 Assume that on a standardized test of 100 questions, a person has a probability of 85% of answering any particular question correctly. Find the probability of answering between 77 and 87 questions, inclusive. (Assume independence, and round your answer to four decimal places.)
P(77 ≤ X ≤ 87) =


To solve this, we need to use the **binomial distribution**, as each question is an independent Bernoulli trial with a success probability \( p = 0.85 \).

Let:
- \( X \) be the number of questions answered correctly,
- \( n = 100 \) (total questions),
- \( p = 0.85 \) (probability of answering correctly).

We want to find the probability \( P(77 \leq X \leq 87) \), which is the sum of the probabilities from \( X = 77 \) to \( X = 87 \).

### Step 1: Approximate with a Normal Distribution
Since \( n \) is large, we can use the **normal approximation to the binomial distribution**. According to the approximation, if \( X \sim \text{Binomial}(n, p) \), then \( X \) can be approximated by \( Y \sim N(\mu, \sigma^2) \), where:
- \( \mu = np = 100 \times 0.85 = 85 \),
- \( \sigma = \sqrt{np(1 - p)} = \sqrt{100 \times 0.85 \times 0.15} \approx 3.5693 \).

### Step 2: Apply the Continuity Correction
To account for the discrete nature of \( X \), we use a continuity correction. We seek:
\[
P(76.5 \leq Y \leq 87.5)
\]
where \( Y \sim N(85, 3.5693^2) \).

### Step 3: Standardize and Calculate
Standardize the values:
\[
P\left( \frac{76.5 - 85}{3.5693} \leq Z \leq \frac{87.5 - 85}{3.5693} \right)
\]
1. For \( 76.5 \): \( \frac{76.5 - 85}{3.5693} \approx -2.38 \)
2. For \( 87.5 \): \( \frac{87.5 - 85}{3.5693} \approx 0.70 \)

Now, we find the probabilities using a standard normal table:
\[
P(-2.38 \leq Z \leq 0.70) = P(Z \leq 0.70) - P(Z \leq -2.38)
\]

From standard normal tables:
- \( P(Z \leq 0.70) \approx 0.7580 \),
- \( P(Z \leq -2.38) \approx 0.0087 \).

Thus:
\[
P(77 \leq X \leq 87) \approx 0.7580 - 0.0087 = 0.7493
\]

### Final Answer
\[
P(77 \leq X \leq 87) \approx 0.7493
\]

Would you like further details or explanations?

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Here are some related questions for further exploration:

1. What is the probability of getting exactly 85 questions correct?
2. How would the probability change if the success rate was 90%?
3. How is the continuity correction applied in normal approximations?
4. Why is the binomial distribution approximated by the normal distribution?
5. How would we calculate \( P(77 \leq X \leq 87) \) without normal approximation?

**Tip:** When using normal approximation for binomial distributions, remember to apply the continuity correction to improve accuracy.

Assume that on a standardized test of 100 questions, a person has a probability of 85% of answering any particular question correctly. Find the probability of answering between 77 and 87 questions, inclusive. (Assume independence, and round your answer to four decimal places.) P(77 ≤ X ≤ 87) =

Learn how to calculate the probability of answering between 77 and 87 questions correctly on a standardized test of 100 questions with a success rate of 85%. This solution uses the normal approximation to the binomial distribution and continuity correction for accurate probability calculation. Ideal for students in Grades 11-12 studying probability theory.

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