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Let's break down and solve each part of **Assignment 5: Problem 4** using the information provided in the image. 

We are given:
- Mean (μ) = 66%
- Standard deviation (σ) = 6%
- The scores are normally distributed.

### (a) Probability that a randomly chosen person scores at most 69%:

We need to find \( P(X \leq 69) \), where \( X \) is a normal variable with mean \( \mu = 66 \) and standard deviation \( \sigma = 6 \).

1. **Z-score formula:**
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   Plugging in \( X = 69 \):
   \[
   Z = \frac{69 - 66}{6} = \frac{3}{6} = 0.5
   \]

2. **Find the probability corresponding to \( Z = 0.5 \)**:
   Using a standard normal distribution table, the probability for \( Z = 0.5 \) is **0.6915**.

   **Final answer (a): 0.6915**

---

### (b) Proportion of people scoring between 70% and 81%:

We need to find \( P(70 \leq X \leq 81) \).

1. **Find the Z-scores for 70 and 81**:
   - For \( X = 70 \):
     \[
     Z_1 = \frac{70 - 66}{6} = \frac{4}{6} = 0.6667
     \]
   - For \( X = 81 \):
     \[
     Z_2 = \frac{81 - 66}{6} = \frac{15}{6} = 2.5
     \]

2. **Find the probabilities for these Z-scores**:
   - From the Z-table, \( P(Z_1 = 0.6667) \approx 0.7475 \)
   - \( P(Z_2 = 2.5) \approx 0.9938 \)

3. **Calculate the proportion**:
   \[
   P(70 \leq X \leq 81) = P(Z_2) - P(Z_1) = 0.9938 - 0.7475 = 0.2463
   \]

   **Final answer (b): 0.2463**

---

### (c) 30th percentile score:

We need to find the score \( X \) such that \( P(X \leq X_{30}) = 0.30 \).

1. **Find the Z-score for the 30th percentile**:
   From the Z-table, the Z-value corresponding to 0.30 is approximately \( Z = -0.5244 \).

2. **Convert Z-score to the actual score**:
   Using the Z-score formula:
   \[
   X = \mu + Z\sigma = 66 + (-0.5244)(6) = 66 - 3.1464 = 62.85
   \]

   **Final answer (c): 62.85%**

---

Would you like further details on any part or have any other questions? 

Here are some related questions to deepen understanding:

1. How do you interpret Z-scores in a normal distribution?
2. What is the importance of the percentile in probability questions?
3. How would you calculate a probability for a non-normally distributed dataset?
4. How can you apply this knowledge in real-world actuarial exams?
5. Why is it important to use a normal distribution for this type of exam score?

**Tip:** Always ensure you're using the correct Z-table for cumulative probabilities to avoid errors in calculations!

The exam scores on a certain Society of Actuaries (SOA) professional examination are normally distributed with a mean score of 66% and a standard deviation of 6%. What is the probability that a randomly chosen person will score at most 69%? What proportion of people will score between 70% and 81%? What is the 30th percentile score for this exam?

Learn how to calculate probabilities and percentiles for SOA exam scores using the normal distribution. This problem covers key statistical concepts, including Z-scores and percentile calculations, suitable for undergraduate students studying statistics or actuarial science.

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