53% of all Americans are home owners. If 30 Americans are randomly selected, find the probability that

a. Exactly 15 of them are are home owners.
b. At most 16 of them are are home owners.
c. At least 17 of them are home owners.
d. Between 11 and 19 (including 11 and 19) of them are home owners.

This problem involves binomial probability since we are given a percentage of Americans who are homeowners, and we are selecting a random sample from that population.

### Key Data:
- Probability of success (homeowner) \( p = 0.53 \)
- Probability of failure (non-homeowner) \( q = 1 - p = 1 - 0.53 = 0.47 \)
- Number of trials (sample size) \( n = 30 \)

For a binomial distribution, the probability of exactly \( k \) successes (homeowners) in \( n \) trials is given by the binomial formula:

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

where \( \binom{n}{k} \) is the binomial coefficient (number of ways to choose \( k \) successes out of \( n \) trials).

Let’s go step by step:

### (a) Probability that exactly 15 are homeowners.

We need to find \( P(X = 15) \):

\[
P(X = 15) = \binom{30}{15} (0.53)^{15} (0.47)^{15}
\]

### (b) Probability that at most 16 are homeowners.

We need to find \( P(X \leq 16) \), which is the cumulative probability:

\[
P(X \leq 16) = P(X = 0) + P(X = 1) + \dots + P(X = 16)
\]

### (c) Probability that at least 17 are homeowners.

We need to find \( P(X \geq 17) \), which is the complement of \( P(X \leq 16) \):

\[
P(X \geq 17) = 1 - P(X \leq 16)
\]

### (d) Probability that between 11 and 19 (inclusive) are homeowners.

We need to find \( P(11 \leq X \leq 19) \), which is the cumulative probability for that range:

\[
P(11 \leq X \leq 19) = P(X = 11) + P(X = 12) + \dots + P(X = 19)
\]

Let me calculate these probabilities for you.Here are the calculated probabilities:

### (a) Probability that exactly 15 of them are homeowners:
\[
P(X = 15) \approx 0.137
\]

### (b) Probability that at most 16 of them are homeowners:
\[
P(X \leq 16) \approx 0.585
\]

### (c) Probability that at least 17 of them are homeowners:
\[
P(X \geq 17) \approx 0.415
\]

### (d) Probability that between 11 and 19 (inclusive) of them are homeowners:
\[
P(11 \leq X \leq 19) \approx 0.883
\]

Would you like any additional details or have any questions?

Here are 5 related questions:
1. What is the probability of selecting exactly 10 homeowners?
2. How would the probabilities change if the probability of homeownership increased to 60%?
3. What is the expected number of homeowners in the group of 30?
4. How does the binomial distribution compare to a normal approximation in this case?
5. How would you compute the probability for fewer than 5 homeowners?

**Tip:** When working with binomial probabilities for large \( n \), using cumulative functions or normal approximations can simplify calculations.

53% of all Americans are homeowners. If 30 Americans are randomly selected, find the probability that a. Exactly 15 of them are homeowners. b. At most 16 of them are homeowners. c. At least 17 of them are homeowners. d. Between 11 and 19 (including 11 and 19) of them are homeowners.

Explore how to calculate binomial probabilities for homeownership in a random sample of 30 Americans. Learn how to find the probability that exactly 15 are homeowners, at most 16, at least 17, and between 11 and 19. Suitable for high school students in Grades 11-12.

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