Assume that on a standardized test of 100 questions, a person has a probability of 75% of answering any particular question correctly. Find the probability of answering between 73 and 83 questions, inclusive. (Assume independence, and round your answer to four decimal places.)

*P*(73 ≤ *X* ≤ 83) =
 





To find the probability that a person answers between 73 and 83 questions correctly, inclusive, we can use a binomial distribution with the following parameters:

- **Number of trials** \( n = 100 \) (since there are 100 questions),
- **Probability of success** \( p = 0.75 \) (since the probability of answering a question correctly is 75%).

### Step 1: Define the Binomial Distribution
In a binomial distribution, the probability mass function is given by:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
where \( X \) represents the number of correct answers out of 100 questions.

However, calculating the exact probability using binomial terms for each integer from 73 to 83 can be computationally intensive. Instead, we can approximate the binomial distribution with a normal distribution, which is suitable here due to the large sample size (\( n = 100 \)).

### Step 2: Approximate with Normal Distribution
For a binomial distribution with parameters \( n \) and \( p \), we can approximate it by a normal distribution with:
- **Mean** \( \mu = n \cdot p = 100 \cdot 0.75 = 75 \),
- **Standard deviation** \( \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{100 \cdot 0.75 \cdot 0.25} = 4.33 \) (rounded to two decimal places).

We want to find \( P(73 \leq X \leq 83) \).

### Step 3: Apply the Continuity Correction
To improve the approximation, we apply a continuity correction by adjusting the range slightly:
\[
P(73 \leq X \leq 83) \approx P(72.5 \leq X \leq 83.5)
\]

### Step 4: Convert to Standard Normal Distribution
Now, we convert this range to the standard normal distribution \( Z \) by using the formula:
\[
Z = \frac{X - \mu}{\sigma}
\]

Calculating for each bound:

1. For \( X = 72.5 \):
   \[
   Z = \frac{72.5 - 75}{4.33} \approx -0.58
   \]

2. For \( X = 83.5 \):
   \[
   Z = \frac{83.5 - 75}{4.33} \approx 1.96
   \]

### Step 5: Find the Probability
Now, we need to find \( P(-0.58 \leq Z \leq 1.96) \).

Using the standard normal table:
- \( P(Z \leq 1.96) \approx 0.9750 \),
- \( P(Z \leq -0.58) \approx 0.2810 \).

Thus:
\[
P(-0.58 \leq Z \leq 1.96) = 0.9750 - 0.2810 = 0.6940
\]

### Final Answer
\[
P(73 \leq X \leq 83) \approx 0.6940
\]

So, the probability of answering between 73 and 83 questions correctly is approximately **0.6940**.

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Would you like further explanation on any part of this solution?

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**Related Questions:**

1. How would the probability change if the success rate were 80% instead of 75%?
2. What is the probability of answering exactly 75 questions correctly?
3. How does the binomial distribution differ from the normal distribution?
4. How can we calculate the probability for a larger range, such as between 60 and 90 questions?
5. How would using a different continuity correction affect the result?

**Tip:** When approximating a binomial distribution with a normal distribution, remember to apply the continuity correction for better accuracy, especially when working with integer ranges.

Assume that on a standardized test of 100 questions, a person has a probability of 75% of answering any particular question correctly. Find the probability of answering between 73 and 83 questions, inclusive. (Assume independence, and round your answer to four decimal places.)

Learn how to calculate the probability of answering between 73 and 83 questions correctly on a standardized test using binomial distribution and normal approximation. This step-by-step solution includes applying a continuity correction and calculating Z-scores. Suitable for students in Grades 10-12.

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