In a school canteen, the number of people joining the noodle stall queue in a period of 30 seconds is a random variable with the distribution Po(1.5).

1. Find the mean number of people joining the queue in a 5-minute period.

A 5-minute period contains $\frac{5 \times 60}{30} = 10$ periods of 30 seconds.
The new mean is $\lambda = 1.5 \times 10 = 15$.

2. Calculate the probability of at least 10 people joining the queue.

Let $X$ be the number of people joining the queue in 5 minutes.
$X \sim \text{Po}(15)$
We need to find $P(X \ge 10)$:
$\begin{aligned} P(X \ge 10) &= 1 - P(X \le 9) \\ &= 1 - 0.06985 \\ &\approx 0.930 \text{ (to 3 s.f.)} \end{aligned}$

1. Find the mean rate for the 5-minute period and define the random variable.

Let $Y$ be the number of people leaving the queue in a 5-minute period.
Since 5 minutes is $10 \times 30$ seconds, the mean rate must be scaled by 10.
$\lambda = 1.9 \times 10 = 19$
$Y \sim \mathrm{Po}(19)$

2. Calculate the probability that at least 10 people left using the complement method.

We want to find $\mathrm{P}(Y \ge 10)$. Using the complement probability:
$\mathrm{P}(Y \ge 10) = 1 - \mathrm{P}(Y \le 9)$
Using a graphing calculator:
$\mathrm{P}(Y \ge 10) \approx 1 - 0.008832 = 0.991168$
$\mathrm{P}(Y \ge 10) \approx 0.991 \text{ (to 3 s.f.)}$

1. Determine the parameters for the number of people joining and leaving in 10 minutes.

Let $X$ be the number of people joining the queue in 10 minutes, and $Y$ be the number of people leaving the queue in 10 minutes.
Since there are $20$ periods of 30 seconds in 10 minutes:
$X \sim \text{Po}(1.5 \times 20) \implies X \sim \text{Po}(30)$
$Y \sim \text{Po}(1.9 \times 20) \implies Y \sim \text{Po}(38)$

2. Find the distribution of the net change in the number of people using a Normal approximation.

Let $W = X - Y$ be the net change in the number of people in the queue.
Since $\lambda_X > 10$ and $\lambda_Y > 10$, we can approximate $X$ and $Y$ with Normal distributions.
$\text{E}(W) = \text{E}(X) - \text{E}(Y) = 30 - 38 = -8$
$\text{Var}(W) = \text{Var}(X) + \text{Var}(Y) = 30 + 38 = 68$
Therefore, $W \sim \text{N}(-8, 68)$ approximately.

3. Formulate the required probability with a continuity correction.

The final queue size is given by $20 + W$.
We are given that the final queue size is $5$:
$20 + W = 5 \implies W = -15$
Because $W$ is a discrete random variable approximated by a continuous Normal distribution, we apply a continuity correction:
$\text{P}(W = -15) \approx \text{P}(-15.5 < W < -14.5)$

4. Calculate the final probability.

Using $W \sim \text{N}(-8, 68)$:
$\text{P}(-15.5 < W < -14.5) = \text{P}\left(\frac{-15.5 - (-8)}{\sqrt{68}} < Z < \frac{-14.5 - (-8)}{\sqrt{68}}\right)$
$= \text{P}\left(-\frac{7.5}{\sqrt{68}} < Z < -\frac{6.5}{\sqrt{68}}\right)$
$\approx \text{P}(-0.9095 < Z < -0.7882)$
$\approx 0.0337 \text{ (to 3 s.f.)}$