In this question you should clearly state the values of the appropriate parameters of any distributions used.

1. Define the appropriate linear combination and find its mean and variance.

Let $B$ and $F$ be the diameters of a basketball and football respectively.
$B \sim \mathrm{N}(\mu, 0.4^2)$ and $F \sim \mathrm{N}(22.0, 0.3^2)$.
Let $X = B - 1.1F$.
$\mathrm{E}(X) = \mathrm{E}(B) - 1.1\mathrm{E}(F) = \mu - 1.1(22.0) = \mu - 24.2$
$\mathrm{Var}(X) = \mathrm{Var}(B) + 1.1^2\mathrm{Var}(F) = 0.4^2 + 1.21(0.3^2) = 0.16 + 0.1089 = 0.2689$

2. Formulate the probability equation and use the inverse normal distribution.

Given $\mathrm{P}(B \ge 1.1F) = 0.35$, we have $\mathrm{P}(X \ge 0) = 0.35$.
Standardising $X$:
$\mathrm{P}\left(Z \ge \frac{0 - (\mu - 24.2)}{\sqrt{0.2689}}\right) = 0.35$
From the standard normal distribution (inverse normal):
$\frac{24.2 - \mu}{\sqrt{0.2689}} \approx 0.38532$

3. Solve the equation for $\mu$.

$24.2 - \mu = 0.38532 \times \sqrt{0.2689}$
$24.2 - \mu \approx 0.19981$
$\mu = 24.2 - 0.19981 = 24.00019$
$\mu \approx 24.0 \text{ (shown)}$

1. Identify the parameters of the normal curve and the significance of the boundary values.

Let the diameter of a basketball be $Y$. From part (a)(i), $Y \sim \mathrm{N}(24.0, 0.4^2)$.
The curve is centered at the mean, $\mu = 24.0$.
The bounds for shading are $22.8$ and $25.2$.
Notice that $22.8 = 24.0 - 3(0.4)$ and $25.2 = 24.0 + 3(0.4)$, corresponding exactly to $\mu - 3\sigma$ and $\mu + 3\sigma$.

2. Describe the features of the sketch and the shaded region.

To draw the sketch:

1. Draw a symmetrical, bell-shaped curve above a horizontal axis.
2. Mark the center of the peak on the axis as $24.0$.
3. Mark $22.8$ and $25.2$ symmetrically on the left and right of $24.0$.
4. Ensure the tails of the curve asymptotically approach the horizontal axis near and beyond these values.
5. Shade the entire area under the curve between $x = 22.8$ and $x = 25.2$.

1. State the distributions of the sample mean of 10 footballs and the basketball diameter.

Let $F$ be the diameter of a football and $B$ be the diameter of a basketball.
From the given information and part (a)(i):
$B \sim \mathrm{N}(24.0, 0.4^2)$
Let $\bar{F}$ be the average diameter of 10 footballs.
$\bar{F} \sim \mathrm{N}\left(22.0, \frac{0.3^2}{10}\right) \implies \bar{F} \sim \mathrm{N}(22.0, 0.009)$

2. Define the difference variable and find its distribution.

Let $W = B - \bar{F}$.
$\mathrm{E}(W) = \mathrm{E}(B) - \mathrm{E}(\bar{F}) = 24.0 - 22.0 = 2.0$
Since $B$ and $\bar{F}$ are independent normal variables, we add their variances:
$\mathrm{Var}(W) = \mathrm{Var}(B) + \mathrm{Var}(\bar{F}) = 0.4^2 + 0.009 = 0.16 + 0.009 = 0.169$
Therefore, $W \sim \mathrm{N}(2.0, 0.169)$.

3. Evaluate the probability that the absolute difference is less than 1.5 cm.

The difference between the diameters is less than $1.5\text{ cm}$, which means the absolute difference $|W| < 1.5$.
$\mathrm{P}(|W| < 1.5) = \mathrm{P}(-1.5 < W < 1.5)$
Using a graphing calculator:
$\mathrm{P}(-1.5 < W < 1.5) \approx 0.11194 \approx 0.112\text{ (to 3 s.f.)}$

1. Set up the inequality using the addition rule and the maximum value of the union of probabilities.

Let $A$ be the event of scoring on the first attempt and $B$ be the event of scoring on the second attempt.
Using the addition rule for probabilities:
$P(A \cup B) = P(A) + P(B) - P(A \cap B)$
Since the probability of any event cannot exceed 1, we know that $P(A \cup B) \le 1$.
Therefore, $P(A) + P(B) - P(A \cap B) \le 1$.

2. Substitute the given probabilities and simplify to show the required minimum value.

Rearranging the inequality gives:
$P(A \cap B) \ge P(A) + P(B) - 1$
Substituting the given values $P(A) = \frac{5}{11}$ and $P(B) = \frac{5}{8}$:
$P(A \cap B) \ge \frac{5}{11} + \frac{5}{8} - 1$
$P(A \cap B) \ge \frac{40}{88} + \frac{55}{88} - \frac{88}{88}$
$P(A \cap B) \ge \frac{95}{88} - \frac{88}{88}$
$P(A \cap B) \ge \frac{7}{88} \text{ (shown)}$

1. Find the probability of scoring on both attempts using the relationship between events.

Let $F$ be the event the player scores on the first attempt and $S$ be the event the player scores on the second attempt.
The probability of scoring on the first attempt but not on the second is $P(F \cap S')$.
We know that $P(F) = P(F \cap S) + P(F \cap S')$.
$P(F \cap S) = P(F) - P(F \cap S')$
$P(F \cap S) = \frac{5}{11} - \frac{15}{88}$
$P(F \cap S) = \frac{40}{88} - \frac{15}{88} = \frac{25}{88}$

2. Check the condition for independence by comparing the product of individual probabilities with the probability of the intersection.

For events $F$ and $S$ to be independent, $P(F \cap S)$ must equal $P(F) \times P(S)$.
$P(F) \times P(S) = \frac{5}{11} \times \frac{5}{8} = \frac{25}{88}$
Since $P(F \cap S) = P(F) \times P(S) = \dfrac{25}{88}$, the events are independent.