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

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



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



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*. 

To solve these problems, we'll use the **normal approximation to the binomial distribution** because we have a large sample size (1,100 Americans) and a binomial probability (10% or 0.10 of Americans are afraid to fly).

The binomial distribution has two parameters:
- \( n = 1100 \), which is the number of trials (i.e., the number of people sampled).
- \( p = 0.10 \), which is the probability that any given American is afraid to fly.

The **mean** \(\mu\) and the **standard deviation** \(\sigma\) for a binomial distribution are given by the formulas:
\[
\mu = n \cdot p
\]
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]

### Step 1: Compute Mean and Standard Deviation

\[
\mu = 1100 \cdot 0.10 = 110
\]
\[
\sigma = \sqrt{1100 \cdot 0.10 \cdot (1 - 0.10)} = \sqrt{1100 \cdot 0.10 \cdot 0.90} = \sqrt{99} \approx 9.95
\]

Now, we'll answer each question using the normal approximation to the binomial.

### Part 1: Probability that 121 or more Americans are afraid to fly

1. Standardize the value \( X = 121 \) using the **z-score** formula:
   \[
   z = \frac{X - \mu}{\sigma}
   \]
   For \( X = 121 \):
   \[
   z = \frac{121 - 110}{9.95} = \frac{11}{9.95} \approx 1.11
   \]
   
2. Find the probability of getting a \( z \)-score of 1.11 or higher. Using a standard normal distribution table or calculator:
   \[
   P(Z \geq 1.11) \approx 0.1335
   \]
   Thus, the probability that 121 or more Americans are afraid to fly is approximately 13.35%.

### Part 2: Probability that 165 or more Americans are afraid to fly

1. Standardize the value \( X = 165 \):
   \[
   z = \frac{165 - 110}{9.95} = \frac{55}{9.95} \approx 5.53
   \]

2. Find the probability of getting a \( z \)-score of 5.53 or higher. This is an extremely high z-score, and the probability is effectively 0%.

   Therefore, the probability that 165 or more Americans are afraid to fly is approximately 0%.

### Part 3: Probability that 8% or less of the Americans are afraid to fly

8% of 1,100 people corresponds to \( 0.08 \times 1100 = 88 \) people.

1. Standardize the value \( X = 88 \):
   \[
   z = \frac{88 - 110}{9.95} = \frac{-22}{9.95} \approx -2.21
   \]

2. Find the probability of getting a \( z \)-score of -2.21 or lower:
   \[
   P(Z \leq -2.21) \approx 0.0136
   \]
   Thus, the probability that 8% or fewer Americans are afraid to fly is approximately 1.36%.

---

### Summary of Results:
1. The probability that **121 or more** Americans are afraid to fly is **13.35%**.
2. The probability that **165 or more** Americans are afraid to fly is **0%**.
3. The probability that **8% or fewer** Americans are afraid to fly is **1.36%**.

---

Do you want further clarification on any part of this? Let me know! Here are 5 related questions:

1. How is the z-score calculated in a binomial distribution?
2. When is it appropriate to use the normal approximation to the binomial distribution?
3. What is the empirical rule in statistics, and how does it relate to z-scores?
4. How does increasing the sample size affect the standard deviation in binomial distributions?
5. Can you explain why the probability for \( Z = 5.53 \) is practically zero?

**Tip**: The normal approximation works best when both \( np \) and \( n(1-p) \) are greater than 5.

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

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

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

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, and 8% or fewer Americans in a sample of 1,100 are afraid to fly using normal approximation to the binomial distribution. This solution covers key concepts such as Z-scores, mean, and standard deviation, with step-by-step explanations. Suitable for students in Grades 11-12.

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