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%.

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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%.

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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.