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Let's solve each part of the problem step-by-step.

Given:
- Mean \(\mu = 63\%\)  
- Standard deviation \(\sigma = 7\%\)

We assume the scores follow a normal distribution.

---

### **(a) Probability that a person scores at most 65%**

We need to find \(P(X \leq 65)\).

1. Convert 65% to a **Z-score**:
   \[
   Z = \frac{X - \mu}{\sigma} = \frac{65 - 63}{7} = \frac{2}{7} \approx 0.2857
   \]

2. Use a Z-table or normal distribution calculator to find the cumulative probability at \(Z = 0.2857\):
   \[
   P(X \leq 65) \approx 0.6112
   \]

**Answer to (a):**  
\(0.6112\)

---

### **(b) Proportion of people scoring between 71% and 81%**

We need to find \(P(71 \leq X \leq 81)\).

1. Convert 71% and 81% to Z-scores:

   - \(Z_{71} = \frac{71 - 63}{7} = \frac{8}{7} \approx 1.1429\)
   - \(Z_{81} = \frac{81 - 63}{7} = \frac{18}{7} \approx 2.5714\)

2. Find the cumulative probabilities:
   - \(P(Z \leq 1.1429) \approx 0.8729\)
   - \(P(Z \leq 2.5714) \approx 0.9949\)

3. Subtract to find the proportion between these Z-scores:
   \[
   P(71 \leq X \leq 81) = 0.9949 - 0.8729 = 0.1220
   \]

**Answer to (b):**  
\(0.1220\)

---

### **(c) 30th percentile score**

We need to find the score corresponding to the 30th percentile, i.e., \(P(X \leq x) = 0.3000\).

1. Find the Z-score for the 30th percentile using a Z-table or calculator:
   \[
   Z_{30} \approx -0.524
   \]

2. Convert the Z-score back to the original score:
   \[
   X = \mu + Z \cdot \sigma = 63 + (-0.524) \cdot 7 = 63 - 3.668 \approx 59.33\%
   \]

**Answer to (c):**  
\(59.33\%\)

---

### Final Answers:
(a) \(0.6112\)  
(b) \(0.1220\)  
(c) \(59.33\%\)

Would you like a more detailed explanation of any part, or do you have any other questions?

---

Here are **five related questions**:
1. How do you interpret Z-scores in a real-world context?
2. What is the cumulative distribution function (CDF) of the normal distribution?
3. What does the 95th percentile represent in this exam setting?
4. How do we calculate probabilities for other non-standard distributions?
5. What would happen to the answers if the standard deviation was larger?

**Tip:** Z-tables are useful tools for understanding standard normal distributions, but software like Python or Excel can quickly automate these calculations.

The exam scores on a certain Society of Actuaries (SOA) professional examination are normally distributed with a mean score of μ = 63% and a standard deviation of σ = 7%. (a) What is the probability that a randomly chosen person who is writing this SOA exam will score at most 65%? (b) What proportion of all people writing this SOA exam will score between 71% and 81%? (c) What is the 30th percentile score for this exam? Enter your answer as a percentage, to two decimal places.

Calculate the probability that a person scores at most 65% on an SOA exam, find the proportion of people scoring between 71% and 81%, and determine the 30th percentile score. This problem covers key concepts in probability and statistics, including normal distribution and Z-scores. Suitable for undergraduate students.

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