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.