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Let's go through the questions in your image one by one.

The problem gives:
- Mean final exam mark, \( \mu = 62\% \)
- Standard deviation, \( \sigma = 11\% \)
- The distribution is normal.

### Part (a): Probability that a randomly chosen final exam mark will be at least 73%
We are looking for \( P(X \geq 73) \) where \( X \) is the exam mark.

First, we compute the z-score:
\[
z = \frac{X - \mu}{\sigma} = \frac{73 - 62}{11} = \frac{11}{11} = 1
\]

Now, using standard normal distribution tables or a calculator, we find the probability corresponding to \( z = 1 \).

From a standard normal distribution table:
\[
P(Z \leq 1) = 0.8413
\]
Thus, the probability that a student scores at least 73% is:
\[
P(X \geq 73) = 1 - P(Z \leq 1) = 1 - 0.8413 = 0.1587
\]
**Answer: 0.1587**

### Part (b): Proportion of students not passing with a 50% threshold
We are looking for \( P(X < 50) \).

First, compute the z-score:
\[
z = \frac{50 - 62}{11} = \frac{-12}{11} \approx -1.0909
\]

Using the standard normal table for \( z = -1.09 \):
\[
P(Z \leq -1.09) = 0.1379
\]
Thus, the proportion of students who do not pass is approximately:
\[
P(X < 50) = 0.1379
\]
**Answer: 0.1379**

### Part (c): Minimum final exam mark needed for top 7%
We are looking for the final exam score corresponding to the top 7%, which means the 93rd percentile in a normal distribution.

From the standard normal distribution table, the z-score corresponding to 0.93 is approximately 1.475.

Now, use the z-score formula to find the exam mark:
\[
X = \mu + z\sigma = 62 + (1.475)(11) \approx 62 + 16.225 = 78.23
\]
**Answer: 78.23%**

### Part (d): Probability that 5 of 28 exams have a mark less than 50%
This is a binomial problem where:
- \( n = 28 \)
- \( p = 0.1379 \) (from part b)
- We want \( P(X = 5) \), where \( X \) is the number of exams with a mark less than 50.

The binomial probability formula is:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
Substitute \( n = 28 \), \( k = 5 \), and \( p = 0.1379 \):

\[
P(X = 5) = \binom{28}{5} (0.1379)^5 (0.8621)^{23}
\]

Let me calculate that for you:

\[
P(X = 5) \approx 0.1885
\]
**Answer: 0.1885**

---

Do you need further details on any of the parts? Here are some related questions you can try:

1. How does changing the standard deviation affect the results in part (a)?
2. What happens to the pass rate if the passing grade is increased to 60%?
3. How would the result in part (c) change if only the top 5% of students earned an A?
4. If a student scores in the top 10%, what is their approximate exam mark?
5. How would you calculate the expected number of students passing the exam out of a group of 50?

**Tip:** Using z-scores is crucial when working with normal distributions, as they allow for easy standardization and reference to standard tables.

The professor of an introductory calculus 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 62% and a standard deviation of 11%. (a) What is the probability that a randomly chosen final exam mark in this course will be at least 73%? Answer to four decimals. (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 final exam? Use four decimals in your answer. (c) The top 7% of students writing the final exam will receive a letter grade of at least A in the course. To two decimal places, find the minimum final exam mark needed to earn a letter grade of at least A in the course. (d) Suppose this professor randomly picked 28 final exams, observing the earned mark on each. What is the probability that 5 of these exams will have a grade of less than 50%? Use four decimals in your answer.

Solve probability questions related to final exam scores with a normal distribution (mean 62%, standard deviation 11%). Includes detailed solutions for finding the probability of scoring at least 73%, proportion of students failing with a 50% threshold, and more.

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