At a department store, shoppers who spend more than $1000 in a single receipt are entitled to a free gift. Each shopper is limited to at most one free gift for the entire promotion period. On average, 8 free gifts are claimed every week. State the distribution of the number of free gifts claimed in one week.

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:
$8m > 10 \implies m > 1.25$