Students seeking admission to Kingdom University and Island University are required to apply using their national examination scores. The scores of the prospective applicants for each university can be modelled by independent normal distributions with means and standard deviations as shown in the table.

1. Define the random variable and formulate the equation for the top $30\%$.

Let $X$ be the score of a prospective applicant for Kingdom University.
$X \sim \text{N}(500, 100^2)$
Let $k$ be the minimum score required for admission.
$\text{P}(X \ge k) = 0.30$

2. Solve for the minimum score using the inverse normal function.

$\text{P}(X < k) = 1 - 0.30 = 0.70$
Using a graphing calculator, $k = 552.440...$
$k \approx 552 \text{ (to 3 s.f.)}$

1. Formulate the distributions of the sample mean scores for both universities.

Let $X$ and $Y$ be the scores of prospective applicants for Kingdom University and Island University respectively.
$X \sim \text{N}(500, 100^2)$ and $Y \sim \text{N}(480, 60^2)$
For random samples of size $5$, the sample means $\bar{X}$ and $\bar{Y}$ are distributed as:
$\bar{X} \sim \text{N}\left(500, \frac{100^2}{5}\right) \implies \bar{X} \sim \text{N}(500, 2000)$
$\bar{Y} \sim \text{N}\left(480, \frac{60^2}{5}\right) \implies \bar{Y} \sim \text{N}(480, 720)$

2. Determine the distribution of the difference between the two sample means.

Let $W = \bar{X} - \bar{Y}$ be the difference in the mean scores.
$\text{E}(W) = \text{E}(\bar{X}) - \text{E}(\bar{Y}) = 500 - 480 = 20$
$\text{Var}(W) = \text{Var}(\bar{X}) + \text{Var}(\bar{Y}) = 2000 + 720 = 2720$
Thus, $W \sim \text{N}(20, 2720)$.

3. Find the probability that the absolute difference exceeds 50.

We require $\text{P}(|W| > 50)$.
$\text{P}(|W| > 50) = \text{P}(W > 50) + \text{P}(W < -50)$
Using a graphing calculator:
$\text{P}(W > 50) \approx 0.28257$
$\text{P}(W < -50) \approx 0.08976$
$\text{P}(|W| > 50) \approx 0.28257 + 0.08976 = 0.37233$
$\text{P}(|W| > 50) \approx 0.372 \text{ (to 3 s.f.)}$

1. Evaluate the probability that a single applicant from each university scores at least 500.

Let $X \sim \mathrm{N}(500, 100^2)$ be the score of a Kingdom University applicant and $Y \sim \mathrm{N}(480, 60^2)$ be the score of an Island University applicant.

For the minimum score of all 10 students to be at least 500, every single student in the sample must score at least 500.
$P(X \ge 500) = 0.5$
$P(Y \ge 500) = P\left(Z \ge \frac{500 - 480}{60}\right) = P\left(Z \ge \frac{1}{3}\right) \approx 0.36944$

2. Calculate the combined probability for all 10 independent students.

Since the sample students' scores are independent, we multiply their individual probabilities:
$\text{Required probability} = [P(X \ge 500)]^5 \times [P(Y \ge 500)]^5$
$= (0.5)^5 \times (0.36944)^5$
$\approx 0.00021503$
$\approx 0.000215 \text{ (to 3 s.f.)}$

1. Define the random variable representing the difference between the total score of 5 Island University students and 5 times the score of a Kingdom University student, and find its expectation.

Let $I \sim \mathrm{N}(480, 60^2)$ and $K \sim \mathrm{N}(500, 100^2)$.
Let the combined random variable be $X = (I_1 + I_2 + I_3 + I_4 + I_5) - 5K$.
$\begin{aligned}\mathrm{E}(X) &= 5\mathrm{E}(I) - 5\mathrm{E}(K)\\&= 5(480) - 5(500)\\&= 2400 - 2500 \\&= -100\end{aligned}$

2. Calculate the variance of the linear combination.

$\begin{aligned}\mathrm{Var}(X) &= 5\mathrm{Var}(I) + 5^2\mathrm{Var}(K)\\&= 5(60^2) + 25(100^2)\\&= 5(3600) + 25(10000)\\&= 18000 + 250000 \\&= 268000\end{aligned}$

3. State the distribution of the linear combination and calculate the required probability.

Since the individual distributions are independent normal distributions, $X \sim \mathrm{N}(-100, 268000)$.
The required probability is $\mathrm{P}(X > 0)$.
Using a graphing calculator:
$\mathrm{P}(X > 0) \approx 0.42342$
$\mathrm{P}(X > 0) = 0.423 \text{ (to 3 s.f.)}$