This problem deals with a normal distribution for final exam scores, where the mean $\mu = 61\%$ and the standard deviation $\sigma = 11\%$. Let's go through each part:

Part (a)

We are asked to find the probability that a randomly chosen exam mark is at least 70%. This requires finding the z-score for 70% and then using a standard normal distribution table.

The z-score formula is: $z = \frac{x - \mu}{\sigma}$ Where:

• $x = 70\%$
• $\mu = 61\%$
• $\sigma = 11\%$

$z = \frac{70 - 61}{11} = \frac{9}{11} \approx 0.8182$

Now, using the standard normal distribution table or a calculator, we can find the area to the left of $z = 0.8182$, which is approximately 0.7939. Therefore, the area to the right, representing the probability of scoring at least 70%, is:

$P(X \geq 70) = 1 - 0.7939 = 0.2061$

Thus, the probability is approximately 0.2061.

Part (b)

We are asked to find the proportion of students who will not pass the final exam if the passing mark is 50%.

We again use the z-score formula, with $x = 50\%$:

$z = \frac{50 - 61}{11} = \frac{-11}{11} = -1$

From the z-table, the area to the left of $z = -1$ is approximately 0.1587. Therefore, the proportion of students who will not pass is 0.1587.

Part (c)

This part asks for the minimum final exam mark that corresponds to the top 3% of students. We are looking for the z-score that corresponds to the 97th percentile (since the top 3% corresponds to a cumulative area of 0.97).

From the z-table, the z-score corresponding to 0.97 is approximately $z = 1.88$.

Using the z-score formula to solve for $x$: $1.88 = \frac{x - 61}{11}$ $x - 61 = 1.88 \times 11 \approx 20.68$ $x = 61 + 20.68 = 81.68$

Thus, the minimum mark needed for an A is approximately 81.68%.

Part (d)

We are asked to find the probability that 4 out of 28 exams will have a grade of less than 50%. The probability of getting less than 50% is $P(X < 50)$, which we found in part (b) to be 0.1587.

This is a binomial probability problem, where:

• $n = 28$,
• $k = 4$,
• $p = 0.1587$.

The binomial probability formula is: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ Substituting in the values: $P(X = 4) = \binom{28}{4} (0.1587)^4 (1 - 0.1587)^{24}$

Let me calculate that for you:

$P(X = 4) \approx 0.1923$

So, the probability that 4 of the 28 exams will have a grade less than 50% is approximately 0.1923.

Would you like further clarifications or more details on any of these steps? Here are some related questions you might consider:

1. How do z-scores relate to normal distributions?
2. Can you explain how to interpret the z-table?
3. What is the difference between cumulative and point probabilities in a normal distribution?
4. How would the answers change if the mean exam score was 65% instead of 61%?
5. How is the binomial distribution related to the normal distribution?

Tip: When solving normal distribution problems, always check if you need to find the area to the left or right of the z-score to avoid mistakes.