某超市出售两种口味（巧克力和草莓）的特定品牌饼干。一天内售出的巧克力饼干盒数的平均值为 2.2。

1. Define the random variable and identify its distribution

Let $X$ be the number of boxes of chocolate biscuits sold in a day.
Assuming the boxes are sold independently and randomly at a constant average rate, $X$ follows a Poisson distribution with mean $2.2$.
$X \sim \text{Po}(2.2)$

2. Calculate the probability that exactly zero boxes were sold

$\text{P}(X = 0) = \frac{e^{-2.2} \cdot 2.2^0}{0!}$
$\text{P}(X = 0) = e^{-2.2} \approx 0.11080$
$\text{P}(X = 0) \approx 0.111 \text{ (to 3 s.f.)}$

1. Determine the probability of no chocolate biscuits being sold in a single day (required from part i).

Let $C$ be the number of boxes of chocolate biscuits sold in a day.
Assuming $C$ follows a Poisson distribution with mean $\lambda = 2.2$:
$C \sim \mathrm{Po}(2.2)$
$\mathrm{P}(C = 0) = e^{-2.2} \approx 0.11080$

2. Model the number of days with no sales in a week as a Binomial distribution and calculate its expected value.

Let $X$ be the number of days in a 7-day week where no boxes of chocolate biscuits are sold.
Assuming the number of boxes sold each day is independent, $X$ follows a Binomial distribution with $n = 7$ and $p = e^{-2.2}$:
$X \sim \mathrm{B}(7, 0.11080)$
The expected number of days is:
$\mathrm{E}(X) = np = 7 \times e^{-2.2}$
$\mathrm{E}(X) \approx 0.77562 \approx 0.776$

1. Set up the equation for $\lambda$ using the Poisson probability distribution.

Let $Y$ be the discrete number of boxes of strawberry biscuits sold in a day. Based on the context, we model this using a Poisson distribution, so $Y \sim \text{Po}(\lambda)$.

The probability of selling less than 2 boxes is:
$P(Y < 2) = P(Y = 0) + P(Y = 1)$

Using the Poisson formula $P(Y = y) = \frac{e^{-\lambda}\lambda^y}{y!}$:
$P(Y < 2) = \frac{e^{-\lambda}\lambda^0}{0!} + \frac{e^{-\lambda}\lambda^1}{1!} = e^{-\lambda} + \lambda e^{-\lambda}$

Given that $P(Y < 2) = 0.6$, the equation is:
$e^{-\lambda}(1 + \lambda) = 0.6$

2. Solve the equation numerically to find $\lambda$.

Using a graphing calculator to find the root of the equation $e^{-\lambda}(1 + \lambda) - 0.6 = 0$:
$\lambda \approx 1.3788$

Rounding to 1 decimal place, we obtain:
$\lambda = 1.4$

1. Set up the Poisson models for a week and combine them

Per day: chocolate $\sim \text{Po}(2.2)$, strawberry $\sim \text{Po}(\lambda)=\text{Po}(1.4)$ (from part (iii)).

Over $7$ days:

• chocolate total $C\sim \text{Po}(7\times2.2)=\text{Po}(15.4)$
• strawberry total $S\sim \text{Po}(7\times1.4)=\text{Po}(9.8)$

Since $C$ and $S$ are independent Poisson variables, the grand total
$T=C+S\sim \text{Po}(15.4+9.8)=\text{Po}(25.2).$

2. Compute the required probability using the complement

Required probability:
$P(T>25)=1-P(T\le 25).$
Using the GC with $T\sim\text{Po}(25.2)$:
$P(T\le 25)\approx 0.524,$
so
$P(T>25)\approx 1-0.524=0.476.$

1. Determine the expected value and variance for the total number of chocolate and strawberry biscuits sold over 30 days.

Let $C$ and $S$ be the number of boxes of chocolate and strawberry biscuits sold in a day. As these represent independent random events in a fixed time interval, they follow Poisson distributions: $C \sim \text{Po}(2.2)$. From part (iii), $S \sim \text{Po}(\lambda)$ where $\lambda \approx 1.37934$ (keeping 5 s.f. for intermediate accuracy).
For a month of 30 days, the total sales are $C_{30}$ and $S_{30}$.
$\text{E}(C_{30}) = \text{Var}(C_{30}) = 30 \times 2.2 = 66$
$\text{E}(S_{30}) = \text{Var}(S_{30}) = 30 \times 1.37934 = 41.380$

2. Apply the Central Limit Theorem to find the approximate normal distribution for the difference in total sales.

Let $W = C_{30} - S_{30}$.
Since the sample size $n = 30$ is sufficiently large, by the Central Limit Theorem, $C_{30}$ and $S_{30}$ can be approximated by Normal distributions.
$\text{E}(W) = \text{E}(C_{30}) - \text{E}(S_{30}) = 66 - 41.380 = 24.620$
$\text{Var}(W) = \text{Var}(C_{30}) + \text{Var}(S_{30}) = 66 + 41.380 = 107.380$
Thus, $W \sim \text{N}(24.620, 107.380)$ approximately.

3. Calculate the probability that more chocolate biscuits were sold than strawberry biscuits.

We require the probability $P(C_{30} > S_{30})$, which is equivalent to $P(W > 0)$.
Using the normal distribution $W \sim \text{N}(24.620, 107.380)$:
$P(W > 0) = P\left(Z > \frac{0 - 24.620}{\sqrt{107.380}}\right)$
$P(W > 0) \approx 0.99124$
$\approx 0.991 \text{ (to 3 s.f.)}$