在一家百货公司，单张收据消费超过 $1000 的顾客有权获得一份免费礼品。在整个促销期间，每位顾客最多只能领取一份免费礼品。平均而言，每周有 8 份免费礼品被领取。 指出一周内领取的免费礼品数量的分布。

1. Identify the distribution of the random variable from the context.

Let $X$ be the number of free gifts claimed in one week.
From the question context, average claims per week is 8, so $X \sim \text{Po}(8)$.

2. Formulate the probability inequality for the stock not running out.

Let $k$ be the number of free gifts kept in stock at the start of the week.
For the stock not to run out, the number of claims $X$ must be at most $k$.
We require $\mathrm{P}(X \le k) \ge 0.95$.

3. Use a graphing calculator to find the least integer $k$ satisfying the inequality.

Using the cumulative probability function on a graphing calculator:
When $k = 12$, $\mathrm{P}(X \le 12) \approx 0.9362 < 0.95$
When $k = 13$, $\mathrm{P}(X \le 13) \approx 0.9658 \ge 0.95$
Thus, the least number of free gifts that must be kept in stock is 13.

1. Define the distribution for the number of free gifts claimed in $m$ weeks and state its Normal approximation.

Let $W$ be the number of free gifts claimed in $m$ weeks.
Since $X \sim \text{Po}(8)$ for one week, $W \sim \text{Po}(8m)$.
For a large mean (typically $\lambda > 10$), the Poisson distribution can be approximated by a Normal distribution. Thus, $W \sim \text{N}(8m, 8m)$ approximately.

2. Formulate the probability inequality with continuity correction.

We are given $\mathrm{P}(W > 50) \le 0.1$.
Applying continuity correction for the discrete to continuous approximation (since $W > 50$ translates to the discrete values $W \ge 51$):
$\mathrm{P}(W_{\text{cont}} > 50.5) \le 0.1$

3. Standardise the Normal variable and form an inequality using the inverse Normal distribution.

$\mathrm{P}\left(Z > \frac{50.5 - 8m}{\sqrt{8m}}\right) \le 0.1$
From the standard Normal distribution, the critical $Z$-value for an upper tail probability of $0.1$ is approximately $1.28155$.
To ensure the probability does not exceed $0.1$, the standardised value must be greater than or equal to this critical value:
$\frac{50.5 - 8m}{\sqrt{8m}} \ge 1.28155$

4. Solve the resulting inequality to find the largest integer value of $m$.

Rearranging the inequality yields a quadratic in terms of $\sqrt{8m}$:
$8m + 1.28155\sqrt{8m} - 50.5 \le 0$
Let $y = \sqrt{8m}$.
$y^2 + 1.28155y - 50.5 \le 0$
Solving the equation $y^2 + 1.28155y - 50.5 = 0$ gives $y \approx 6.494$ (rejecting the negative root).
Therefore, $\sqrt{8m} \le 6.494 \implies 8m \le 42.17 \implies m \le 5.27$
The largest integer value satisfying this is $m = 5$.

5. State the inequality satisfied by $m$ for the Normal approximation to be suitable.

For the Normal approximation to a Poisson distribution to be suitable, the mean $\lambda$ must be greater than $10$.
Therefore, the inequality is:
$8m > 10 \implies m > 1.25$