64% of all Americans live in cities with population greater than 100,000 people. If 31 Americans are randomly selected, find the probability that

a. Exactly 21 of them live in cities with population greater than 100,000 people. 
b. At most 19 of them live in cities with population greater than 100,000 people. 
c. At least 18 of them live in cities with population greater than 100,000 people. 
d. Between 16 and 20 (including 16 and 20) of them live in cities with population greater than 100,000 people. 

This problem involves a **binomial distribution**, where each trial (an American randomly selected) has two possible outcomes:

- Success: The person lives in a city with a population greater than 100,000 (with probability \( p = 0.64 \)).
- Failure: The person lives in a city with a population of 100,000 or less (with probability \( q = 1 - p = 0.36 \)).

The number of trials (31) is fixed, and we are asked to find the probability of certain outcomes.

The probability mass function (PMF) for a binomial distribution is given by:

\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

where:
- \( n = 31 \) (the number of trials),
- \( k \) is the number of successes,
- \( p = 0.64 \) (the probability of success on each trial).

Now, let’s calculate each part:

### a. Exactly 21 of them live in cities with population greater than 100,000 people.

We need to find \( P(X = 21) \), where \( X \) follows a binomial distribution with parameters \( n = 31 \) and \( p = 0.64 \).

\[
P(X = 21) = \binom{31}{21} (0.64)^{21} (0.36)^{10}
\]

### b. At most 19 of them live in cities with population greater than 100,000 people.

We need to find \( P(X \leq 19) \), which is the cumulative probability from \( X = 0 \) to \( X = 19 \):

\[
P(X \leq 19) = \sum_{k=0}^{19} P(X = k)
\]

### c. At least 18 of them live in cities with population greater than 100,000 people.

We need to find \( P(X \geq 18) \), which is the complement of \( P(X \leq 17) \):

\[
P(X \geq 18) = 1 - P(X \leq 17)
\]

### d. Between 16 and 20 (including 16 and 20) of them live in cities with population greater than 100,000 people.

We need to find \( P(16 \leq X \leq 20) \):

\[
P(16 \leq X \leq 20) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)
\]

Let's calculate each of these probabilities using the binomial formula.Here are the calculated probabilities:

a. The probability that exactly 21 of them live in cities with a population greater than 100,000 is approximately \( 0.1379 \) (13.79%).

b. The probability that at most 19 of them live in cities with a population greater than 100,000 is approximately \( 0.4426 \) (44.26%).

c. The probability that at least 18 of them live in cities with a population greater than 100,000 is approximately \( 0.8104 \) (81.04%).

d. The probability that between 16 and 20 of them (inclusive) live in cities with a population greater than 100,000 is approximately \( 0.5364 \) (53.64%).

Would you like more details on any of these calculations or explanations?

Here are 5 related questions to expand on this topic:
1. What is the expected number of Americans living in cities with populations greater than 100,000 in this scenario?
2. What is the variance and standard deviation of the number of Americans living in large cities in this sample?
3. How would the probabilities change if the percentage of Americans living in large cities increased to 70%?
4. How would you calculate the probability that fewer than 15 of them live in cities with populations greater than 100,000?
5. How can the binomial distribution be approximated by a normal distribution for large \(n\)?

**Tip**: Binomial distributions can often be approximated using the normal distribution when the sample size is large enough.

64% of all Americans live in cities with population greater than 100,000 people. If 31 Americans are randomly selected, find the probability that a) Exactly 21 of them live in cities with population greater than 100,000 people. b) At most 19 of them live in cities with population greater than 100,000 people. c) At least 18 of them live in cities with population greater than 100,000 people. d) Between 16 and 20 (including 16 and 20) of them live in cities with population greater than 100,000 people.

Explore a detailed solution to a binomial probability problem where 64% of Americans live in cities with populations greater than 100,000. Learn how to calculate probabilities for exact, at most, at least, and between certain numbers in a sample of 31 Americans. Suitable for Grades 10-12.

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