https://file.aiquickdraw.com/mathbot/file/1730745395487o4q6yk38.jpg

Let's work through each question step-by-step.

The distribution of final exam marks is given to be normally distributed with:
- Mean (\(\mu\)) = 60%
- Standard deviation (\(\sigma\)) = 9%

### Part (a)
**Question**: What is the probability that a randomly chosen final exam mark will be at least 72%?

**Solution**:
1. Convert the score of 72% to a Z-score using the formula:
   \[
   Z = \frac{X - \mu}{\sigma}
   \]
   where \( X = 72 \), \( \mu = 60 \), and \( \sigma = 9 \).

   \[
   Z = \frac{72 - 60}{9} = \frac{12}{9} = 1.3333
   \]

2. Look up the Z-score of 1.3333 in a standard normal table, or use a calculator to find the area to the left of this Z-score.
   
   This area is approximately 0.9082.

3. Since we want the probability of scoring **at least** 72%, we need the area to the **right**:
   \[
   P(X \geq 72) = 1 - 0.9082 = 0.0918
   \]

4. So, the answer to part (a) is:
   \[
   \boxed{0.0918}
   \]

---

### Part (b)
**Question**: What proportion of students will not pass the final exam if the passing mark is 50%?

**Solution**:
1. Calculate the Z-score for 50%:
   \[
   Z = \frac{50 - 60}{9} = \frac{-10}{9} = -1.1111
   \]

2. Look up the Z-score of -1.1111 in the standard normal table, or use a calculator. The area to the left of -1.1111 is approximately 0.1335.

3. Since we want the proportion of students **not passing** (scoring below 50%), this is directly given by this area.

4. So, the answer to part (b) is:
   \[
   \boxed{0.1335}
   \]

---

### Part (c)
**Question**: Find the minimum final exam mark needed to be in the top 3% of students.

**Solution**:
1. For the top 3%, we need the Z-score that corresponds to the 97th percentile (since the top 3% is above 97%).

2. The Z-score corresponding to 0.97 in a standard normal distribution is approximately 1.8808.

3. Convert this Z-score back to the original score using the formula:
   \[
   X = Z \cdot \sigma + \mu
   \]
   \[
   X = 1.8808 \cdot 9 + 60 = 16.9272 + 60 = 76.9272
   \]

4. Rounding to two decimal places, the minimum mark needed is:
   \[
   \boxed{76.93\%}
   \]

---

### Part (d)
**Question**: If the professor randomly picked 28 exams, what is the probability that 4 of these exams will have a grade of less than 50%?

**Solution**:
1. From part (b), we know the probability of scoring less than 50% is \( p = 0.1335 \).

2. This situation follows a **binomial distribution** with parameters \( n = 28 \) and \( p = 0.1335 \). We want the probability that exactly 4 exams score below 50%.

3. The probability of exactly \( k = 4 \) exams scoring below 50% is given by the binomial formula:
   \[
   P(X = 4) = \binom{28}{4} p^4 (1 - p)^{28 - 4}
   \]
   \[
   = \binom{28}{4} (0.1335)^4 (1 - 0.1335)^{24}
   \]

4. Calculate the binomial coefficient \( \binom{28}{4} = \frac{28!}{4!(28-4)!} = 20475 \).

5. Substitute values into the formula and solve:
   \[
   P(X = 4) \approx 20475 \times (0.1335)^4 \times (0.8665)^{24}
   \]

6. After calculation, this yields approximately:
   \[
   \boxed{0.2061}
   \]

---

Do you want more details on any part? Here are five related questions to consider:

1. How would you find the probability of scoring below a certain percentage, such as 40%?
2. If the passing mark increased to 55%, how would the answer to part (b) change?
3. What would the Z-score be for a mark of 65%?
4. If the standard deviation increased to 12%, how would it affect the answer to part (c)?
5. How does the binomial distribution used in part (d) compare to a Poisson distribution?

**Tip**: In normally distributed data, around 68% of data falls within one standard deviation from the mean, about 95% within two, and 99.7% within three. This is useful for quick approximations.

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 μ = 60% and a standard deviation of σ = 9%. (a) What is the probability that a randomly chosen final exam mark in this course will be at least 72%? (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? (c) The top 3% 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 4 of these exams will have a grade of less than 50%?

Explore the probability calculations for final exam grades in an introductory calculus class with a normal distribution (mean 60%, standard deviation 9%). Solve for probabilities of scoring at least 72%, not passing with less than 50%, achieving an A grade in the top 3%, and binomial probability of 4 out of 28 exams scoring below 50%.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation