This problem involves a binomial distribution, where the number of correctly answered questions follows:

$X \sim \text{Binomial}(n=100, p=0.75)$

We need to calculate:

$P(79 \leq X \leq 89)$

Since $n$ is large, we can approximate the binomial distribution using the normal distribution with the following parameters:

• Mean:
  $\mu = n p = 100(0.75) = 75$
• Standard deviation:
  $\sigma = \sqrt{n p (1 - p)} = \sqrt{100(0.75)(0.25)} = \sqrt{18.75} \approx 4.33$

Using the normal approximation, we apply the continuity correction and transform the probability into the standard normal distribution:

$P(78.5 \leq X \leq 89.5)$

Standardizing to Z-Scores:

$Z = \frac{X - \mu}{\sigma}$

For $X = 78.5$:

$Z = \frac{78.5 - 75}{4.33} = \frac{3.5}{4.33} \approx 0.81$

For $X = 89.5$:

$Z = \frac{89.5 - 75}{4.33} = \frac{14.5}{4.33} \approx 3.35$

Finding Probabilities:

From standard normal tables:

$P(Z \leq 3.35) \approx 0.9996$

$P(Z \leq 0.81) \approx 0.7910$

Thus, the probability is:

$P(79 \leq X \leq 89) \approx 0.9996 - 0.7910 = 0.2086$

Final Answer:

$\boxed{0.2086}$

Would you like a deeper explanation of any step? Here are some related questions:

1. How do you derive the normal approximation for a binomial distribution?
2. Why is a continuity correction necessary in this approximation?
3. How can we calculate binomial probabilities without normal approximation?
4. What are other methods to approximate binomial probabilities?
5. How do standard normal tables work in probability calculations?

Tip: When using normal approximation for a binomial distribution, always check that $np$ and $n(1 - p)$ are both at least 5 to ensure accuracy.