To solve these problems, we will use the properties of the normal distribution, specifically the standard normal distribution (Z-scores).

Given:

• Mean ($\mu$) = 66%
• Standard deviation ($\sigma$) = 13%

(a) Probability of scoring at least 71%

1. Calculate the Z-score for 71%:

   $Z = \frac{X - \mu}{\sigma} = \frac{71 - 66}{13} = \frac{5}{13} \approx 0.3846$

2. Find the probability of $Z \geq 0.3846$. This is done by looking up the Z-score in the standard normal distribution table or using a calculator.

   $P(Z \geq 0.3846) = 1 - P(Z \leq 0.3846)$

   From standard normal distribution tables or calculators:

   $P(Z \leq 0.3846) \approx 0.6504$

   Thus,

   $P(Z \geq 0.3846) \approx 1 - 0.6504 = 0.3496$

So, the probability that a randomly chosen final exam mark is at least 71% is approximately 0.3496 or 34.96%.

(b) Proportion of students who will not pass (final exam mark < 50%)

1. Calculate the Z-score for 50%:

   $Z = \frac{50 - 66}{13} = \frac{-16}{13} \approx -1.2308$

2. Find the probability of $Z < -1.2308$:

   $P(Z < -1.2308) \approx 0.1096$

So, approximately 10.96% of students will not pass the statistics final exam.

(c) Minimum final exam mark needed for at least an A (top 5.9%)

1. To find the Z-score corresponding to the top 5.9%, we look for $P(Z \leq z) = 0.941$ (since 1 - 0.059 = 0.941).

   From Z-tables or calculators, we find:

   $Z \approx 1.555$

2. Convert the Z-score back to the exam mark:

   $X = \mu + Z \cdot \sigma = 66 + 1.555 \cdot 13 \approx 66 + 20.215 \approx 86.215$

Thus, the minimum final exam mark needed to earn at least an A in the course is approximately 86.22%.

Summary of Answers:

• (a) Probability of at least 71%: 34.96%
• (b) Proportion not passing: 10.96%
• (c) Minimum mark for an A: 86.22%

Would you like any further details or explanations on these calculations? Here are some questions for further exploration:

1. How does the normal distribution apply to different types of exam grading?
2. What is the significance of the standard deviation in this context?
3. How can one estimate the mean and standard deviation from a set of grades?
4. What other statistical methods can be used to analyze exam results?
5. How does the concept of Z-scores help in comparing different distributions?

Tip: Always visualize the normal distribution curve when calculating probabilities to better understand the areas under the curve corresponding to Z-scores.