Packets of a particular brand of potato chips are delivered to a supermarket in boxes of 60. On average, 1.8 packets in a box are underweight. The number of underweight packets from a randomly chosen box is the random variable $X$. Assume that $X$ has a binomial distribution.

1. Determine the exact distribution of the number of underweight packets in two boxes.

Let $X$ be the number of underweight packets in a box of 60. We are given $X \sim \text{B}(60, p)$.
Since the mean is $1.8$, we have $60p = 1.8 \implies p = 0.03$.
Let $Y$ be the number of underweight packets in two randomly chosen boxes ($n = 120$).
Assuming the boxes are independent, the exact distribution is $Y \sim \text{B}(120, 0.03)$.

2. State the suitable approximation and its parameter.

For $Y \sim \text{B}(120, 0.03)$, since $n = 120$ is large ($n \ge 50$) and $p = 0.03$ is small such that $np = 3.6 < 5$, the Poisson approximation is a mathematically suitable approximation.
Thus, $Y \approx \text{Po}(3.6)$.
The parameter of the distribution used is the mean $\lambda = 3.6$.

3. Calculate the required probability using the approximation.

We want to find the probability that there are more than 6 underweight packets, which is $P(Y > 6)$.
$P(Y > 6) = 1 - P(Y \le 6)$
Using a graphing calculator for the Poisson distribution with $\lambda = 3.6$:
$P(Y \le 6) \approx 0.92673$
$P(Y > 6) = 1 - 0.92673 = 0.07327 \approx 0.0733 \text{ (to 3 s.f.)}$

1. Find the variance of the number of underweight packets in a box, $X$.

Since $E(X) = 1.8$ and $X \sim \text{B}(60, p)$,
$60p = 1.8 \implies p = 0.03$
$\text{Var}(X) = np(1-p) = 1.8(1 - 0.03) = 1.746$

2. State the approximate distribution of the sample mean $\bar{X}$ using the Central Limit Theorem.

Let $\bar{X}$ be the sample mean number of underweight packets per box for 50 boxes.
Since the sample size $n = 50$ is large ($n \ge 30$), by the Central Limit Theorem,
$\bar{X} \sim \text{N}\left(1.8, \frac{1.746}{50}\right)$ approximately.
$\bar{X} \sim \text{N}(1.8, 0.03492)$

3. Calculate the required probability.

$P(\bar{X} > 2) \approx 0.14225 \approx 0.142 \text{ (to 3 s.f.)}$