In this question, you should state clearly all the distributions that you use, together with the values of the appropriate parameters. A cup of cheese tea is prepared with two ingredients, cream cheese foam and tea. The barista preparing the cheese tea is not allowed to remove cream cheese foam nor tea once they have been poured into the cup. The amounts, in ml, of cream cheese foam and tea in one cup of cheese tea made by an experienced barista have independent normal distributions with means and standard deviations as shown in the table.

	Mean amount (ml)	Standard deviation (ml)
Cream cheese foam	72	5
Tea	421	4

1. Define the random variables and find the distribution of the total amount of cheese tea in a cup.

Let $C$ and $T$ be the amount of cream cheese foam and tea in a cup respectively.
$C \sim \mathrm{N}(72, 5^2)$ and $T \sim \mathrm{N}(421, 4^2)$

Let $X = C + T$ be the total amount of cheese tea in a cup.
$\mathrm{E}(X) = \mathrm{E}(C) + \mathrm{E}(T) = 72 + 421 = 493$
$\mathrm{Var}(X) = \mathrm{Var}(C) + \mathrm{Var}(T) = 5^2 + 4^2 = 25 + 16 = 41$

Therefore, the distribution is $X \sim \mathrm{N}(493, 41)$.

2. Calculate the probability that the amount of cheese tea is at least 490 ml.

We require $\mathrm{P}(X \ge 490)$.
Using a graphing calculator,
$\mathrm{P}(X \ge 490) = 0.68033\dots \approx 0.680 \text{ (to 3 s.f.)}$

1. Define the normal distributions and calculate the probability of a single cup being well-made.

Let $X$ and $Y$ be the amounts (in ml) of cream cheese foam and tea respectively.
$X \sim \mathrm{N}(72, 5^2)$
$Y \sim \mathrm{N}(421, 4^2)$

Since the amounts are independent, the probability of a cup being well-made is:
$\mathrm{P}(\text{well-made}) = \mathrm{P}(65 < X < 75) \times \mathrm{P}(416 < Y < 424)$
$= 0.64499 \times 0.66772$
$= 0.43067 \text{ (to 5 s.f.)}$

2. Define the binomial distribution and find the probability that more than 5 out of 10 cups are well-made.

Let $W$ be the number of well-made cups out of $10$.
$W \sim \mathrm{B}(10, 0.43067)$

$\mathrm{P}(W > 5) = 1 - \mathrm{P}(W \le 5)$
$= 1 - 0.77381$
$= 0.22619 \approx 0.226 \text{ (to 3 s.f.)}$

1. Formulate the profit for a single cup and find its expected value.

Let $C$ and $T$ be the amounts, in ml, of cream cheese foam and tea in one cup respectively.
The profit for one cup, $P$, is given by subtracting the cost of the ingredients from the selling price:
$P = 4.90 - \left(\frac{20}{1000}\right)C - \left(\frac{1.20}{1000}\right)T = 4.90 - 0.02C - 0.0012T$

The expected profit for one cup is:
$\mathrm{E}(P) = 4.90 - 0.02\mathrm{E}(C) - 0.0012\mathrm{E}(T)$
$\mathrm{E}(P) = 4.90 - 0.02(72) - 0.0012(421) = 2.9548$

2. Calculate the variance of the profit for a single cup.

Since $C$ and $T$ are independent,
$\mathrm{Var}(P) = (-0.02)^2\mathrm{Var}(C) + (-0.0012)^2\mathrm{Var}(T)$
$\mathrm{Var}(P) = 0.0004(5^2) + 0.00000144(4^2)$
$\mathrm{Var}(P) = 0.01 + 0.00002304 = 0.01002304$

3. Define the distribution for the total profit of 100 cups and state the necessary assumption.

Let $W = P_1 + P_2 + \dots + P_{100}$ be the total profit for 100 randomly chosen cups.
$\mathrm{E}(W) = 100 \times 2.9548 = 295.48$
$\mathrm{Var}(W) = 100 \times 0.01002304 = 1.002304$

Since a linear combination of independent normal variables is also normally distributed,
$W \sim \mathrm{N}(295.48, 1.002304)$

Assumption: The amounts of cream cheese foam and tea from cup to cup (or the profits of each cup) are independent of one another.

4. Calculate the probability that the total profit is at least $295.

We need to find $\mathrm{P}(W \ge 295)$.
Using a graphing calculator:
$\mathrm{P}(W \ge 295) \approx 0.68419 \approx 0.684 \text{ (to 3 s.f.)}$

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

Let $\bar{X}$ be the sample mean amount of cheese tea in $40$ cups.
Since the sample size $n = 40 \ge 30$ is large, by the Central Limit Theorem, $\bar{X}$ is approximately normally distributed.
$\bar{X} \sim \text{N}\left(498, \frac{30^2}{40}\right)$
$\bar{X} \sim \text{N}(498, 22.5)$

2. Calculate the probability that the sample mean does not exceed $500$ ml.

Using a graphing calculator:
$\text{P}(\bar{X} \le 500) = 0.66336...$
$\text{P}(\bar{X} \le 500) \approx 0.663 \text{ (to 3 s.f.)}$