In this question you should state clearly all the distributions that you use, together with the values of the appropriate parameters. A petrol station sells two different grades of petrol. The volume, in litres, of the two different grades of petrol sold in an hour are normally distributed with means and standard deviations as shown in the table below.

	Mean	Standard deviation
Grade $X$	$\mu$	$\sigma$
Grade $Y$	200	5

1. Contrast the unimodal property of a normal distribution with the likely bimodal nature of the combined distribution.

A normal distribution curve is unimodal (has a single peak) and symmetrical.
Since the two different grades of petrol are likely to have different means, the combined distribution of a randomly chosen grade of petrol is likely to have two peaks (bimodal). Therefore, it is not appropriate to model it with a normal distribution.

1. Formulate the given probabilities in terms of the random variable $X$ and standardise them to the standard normal distribution $Z$.

Let $X \sim \mathrm{N}(\mu, \sigma^2)$ be the volume of Grade $X$ petrol sold in an hour.
Given $\mathrm{P}(X \leqslant 170) = 0.023$ and $\mathrm{P}(X > 180) = 0.16$.

Standardising gives:
$\mathrm{P}\left(Z \leqslant \frac{170 - \mu}{\sigma}\right) = 0.023$
$\mathrm{P}\left(Z > \frac{180 - \mu}{\sigma}\right) = 0.16 \implies \mathrm{P}\left(Z \leqslant \frac{180 - \mu}{\sigma}\right) = 1 - 0.16 = 0.84$

2. Use the inverse normal distribution to form simultaneous equations in terms of $\mu$ and $\sigma$.

Using the inverse normal function on a graphing calculator:
$\frac{170 - \mu}{\sigma} \approx -1.99539 \quad \text{--- (1)}$
$\frac{180 - \mu}{\sigma} \approx 0.99446 \quad \text{--- (2)}$

3. Solve the simultaneous equations to find the values of $\mu$ and $\sigma$.

Rearranging the equations into linear forms:
$170 - \mu = -1.99539\sigma$
$180 - \mu = 0.99446\sigma$

Subtracting the first equation from the second:
$10 = (0.99446 - (-1.99539))\sigma$
$10 = 2.98985\sigma$
$\sigma = 3.3446 \approx 3.34 \text{ (to 3 s.f.)}$

Substituting $\sigma$ back into the second equation:
$180 - \mu = 0.99446(3.3446)$
$\mu = 180 - 3.3261$
$\mu = 176.67 \approx 177 \text{ (to 3 s.f.)}$

1. Determine the mean $\mu$ and standard deviation $\sigma$ of Grade $X$ petrol (required from part ii).

Let $X \sim \mathrm{N}(\mu, \sigma^2)$ be the hourly volume of Grade $X$.
From part (ii):
$\mathrm{P}(X \le 170) = 0.023 \implies \dfrac{170 - \mu}{\sigma} \approx -1.99539$
$\mathrm{P}(X > 180) = 0.16 \implies \mathrm{P}(X \le 180) = 0.84 \implies \dfrac{180 - \mu}{\sigma} \approx 0.99446$
Subtracting the first equation from the second:
$\dfrac{10}{\sigma} = 0.99446 - (-1.99539) = 2.98985 \implies \sigma \approx 3.3446$
Substituting $\sigma$ back:
$\mu \approx 180 - 0.99446(3.3446) \approx 176.67$
(Keep exact values $\mu = 176.67389\dots$ and $\sigma = 3.344648\dots$ in the calculator for subsequent parts.)

2. Define the linear combination $D$ for the difference and calculate its expectation and variance.

Let $X_1, X_2, X_3$ be the independent volumes of Grade $X$ sold in three hours, and $Y$ be the volume of Grade $Y$ sold in 1 hour where $Y \sim \mathrm{N}(200, 5^2)$.
We require the probability involving $D = (X_1 + X_2 + X_3) - 3Y$.
$\mathrm{E}(D) = 3\mathrm{E}(X) - 3\mathrm{E}(Y) = 3(176.67389) - 3(200) = -69.978$
$\mathrm{Var}(D) = 3\mathrm{Var}(X) + 3^2\mathrm{Var}(Y) = 3(3.344648^2) + 9(5^2)$
$\mathrm{Var}(D) = 33.560 + 225 = 258.560$
Thus, $D \sim \mathrm{N}(-69.978, 258.560)$.

