Let's solve each part of this problem step-by-step.

Given Information

The problem states that the final exam marks are normally distributed with:

• Mean $\mu = 63$%
• Standard deviation $\sigma = 10$%

Solution

(a) Probability that a random chosen final exam mark will be at most 71.3%

To find this probability, we calculate the z-score for 71.3% and use the standard normal distribution table.

1. Calculate the z-score: $z = \frac{X - \mu}{\sigma} = \frac{71.3 - 63}{10} = \frac{8.3}{10} = 0.83$

2. Find the probability: Using the standard normal distribution table or calculator, we look up the probability for $z = 0.83$. $P(Z \leq 0.83) \approx 0.79673$

   Answer: The probability that a random final exam mark will be at most 71.3% is approximately 0.79673.

(b) Proportion of students who will not pass the statistics final exam (if passing mark is 50%)

Here, we need to find the probability that a student scores less than 50%.

1. Calculate the z-score: $z = \frac{X - \mu}{\sigma} = \frac{50 - 63}{10} = \frac{-13}{10} = -1.3$

2. Find the probability: Using the standard normal distribution table, we look up the probability for $z = -1.3$. $P(Z \leq -1.3) \approx 0.09680$

   Answer: The proportion of students who will not pass the statistics final exam is approximately 0.09680 (or 9.68%).

(c) Maximum final exam mark needed to receive a letter grade of at most an F (bottom 3.5% of students)

In this part, we need to find the score corresponding to the bottom 3.5% of the distribution, which means finding the z-score that corresponds to the 3.5th percentile.

1. Find the z-score for the 3.5th percentile: From the standard normal distribution table, the z-score corresponding to the 3.5th percentile is approximately $z = -1.81$.

2. Calculate the exam mark: Using the z-score formula, solve for $X$: $X = \mu + z \cdot \sigma = 63 + (-1.81) \cdot 10 = 63 - 18.1 = 44.9$

   Answer: The maximum final exam mark needed to earn a letter grade of at most an F is approximately 44.9%.

Summary of Answers

• (a) Probability of scoring at most 71.3%: 0.79673
• (b) Proportion of students not passing (below 50%): 0.09680
• (c) Maximum mark for bottom 3.5% (F grade): 44.9%

Would you like a deeper breakdown of any part of these calculations?

Additional Related Questions

1. How would the probability in (a) change if the mean was higher?
2. What score corresponds to the top 10% of students?
3. How would you calculate the probability for any given percentage using the cumulative distribution function?
4. What is the z-score formula and when is it typically used in statistics?
5. How can standard deviation affect the distribution of grades?

Tip

When solving normal distribution problems, always start by calculating the z-score to standardize your value. This makes it easier to find probabilities using the standard normal table or calculator.