在本题中，你应该写出你所使用的任何正态分布的参数。 加油站（站点 $A$）发现其日销售量（以升为单位）服从均值为 5400、标准差为 240 的正态分布。 另一个加油站（站点 $B$）的日销售量服从均值为 6200 升、标准差为 $s$ 升的正态分布。

1. Identify the key parameters and coordinates for the curve

Let $X$ be the daily sales at Station $A$.
$X \sim \mathrm{N}(5400, 240^2)$
Mean, $\mu = 5400$
Standard deviation, $\sigma = 240$
The distribution curve is symmetrical and peaks at the mean, $x = 5400$.
The extremities of the required range are $4500$ and $6300$.
Observe that $5400 - 4500 = 900 = 3.75\sigma$ and $6300 - 5400 = 900 = 3.75\sigma$.

2. State the necessary features to be shown on the sketch

Since the bounds $4500$ and $6300$ are $3.75$ standard deviations away from the mean, the curve will be practically touching the $x$-axis at these points.
The sketch must show:

• A bell-shaped curve above the horizontal axis.
• Line of symmetry at $x = 5400$ (with the peak explicitly labelled).
• The $x$-axis marked with the values $4500$, $5400$, and $6300$.

1. State the distribution of daily sales at Station A and find the probability that it exceeds 5950 litres.

Let $A$ be the daily sales, in litres, of petrol at Station A.
$A \sim \mathrm{N}(5400, 240^2)$
$\mathrm{P}(A > 5950) = 0.0109599$

2. Calculate the expected number of days in a 365-day year.

$\text{Expected number of days} = 365 \times \mathrm{P}(A > 5950)$
$= 365 \times 0.0109599$
$= 4.00037$
$\approx 4.00$ (to 3 s.f.)

1. Define the random variable for the total sales over 6 days and find its normal distribution parameters.

Let $X$ be the daily sales at Station $A$. We are given $X \sim \mathrm{N}(5400, 240^2)$.
Let $T = X_1 + X_2 + X_3 + X_4 + X_5 + X_6$ be the total sales on 6 randomly chosen days.
$\mathrm{E}(T) = 6 \times 5400 = 32400$
$\mathrm{Var}(T) = 6 \times 240^2 = 345600$
Hence, $T \sim \mathrm{N}(32400, 345600)$.

2. Calculate the probability that the total sales are less than 32000 litres.

$\mathrm{P}(T < 32000) \approx 0.24812$
$\mathrm{P}(T < 32000) = 0.248$ (to 3 s.f.)

1. Set up the probability equation using the given information.

Let $Y$ be the daily sales at Station $B$. We are given $Y \sim \mathrm{N}(6200, s^2)$.
$\mathrm{P}(Y > 5000) = 0.99$

2. Standardise the variable and find the corresponding $z$-value.

$\mathrm{P}\left(Z > \frac{5000 - 6200}{s}\right) = 0.99$
$\mathrm{P}\left(Z < \frac{-1200}{s}\right) = 0.01$
Using the inverse standard normal distribution:
$\frac{-1200}{s} = -2.3263$

3. Solve for the standard deviation $s$.

$s = \frac{-1200}{-2.3263}$
$s = 515.84$
$s \approx 516 \text{ (to 3 s.f.)}$

1. Determine the standard deviation $s$ for Station $B$ using information from part (iv).

From part (iv), we are given $\mathrm{P}(B > 5000) = 0.99$.
Standardising the variable:
$\mathrm{P}\left(Z > \frac{5000 - 6200}{s}\right) = 0.99$
$\frac{-1200}{s} = \Phi^{-1}(0.01) \approx -2.32635$
$s = \frac{-1200}{-2.32635} \approx 515.83$

2. Formulate the linear combination representing the condition, state the necessary assumption, and find its mean and variance.

Let $W = B - 0.5A$. We want to find the probability that $W \ge 2000$.
For the variance of the sum to be the sum of the variances, we must assume that the daily sales at Station $A$ and Station $B$ are independent.
$\mathrm{E}(W) = \mathrm{E}(B) - 0.5\mathrm{E}(A) = 6200 - 0.5(5400) = 3500$
$\mathrm{Var}(W) = \mathrm{Var}(B) + 0.5^2\mathrm{Var}(A)$
$\mathrm{Var}(W) \approx 515.83^2 + 0.25(240^2) \approx 266080.5 + 14400 = 280480.5$
Thus, $W \sim \mathrm{N}(3500, 280480.5)$.

3. Calculate the probability that Station $B$ sales exceed half of Station $A$ by at least 2000 litres.

$\mathrm{P}(B > 0.5A + 2000) = \mathrm{P}(W \ge 2000)$
Using the normal distribution on a graphing calculator:
$\mathrm{P}(W \ge 2000) \approx 0.99769 \approx 0.998 \text{ (to 3 s.f.)}$