3. Calculate the probability that the difference lies within 72 litres.

We require $\mathrm{P}(|D| < 72) = \mathrm{P}(-72 < D < 72)$.
Using the graphing calculator's normal cumulative distribution function with $\mu = -69.978$ and $\sigma = \sqrt{258.560} \approx 16.080$:
$\mathrm{P}(-72 < D < 72) \approx 0.550024$
Rounding to 5 significant figures, the probability is $0.55002$.

1. State the necessary assumption of independence for the variance of the linear combination of random variables to be valid.

For the variance of the sum $(X_1 + X_2 + X_3 - 3Y)$ to be calculated as $\operatorname{Var}(X_1) + \operatorname{Var}(X_2) + \operatorname{Var}(X_3) + 3^2\operatorname{Var}(Y)$, the individual random variables must be independent. Thus, the volume of Grade $X$ petrol sold in each hour must be independent of one another, and independent of Grade $Y$.

1. Find the distribution of the hourly volume of Grade $Y$ petrol sold in June.

Let $Y$ be the hourly volume of Grade $Y$ petrol sold in May.
$Y \sim \mathrm{N}(200, 5^2)$
With a $10\%$ increase, the hourly volume in June is modelled by $Y_J = 1.1Y$.
$\mathrm{E}(Y_J) = 1.1(200) = 220$
$\mathrm{Var}(Y_J) = 1.1^2(5^2) = 1.21(25) = 30.25$
$Y_J \sim \mathrm{N}(220, 30.25)$

2. Find the distribution of the mean hourly volume for a 24-hour day in May and in June.

Let $\bar{Y}$ and $\bar{Y}_J$ be the mean hourly volume in a 24-hour day in May and June, respectively.
Since the daily operational hours are 24, the sample size is $n=24$.
$\bar{Y} \sim \mathrm{N}\left(200, \frac{25}{24}\right)$
$\bar{Y}_J \sim \mathrm{N}\left(220, \frac{30.25}{24}\right)$

3. Find the distribution of the difference in mean hourly volumes.

Let $W = \bar{Y}_J - \bar{Y}$. Since the days are independent,
$\mathrm{E}(W) = \mathrm{E}(\bar{Y}_J) - \mathrm{E}(\bar{Y}) = 220 - 200 = 20$
$\mathrm{Var}(W) = \mathrm{Var}(\bar{Y}_J) + \mathrm{Var}(\bar{Y}) = \frac{30.25}{24} + \frac{25}{24} = \frac{55.25}{24}$
$W \sim \mathrm{N}\left(20, \frac{55.25}{24}\right)$

4. Formulate the probability inequality for $k$ and standardise.

We are given $\mathrm{P}(\bar{Y}_J - \bar{Y} \ge k) \le 0.2$, which translates to $\mathrm{P}(W \ge k) \le 0.2$.
Standardising the inequality:
$\mathrm{P}\left(Z \ge \frac{k - 20}{\sqrt{55.25/24}}\right) \le 0.2$
Using a graphing calculator or standard normal table, the critical value $z$ for which $\mathrm{P}(Z \ge z) = 0.2$ is $z \approx 0.84162$.
Therefore, $\frac{k - 20}{\sqrt{55.25/24}} \ge 0.84162$.

5. Solve for the minimum integer value of $k$.

$k - 20 \ge 0.84162 \times \sqrt{\frac{55.25}{24}}$
$k \ge 20 + 0.84162 \times 1.51726$
$k \ge 21.277$
Since $k$ must be an integer, the minimum value of $k$ is $22$.