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.

	Mean Score	Standard Deviation
Kingdom University	500	100
Island University	480	60

A random sample of 5 prospective applicants of Kingdom University and a random sample of 5 prospective applicants of Island University are chosen. Find the probability that

1. Set up the probability equation based on the top 30% admission condition.

Let $X$ be the score of a randomly chosen prospective applicant for Kingdom University.
Given $X \sim \text{N}(500, 100^2)$.

The university admits the top $30\%$ of applicants. Let $x$ be the minimum score required.
$\text{P}(X \ge x) = 0.3$
$\text{P}(X < x) = 1 - 0.3 = 0.7$

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

Using the inverse normal function on a graphing calculator:
$x = 552.44$

The minimum score a student must achieve is $552$ (to 3 s.f.).

1. Find the distribution of the difference between the two sample means.

Let $\bar{K}$ and $\bar{I}$ be the sample mean scores of the 5 applicants from Kingdom University and Island University respectively.

Since the populations are normally distributed, the sample means follow:
$\bar{K} \sim \mathrm{N}\left(500, \frac{100^2}{5}\right) \Rightarrow \bar{K} \sim \mathrm{N}(500, 2000)$
$\bar{I} \sim \mathrm{N}\left(480, \frac{60^2}{5}\right) \Rightarrow \bar{I} \sim \mathrm{N}(480, 720)$

Let the difference in sample means be $D = \bar{K} - \bar{I}$.
$\mathrm{E}(D) = \mathrm{E}(\bar{K}) - \mathrm{E}(\bar{I}) = 500 - 480 = 20$
$\mathrm{Var}(D) = \mathrm{Var}(\bar{K}) + \mathrm{Var}(\bar{I}) = 2000 + 720 = 2720$
Thus, $D \sim \mathrm{N}(20, 2720)$.

2. Formulate the probability for the absolute difference exceeding 50.

The phrase "difference in mean scores" implies the absolute difference between the two variables. Therefore, we need to find $\mathrm{P}(|D| > 50)$.

$\mathrm{P}(|D| > 50) = \mathrm{P}(D < -50) + \mathrm{P}(D > 50)$

3. Calculate the required probability using a graphing calculator.

Using the normal cumulative distribution function on a graphing calculator:
$\mathrm{P}(D < -50) \approx 0.089771$
$\mathrm{P}(D > 50) \approx 0.282565$

$\mathrm{P}(|D| > 50) \approx 0.089771 + 0.282565 = 0.372336 \approx 0.372 \text{ (to 3 s.f.)}$

1. Express the probability of the minimum score being at least 500 in terms of individual student scores.

For the minimum score of the 10 students to be at least 500, every single student in both samples must score at least 500. Since the students' scores are independent:
$P(\text{min} \ge 500) = [P(X \ge 500)]^5 \times [P(Y \ge 500)]^5$

2. Find the individual probabilities for students from each university.

Let $X \sim \mathrm{N}(500, 100^2)$ be the score for Kingdom University and $Y \sim \mathrm{N}(480, 60^2)$ be the score for Island University.
$P(X \ge 500) = 0.5$
$P(Y \ge 500) \approx 0.36944$

3. Compute the final probability.

$\text{Required probability} = (0.5)^5 \times (0.36944)^5$
$\approx 0.00021908$
$\approx 0.000219 \text{ (to 3 s.f.)}$

1. Define the random variables and the combined variable to investigate.

Let $X$ be the score of a randomly chosen prospective student from Kingdom University, and $Y$ be the score of a randomly chosen prospective student from Island University.
We are given $X \sim \mathrm{N}(500, 100^2)$ and $Y \sim \mathrm{N}(480, 60^2)$.
Let $W = (Y_1 + Y_2 + Y_3 + Y_4 + Y_5) - 5X$ be the difference between the total score of the 5 Island University students and 5 times the score of a Kingdom University student.
We want to find $\mathrm{P}(W > 0)$.

2. Calculate the expectation and variance of the combined variable.

$\mathrm{E}(W) = 5\mathrm{E}(Y) - 5\mathrm{E}(X)$
$\mathrm{E}(W) = 5(480) - 5(500) = -100$

$\mathrm{Var}(W) = 5\mathrm{Var}(Y) + 5^2\mathrm{Var}(X)$
$\mathrm{Var}(W) = 5(60^2) + 25(100^2) = 18000 + 250000 = 268000$
So, $W \sim \mathrm{N}(-100, 268000)$.

3. Calculate the required probability using the standard normal distribution.

$\mathrm{P}(W > 0) = \mathrm{P}\left(Z > \frac{0 - (-100)}{\sqrt{268000}}\right)$
$\mathrm{P}(W > 0) = \mathrm{P}(Z > 0.193167) \approx 0.423415$
Correcting to 3 significant figures, the probability is $0.423$.