Let's solve the problem step by step.

Given:

• $X \sim N(50, 10^2)$, where $X$ is the percentage score.
• $X$ is considered "outstanding" if $X \geq 65$.
• A group of 8 applicants is being assessed.

Part (a):

(i) At least one is scored as outstanding:

Step 1: Calculate $P(X \geq 65)$.

Convert to a standard normal variable: $Z = \frac{X - \mu}{\sigma} = \frac{65 - 50}{10} = 1.5$ Using the standard normal table: $P(X \geq 65) = P(Z \geq 1.5) = 1 - P(Z \leq 1.5) = 1 - 0.9332 = 0.0668$

So, the probability of an applicant being "outstanding" is $0.0668$.

Step 2: Use the complement rule for "at least one."

The probability that none are "outstanding" is: $P(\text{None outstanding}) = (1 - P(X \geq 65))^8 = (1 - 0.0668)^8 = 0.5662$

Thus, the probability that at least one is "outstanding" is: $P(\text{At least one}) = 1 - P(\text{None outstanding}) = 1 - 0.5662 = 0.4338$

(ii) Exactly two are scored as outstanding:

This is a binomial distribution: $P(\text{Exactly 2}) = \binom{8}{2} (P(X \geq 65))^2 (1 - P(X \geq 65))^6$

Substitute the values: $P(\text{Exactly 2}) = \binom{8}{2} (0.0668)^2 (1 - 0.0668)^6$

$P(\text{Exactly 2}) = 28 \cdot (0.0668)^2 \cdot (0.9332)^6$

$P(\text{Exactly 2}) = 28 \cdot 0.00446224 \cdot 0.6082 = 0.0761$

(iii) Fewer than three are scored as outstanding:

This includes the probabilities of 0, 1, or 2 outstanding applicants: $P(\text{Fewer than 3}) = P(\text{0}) + P(\text{1}) + P(\text{2})$

Step 1: Calculate $P(\text{0})$: $P(\text{0}) = \binom{8}{0} (P(X \geq 65))^0 (1 - P(X \geq 65))^8 = (0.9332)^8 = 0.5662$

Step 2: Calculate $P(\text{1})$: $P(\text{1}) = \binom{8}{1} (P(X \geq 65))^1 (1 - P(X \geq 65))^7$ $P(\text{1}) = 8 \cdot 0.0668 \cdot (0.9332)^7 = 8 \cdot 0.0668 \cdot 0.6533 = 0.3491$

Step 3: Add probabilities: $P(\text{Fewer than 3}) = P(\text{0}) + P(\text{1}) + P(\text{2})$ $P(\text{Fewer than 3}) = 0.5662 + 0.3491 + 0.0761 = 0.9914$

Part (b):

Evaluate the interviewer’s claim:

The model is $X \sim N(50, 10^2)$, which implies $X$ is normally distributed.

A normal distribution theoretically has no strict bounds, so values above 100 (or even below 0) are possible, although extremely rare. To evaluate the probability:

$P(X > 100) = P\left(Z > \frac{100 - 50}{10}\right) = P(Z > 5)$

From the standard normal table: $P(Z > 5) \approx 0$

Thus, while the model technically allows scores above 100, their probability is negligible. The claim is overly cautious but not entirely incorrect.

Would you like detailed steps or clarifications on any part of this solution? Let me know!

Follow-Up Questions:

1. How do we derive probabilities from normal distributions?
2. What does "binomial distribution" mean in this context?
3. How can we compute probabilities for more complex conditions in distributions?
4. Why is the interviewer’s claim valid or invalid?
5. How can normal distribution be adjusted to fit real-world bounds?

Tip: Always visualize the distribution to understand its behavior better!