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Let's solve each part of the problem:

The problem states:
- The distribution of final exam grades is normally distributed.
- Mean (μ) = 67%
- Standard deviation (σ) = 14%

We need to solve parts (a), (b), and (c) of the question based on this information.

### Part (a)
**Question:** What is the probability that a random chosen final exam mark in this course will be at most 73.8%?

1. Convert the score to a z-score using the formula:
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   where \(X = 73.8\%\), \(\mu = 67\%\), and \(\sigma = 14\%\).

2. Calculate:
   \[
   Z = \frac{73.8 - 67}{14} = \frac{6.8}{14} \approx 0.4857
   \]

3. Use the z-table or a calculator to find the cumulative probability for \(Z = 0.4857\).

### Part (b)
**Question:** In order to pass this course, a student must have a final exam mark of at least 50%. What proportion of students will not pass the statistics final exam?

1. Convert 50% to a z-score:
   \[
   Z = \frac{50 - 67}{14} = \frac{-17}{14} \approx -1.2143
   \]

2. Find the cumulative probability for \(Z = -1.2143\). This will give the proportion of students scoring below 50%.

### Part (c)
**Question:** The top 5.6% of students writing the final exam will receive a letter grade of at least an A. Find the minimum final exam mark needed on the statistics final to earn a letter grade of at least an A.

1. Find the z-score that corresponds to the top 5.6% (or the 94.4th percentile, as it is 100% - 5.6%).

2. Using the z-table, the z-score for the 94.4th percentile is approximately \(Z \approx 1.5849\).

3. Convert this z-score to the corresponding grade using the formula:
   \[
   X = Z \cdot \sigma + \mu
   \]
   Substituting values:
   \[
   X = 1.5849 \cdot 14 + 67
   \]

Let's compute these values for precise answers.Here are the solutions:

### Part (a)
The probability that a randomly chosen final exam mark will be at most 73.8% is approximately **0.68642** or **68.642%**.

### Part (b)
The proportion of students who will not pass the statistics final exam (i.e., scoring below 50%) is approximately **0.11232** or **11.232%**.

### Part (c)
The minimum final exam mark needed to be in the top 5.6% and earn an A is approximately **89.24975%**.

Would you like a more detailed explanation on any part of this? 

---

Here are 5 additional questions for deeper exploration:
1. What would be the probability of scoring above 80% on the exam?
2. If the passing mark was raised to 60%, what proportion would not pass?
3. How would the distribution change if the standard deviation increased to 20%?
4. What score would place a student in the top 10%?
5. What is the probability of scoring between 60% and 80%?

**Tip:** Understanding how to convert scores to z-scores and using z-tables is essential for working with normally distributed data.

The professor of a Statistics class has stated that, historically, the distribution of final exam grades in the course resembles a Normal distribution with a mean final exam mark of μ = 67% and a standard deviation of σ = 14%.

(a) What is the probability that a random chosen final exam mark in this course will be at most 73.8%?
(b) In order to pass this course, a student must have a final exam mark of at least 50%. What proportion of students will not pass the statistics final exam?
(c) The top 5.6% of students writing the final exam will receive a letter grade of at least an A in the course. Find the minimum final exam mark needed on the statistics final to earn a letter grade of at least an A.

Solve probability questions involving a normally distributed set of final exam grades in a Statistics course with a mean of 67% and a standard deviation of 14%. Calculate probabilities, passing thresholds, and the minimum mark required to be in the top 5.6%.

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