probability in context

The probability that John receives at least one SMS is constant for each of the 14 30-minute periods in a school day.

2. State the assumption of independence in context

Whether John receives at least one SMS in a 30-minute period is independent of whether he receives at least one SMS in any other 30-minute period.

1. Identify a binomial assumption and explain why it might fail in the given context.

Identify the 'constant probability' assumption. Explain its limitation contextually: The probability of $0.95$ is unlikely to be constant throughout the 14 periods, as a student's phone usage and the likelihood of receiving messages typically fluctuate between lesson times and break times.

1. Identify the distribution parameters for the random variable $X$.

Given $n = 14$ and $p = 0.95$,
$X \sim \mathrm{B}(14, 0.95)$

2. Calculate the probability of at most 10 periods using a graphing calculator.

$\mathrm{P}(X \le 10) = 0.0041732...$
$\mathrm{P}(X \le 10) \approx 0.00417 \text{ (to 3 s.f.)}$

1. Formulate the exact distribution for the number of periods with no SMS in a week.

Let $Y$ be the number of periods in $5$ school days in which John does not receive any SMS.
Total number of periods, $n = 14 \times 5 = 70$.
The probability of not receiving any SMS in a single period is $p = 1 - 0.95 = 0.05$.
Exactly, the distribution is $Y \sim \text{B}(70, 0.05)$.

2. Determine and state the appropriate approximation.

Since $n = 70 \ge 50$ and $p = 0.05 \le 0.1$ (such that $np = 3.5 < 5$), the binomial distribution can be appropriately approximated by a Poisson distribution.
Mean $\lambda = np = 70 \times 0.05 = 3.5$.
Approximate distribution: $Y \sim \text{Po}(3.5)$.

3. Find the most likely number of periods by checking probabilities near the mean.

The most likely number of periods is the value of $y$ that gives the highest probability, which occurs around the mean $\lambda = 3.5$. We calculate the probabilities for the nearest integers ($y=3$ and $y=4$):
$P(Y = 3) = \frac{e^{-3.5}(3.5)^3}{3!} \approx 0.2158$
$P(Y = 4) = \frac{e^{-3.5}(3.5)^4}{4!} \approx 0.1888$
Since $P(Y = 3) > P(Y = 4)$, the most likely number of periods is $3$.