It appears that the question asks to calculate the probability of a person answering between 74 and 84 questions correctly, assuming a standardized test of 100 questions, where each question has an 80% probability of being answered correctly.

This is a binomial probability problem, where:

• $n = 100$ (number of trials, or questions),
• $p = 0.80$ (probability of success, or answering correctly),
• $k$ is the number of successes (correct answers),
• We are interested in finding $P(74 \leq X \leq 84)$.

However, this problem is easier to solve using the normal approximation to the binomial distribution, since $n$ is large. The normal approximation is given by:

$\mu = n \times p = 100 \times 0.8 = 80$ $\sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{100 \times 0.8 \times 0.2} = \sqrt{16} = 4$

Using the normal approximation:

$P(74 \leq X \leq 84) \approx P\left(\frac{73.5 - 80}{4} \leq Z \leq \frac{84.5 - 80}{4}\right)$

This becomes:

$P\left(\frac{-6.5}{4} \leq Z \leq \frac{4.5}{4}\right) = P(-1.625 \leq Z \leq 1.125)$

Using a Z-table or calculator:

$P(Z \leq 1.125) \approx 0.8697$ $P(Z \leq -1.625) \approx 0.0524$

Thus, the probability is:

$P(74 \leq X \leq 84) = 0.8697 - 0.0524 = 0.8173$

Your previous answer of 0.8186 is close but slightly off. The correct rounded answer should be approximately 0.8173.

Would you like further clarification or details on this calculation?

Here are five follow-up questions you could consider:

1. How does the normal approximation to the binomial distribution work?
2. What happens when you apply a continuity correction in normal approximation?
3. Can you use other methods to calculate binomial probabilities when $n$ is large?
4. What are the differences between the exact binomial probability and the normal approximation?
5. How can you use technology (like calculators or software) to find binomial probabilities efficiently?

Tip: For large values of $n$, it's often practical to use the normal approximation to the binomial distribution to save calculation time.