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?

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.