https://file.aiquickdraw.com/mathbot/file/1728890339450c5tr7x85.jpg

The problem involves a binomial distribution, where:

- The probability \( p \) that an individual is afraid to fly is 0.10 (10%).
- The number of trials \( n \) is 1,100 (Americans sampled).

We need to compute the probabilities for different scenarios. Since the number of trials \( n \) is large, we can approximate the binomial distribution using a normal distribution. Here are the steps:

### 1. Normal Approximation to the Binomial:
- The mean \( \mu \) and standard deviation \( \sigma \) for a binomial distribution can be approximated as:
  \[
  \mu = np = 1100 \times 0.10 = 110
  \]
  \[
  \sigma = \sqrt{np(1 - p)} = \sqrt{1100 \times 0.10 \times 0.90} = \sqrt{99} \approx 9.95
  \]

### (a) Probability that 121 or more Americans are afraid to fly:
- We convert this to a normal distribution problem:
  \[
  z = \frac{121 - 110}{9.95} \approx 1.11
  \]
  Using the normal distribution table or a calculator, we find the probability for \( z = 1.11 \) and subtract it from 1 to find the area to the right:
  \[
  P(Z > 1.11) \approx 0.1335
  \]
  So the probability is approximately 13.35%.

### (b) Probability that 165 or more Americans are afraid to fly:
- Convert to a z-score:
  \[
  z = \frac{165 - 110}{9.95} \approx 5.52
  \]
  The probability of \( Z > 5.52 \) is extremely small, close to 0%. So, the probability is approximately 0.00%.

### (c) Probability that 8% or less of the Americans are afraid to fly:
- If 8% or fewer Americans are afraid to fly, that corresponds to \( 0.08 \times 1100 = 88 \).
  Convert to a z-score:
  \[
  z = \frac{88 - 110}{9.95} \approx -2.21
  \]
  Using the z-table, we find the probability for \( Z < -2.21 \):
  \[
  P(Z < -2.21) \approx 0.0136
  \]
  So the probability is approximately 1.36%.

### Summary of Answers:
- (a) 13.35%
- (b) 0.00%
- (c) 1.36%

Let me know if you would like further details or clarification. Here are five follow-up questions for further exploration:

1. What is the probability that exactly 121 Americans are afraid to fly?
2. How would the results change if 2,000 Americans were sampled instead of 1,100?
3. Can you calculate the expected number of Americans who are afraid to fly?
4. How would the results differ if the percentage of Americans afraid to fly was 15% instead of 10%?
5. What would the probability be if 5% or less of Americans were afraid to fly?

**Tip:** For large sample sizes in binomial problems, the normal approximation is often used for faster computation.

According to a survey, 10% of Americans are afraid to fly. Suppose 1,100 Americans are sampled.

a. What is the probability percentage that 121 or more Americans in the survey are afraid to fly? Round the percent to two decimal places.

b. What is the probability percentage that 165 or more Americans in the survey are afraid to fly? Round the percent to two decimal places.

c. What is the probability percentage that 8% or less of the Americans surveyed answered they were afraid to fly? Round the percent to two decimal places.

Learn how to calculate the probability that 121 or more, 165 or more, or 8% or less of 1,100 Americans are afraid to fly using normal approximation to the binomial distribution. This detailed guide covers key statistical concepts, including probability, mean, standard deviation, and Z-score calculations. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation