The professor of a Statistics class has stated that, historically, the distribution of final exam grades in the course resemble a Normal distribution with a mean final exam mark of  66% and a standard deviation of 13%. (a) What is the probability that a random chosen final exam mark in this course will be at least 71%?

(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 5.9% 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 in the course.

%

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.

What is the probability that a random chosen final exam mark in this course will be at least 71%? What proportion of students will not pass the statistics final exam? Find the minimum final exam mark needed on the statistics final to earn a letter grade of at least an A in the course.

Explore how to calculate the probability of scoring at least 71% and the proportion of students who will not pass the statistics final exam. Learn to find the minimum mark needed for an A grade in a course with a Normal distribution of scores. This problem integrates key concepts of normal distribution, Z-scores, and statistical probability.

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