Let's solve each part of this problem step-by-step:

Given Information:

• Mean score ($\mu$) = 64%
• Standard deviation ($\sigma$) = 5%
• The distribution of scores is normal.

Part (a)

Question: What is the probability that a randomly chosen person who is writing this SOA exam will score at most 66%?

To find this probability, we need to calculate the Z-score for 66%.

1. Calculate the Z-score: $Z = \frac{X - \mu}{\sigma} = \frac{66 - 64}{5} = \frac{2}{5} = 0.4$

2. Find the Probability: Using the Z-score of 0.4, we look up the cumulative probability in a Z-table or use statistical software. The cumulative probability for $Z = 0.4$ is approximately 0.6554.

Answer to (a): 0.6554

Part (b)

Question: What proportion of all people writing this SOA exam will score between 74% and 83%?

To find this proportion, we calculate the Z-scores for both 74% and 83%, then find the probability between them.

1. Calculate the Z-score for 74%: $Z = \frac{74 - 64}{5} = \frac{10}{5} = 2.0$

2. Calculate the Z-score for 83%: $Z = \frac{83 - 64}{5} = \frac{19}{5} = 3.8$

3. Find the Probability:

   • For $Z = 2.0$, the cumulative probability is approximately 0.9772.
   • For $Z = 3.8$, the cumulative probability is almost 1.0000.

4. Calculate the Proportion: $\text{Proportion} = 1.0000 - 0.9772 = 0.0228$

Answer to (b): 0.0228

Part (c)

Question: What is the 36th percentile score for this exam?

The 36th percentile score corresponds to the score below which 36% of the exam scores fall.

1. Find the Z-score for the 36th Percentile: The Z-score corresponding to the 36th percentile is approximately -0.36 (from Z-tables or statistical software).

2. Convert Z-score to Original Score: $X = \mu + Z \cdot \sigma = 64 + (-0.36) \cdot 5 = 64 - 1.8 = 62.2$

Answer to (c): 62.2%

Summary of Answers:

• (a): 0.6554
• (b): 0.0228
• (c): 62.2%

Would you like more details or have any questions about these steps?

Additional Related Questions:

1. How would the answers change if the standard deviation were 6% instead of 5%?
2. What is the probability that a student scores more than 70%?
3. What score corresponds to the 90th percentile?
4. If the mean score shifted to 65%, how would that affect the probability in part (a)?
5. What is the probability of scoring between 60% and 68%?

Tip:

In normal distribution problems, using Z-scores simplifies finding probabilities and percentiles across different datasets.