The problem involves a normal distribution of final exam grades in an introductory calculus class with a mean $\mu = 64\%$ and a standard deviation $\sigma = 10\%$. Let's solve each part using this information.

(a) Probability that a randomly chosen final exam mark will be at least 71%:

We need to calculate $P(X \geq 71\%)$ where $X$ follows a normal distribution with $\mu = 64\%$ and $\sigma = 10\%$.

Step 1: Convert to a z-score using the formula: $z = \frac{X - \mu}{\sigma}$ $z = \frac{71 - 64}{10} = \frac{7}{10} = 0.7$

Step 2: Look up the z-score of 0.7 in a standard normal distribution table (or use a calculator). The cumulative probability for $z = 0.7$ is approximately 0.7580.

Step 3: Since we need $P(X \geq 71)$, we subtract the cumulative probability from 1: $P(X \geq 71) = 1 - 0.7580 = 0.2420$ Answer: 0.2420

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

We are asked to find $P(X < 50\%)$.

Step 1: Convert to a z-score: $z = \frac{50 - 64}{10} = \frac{-14}{10} = -1.4$

Step 2: The cumulative probability for $z = -1.4$ is approximately 0.0808.

Thus, the proportion of students who will not pass is 0.0808. Answer: 0.0808

(c) Minimum final exam mark needed for the top 3% to earn an A:

We need to find the value of $X$ such that $P(X \geq X_A) = 0.03$, or equivalently, $P(X \leq X_A) = 0.97$.

Step 1: Look up the z-score corresponding to the cumulative probability of 0.97. This gives $z \approx 1.88$.

Step 2: Convert the z-score back to the original exam score: $X_A = \mu + z \cdot \sigma = 64 + 1.88 \cdot 10 = 64 + 18.8 = 82.8$ Thus, the minimum score needed for an A is 82.8%. Answer: 82.80%

(d) Probability that 5 out of 26 exams have a grade of less than 50%:

This follows a binomial distribution with $n = 26$, $p = P(X < 50\%) = 0.0808$ (from part b), and we are looking for $P(X = 5)$.

The binomial probability formula is: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ Substitute $n = 26$, $k = 5$, and $p = 0.0808$: $P(X = 5) = \binom{26}{5} (0.0808)^5 (1 - 0.0808)^{21}$ Using a calculator or software, this gives approximately 0.0022.

Answer: 0.0022

Let me know if you need more details on any part! Here are some follow-up questions you could consider:

1. How can you interpret z-scores in a normal distribution?
2. Why do we subtract from 1 when calculating probabilities for "at least" questions?
3. Can you describe the difference between binomial and normal distributions?
4. How do you find critical values for different percentages in a normal distribution?
5. What real-world examples can you think of where normal distributions are applied?

Tip: Always remember that probabilities in a normal distribution are cumulative—this is why we often use tables or software to calculate the areas under the curve.