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The problem asks to approximate a binomial distribution with a normal distribution, using a sample of 1000 people, where 19.3% of them have a disability. Let's solve the two parts.

### Step 1: Setup
We are dealing with a binomial distribution with:
- \( n = 1000 \) (sample size),
- \( p = 0.193 \) (probability of a person having a disability),
- \( q = 1 - p = 0.807 \) (probability of not having a disability).

The mean and standard deviation of a binomial distribution can be approximated by:
- **Mean** \( \mu = n \cdot p = 1000 \times 0.193 = 193 \),
- **Standard deviation** \( \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{1000 \times 0.193 \times 0.807} \approx 12.4545 \).

### Step 2: Normal Approximation
We can use the normal approximation to solve both questions.

#### Part (a): Probability that more than 204 people in the sample have a disability.

For this, we calculate \( P(X > 204) \).

1. Convert \( X = 204 \) to the standard normal variable \( Z \):
   \[
   Z = \frac{204 - 193}{12.4545} \approx 0.8837
   \]

2. Using the standard normal table, the probability corresponding to \( Z = 0.8837 \) is about 0.8127.

3. Therefore, the probability that more than 204 people have a disability is:
   \[
   P(X > 204) = 1 - 0.8127 = 0.1873
   \]

#### Part (b): Probability that more than 177 but less than 209 people in the sample have a disability.

For this, we calculate \( P(177 < X < 209) \).

1. Convert \( X = 177 \) and \( X = 209 \) to standard normal variables:
   \[
   Z_1 = \frac{177 - 193}{12.4545} \approx -1.2854
   \]
   \[
   Z_2 = \frac{209 - 193}{12.4545} \approx 1.2867
   \]

2. Using the standard normal table:
   - For \( Z_1 = -1.2854 \), the probability is approximately 0.0992.
   - For \( Z_2 = 1.2867 \), the probability is approximately 0.9008.

3. Therefore, the probability that between 177 and 209 people have a disability is:
   \[
   P(177 < X < 209) = 0.9008 - 0.0992 = 0.8016
   \]

### Final Answers:
- **(a)** \( P(X > 204) \approx 0.1873 \),
- **(b)** \( P(177 < X < 209) \approx 0.8016 \).

Would you like further details or have any questions?

Here are five follow-up questions to expand your understanding:
1. How do we determine when to use normal approximation for a binomial distribution?
2. What conditions must be met for the normal approximation to be valid?
3. How do you interpret the Z-scores in the context of this problem?
4. Why do we subtract the normal probability for the upper tail in part (a)?
5. How would the results change if the sample size were smaller?

**Tip**: When approximating binomial distributions, always check if \( np \) and \( nq \) are both greater than 5 to validate the normal approximation method.

There were 49.7 million people with some type of long-lasting condition or disability living in the United States in 2000. This represented 19.3 percent of civilians aged five and over. A sample of 1000 people was selected at random. Use normal approximation to answer the following questions: (a) Approximate the probability that more than 204 people in the sample have a disability. (b) Approximate the probability that more than 177 but less than 209 people in the sample have a disability.

Solve a normal approximation problem with binomial distribution using a sample of 1000 people where 19.3% have a disability. Learn how to approximate the probability that more than 204 or between 177 and 209 people in the sample have a disability.

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