S and T are independent random variables with the distributions $N(18,3^2)$ and $N(\mu,\sigma^2)$ respectively. It is given that $P(T<4)=P(T>9)$ and $P(S<3T)=0.65$. Calculate the values of $\mu$ and $\sigma$. A fruit stall sells grapes that are packed in packets with masses in grams that follow the distribution $N(850,30^2)$. The grapes are sold at $18 per kilogram. Find the probability that a customer pays more than $30 for two packets of grapes. The fruit stall accepts payment by cash or PayNow. The number of customers who pay by PayNow in a day is a random variable with mean 12 and variance 4.8. In a month of 30 days, find the probability that the average number of customers per day who pay by PayNow is more than 12.3.

1. Find the mean $\mu$ using the symmetry of the normal distribution.

We are given $P(T < 4) = P(T > 9)$.
By the symmetry of the normal curve, the mean $\mu$ lies exactly midway between 4 and 9:
$\mu = \frac{4 + 9}{2} = 6.5$

2. Define a new random variable for the linear combination and find its mean and variance.

Let $W = S - 3T$.
$\mathrm{E}(W) = \mathrm{E}(S) - 3\mathrm{E}(T) = 18 - 3(6.5) = -1.5$
$\mathrm{Var}(W) = \mathrm{Var}(S) + 3^2 \mathrm{Var}(T) = 3^2 + 9\sigma^2 = 9 + 9\sigma^2$
Since $S$ and $T$ are independent, $W \sim \mathrm{N}(-1.5, 9 + 9\sigma^2)$.

3. Standardise $W$ and form an equation using the given probability and the inverse normal distribution.

We are given $P(S < 3T) = 0.65$, which is equivalent to $P(W < 0) = 0.65$.
Standardising the normal variable:
$P\left(Z < \frac{0 - (-1.5)}{\sqrt{9 + 9\sigma^2}}\right) = 0.65$
From the graphing calculator, $\Phi^{-1}(0.65) \approx 0.38532$.
$\frac{1.5}{\sqrt{9 + 9\sigma^2}} = 0.38532$

4. Solve the algebraic equation for $\sigma$.

$\sqrt{9 + 9\sigma^2} = \frac{1.5}{0.38532} = 3.89286$
Squaring both sides:
$9 + 9\sigma^2 = 15.154$
$9\sigma^2 = 6.154 \\ \sigma^2 = 0.6838 \\ \sigma = \sqrt{0.6838} \approx 0.827 \text{ (to 3 s.f.)}$

1. Define the random variable for the total cost of two packets and calculate its expectation and variance.

Let $X$ be the mass of one packet of grapes in grams. $X \sim N(850, 30^2)$.
The total cost of two independent packets is $C = 0.018X_1 + 0.018X_2$.
$\text{E}(C) = 0.018(850) + 0.018(850) = 30.6$
$\text{Var}(C) = 0.018^2(30^2) + 0.018^2(30^2) = 0.5832$
Thus, $C \sim N(30.6, 0.5832)$.

2. Calculate the probability that the total cost exceeds $30.

$\text{P}(C > 30) = 0.783965\dots$
$\text{P}(C > 30) \approx 0.784 \text{ (to 3 s.f.)}$

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

Let $X$ be the number of customers who pay by PayNow in a day, and $\bar{X}$ be the average number of such customers per day over 30 days.

Given $\mu = 12$ and $\sigma^2 = 4.8$.
Since the sample size $n = 30$ is sufficiently large ($n \ge 30$), by the Central Limit Theorem, $\bar{X}$ is approximately normally distributed:
$\bar{X} \sim N\left(\mu, \frac{\sigma^2}{n}\right)$
$\bar{X} \sim N\left(12, \frac{4.8}{30}\right)$
$\bar{X} \sim N(12, 0.16)$

2. Calculate the probability that the average number of customers is more than 12.3.

We want to find $P(\bar{X} > 12.3)$.
Using the standard normal variable $Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}$:
$P(\bar{X} > 12.3) = P\left(Z > \frac{12.3 - 12}{\sqrt{0.16}}\right)$
$P(\bar{X} > 12.3) = P\left(Z > \frac{0.3}{0.4}\right)$
$P(\bar{X} > 12.3) = P(Z > 0.75)$
$P(\bar{X} > 12.3) \approx 0.226627$
$P(\bar{X} > 12.3) \approx 0.227 \text{ (to 3 s.f.)}$