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The problem asks to find the probability that a person answers between 79 and 89 questions (inclusive) correctly, out of 100, assuming each question has a probability of 80% (0.8) of being answered correctly. This follows a **binomial distribution** where:

- \( n = 100 \) (the number of trials),
- \( p = 0.8 \) (the probability of success on each trial),
- \( X \) is the number of correct answers.

We are asked to find the probability:

\[
P(79 \leq X \leq 89)
\]

For a binomial distribution, the probability mass function is given by:

\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]

However, since we are asked for the probability of a range, it's more efficient to use a **normal approximation to the binomial distribution**. The approximation works because the number of trials is large. Here's how:

1. **Mean** (\( \mu \)):  
   \[
   \mu = n \cdot p = 100 \cdot 0.8 = 80
   \]

2. **Standard deviation** (\( \sigma \)):  
   \[
   \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{100 \cdot 0.8 \cdot 0.2} = \sqrt{16} = 4
   \]

3. **Convert the range \( 79 \leq X \leq 89 \) to z-scores**:  
   Use the formula:
   \[
   z = \frac{x - \mu}{\sigma}
   \]
   For \( X = 79 \):
   \[
   z_1 = \frac{79 - 80}{4} = \frac{-1}{4} = -0.25
   \]
   For \( X = 89 \):
   \[
   z_2 = \frac{89 - 80}{4} = \frac{9}{4} = 2.25
   \]

4. **Find the corresponding probabilities using a standard normal distribution table**:
   - \( P(Z \leq -0.25) \) corresponds to 0.4013.
   - \( P(Z \leq 2.25) \) corresponds to 0.9878.

5. **Calculate the probability**:  
   \[
   P(79 \leq X \leq 89) = P(Z \leq 2.25) - P(Z \leq -0.25) = 0.9878 - 0.4013 = 0.5865
   \]

So, the probability is approximately **0.5865**.

Would you like more details on any step, or do you have any questions?

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Here are five related questions:
1. What is the binomial probability formula for a single value of \(X\)?
2. Why is the normal approximation used for large sample sizes in binomial distributions?
3. How do you compute z-scores for a given range?
4. What is the relationship between the mean and standard deviation in a normal distribution?
5. How does changing the probability \(p\) affect the binomial distribution?

**Tip:** When using normal approximation for binomial distributions, remember the rule of thumb: \( n \cdot p \) and \( n \cdot (1 - p) \) should both be at least 5 for the approximation to be accurate.

Assume that on a standardized test of 100 questions, a person has a probability of 80% of answering any particular question correctly. Find the probability of answering between 79 and 89 questions (inclusive). Assume independence, and round your answer to four decimal places.

This problem explores the probability of answering between 79 and 89 questions correctly on a 100-question test with an 80% success rate per question. The solution uses a normal approximation to the binomial distribution, highlighting key probability concepts and theorems. Suitable for Grades 10-12.

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