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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.

Assignment 5: Problem 6, involving a normal distribution with mean 64% and standard deviation 10%. Four parts include calculating probabilities and minimum marks based on various criteria.

Solve a problem involving normal distribution with mean 64% and standard deviation 10%. This problem includes calculating probabilities for final exam marks, minimum score for an A grade, and more, suitable for university-level statistics students.

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