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.