Answer:
a. Exactly 4 of them live in poverty:
To find the probability that exactly 4 of the 30 Americans live in poverty, we need to calculate the probability of 4 specific events:
P(exactly 4 in poverty) = P(1st person in poverty) × P(2nd person in poverty) × P(3rd person in poverty) × P(4th person in poverty)
According to the U.S. Census Bureau's 2020 American Community Survey (ACS), the poverty rate in the United States is 18% or 18/100, so the probability of the 1st person being in poverty is:
P(1st person in poverty) = 18/100 ≈ 0.18
Similarly, the probability of the 2nd, 3rd, and 4th persons being in poverty is:
P(2nd person in poverty) = 18/100 ≈ 0.18
P(3rd person in poverty) = 18/100 ≈ 0.18
P(4th person in poverty) = 18/100 ≈ 0.18
Now, we can multiply these probabilities to get the probability of exactly 4 people being in poverty:
P(exactly 4 in poverty) = 0.18 × 0.18 × 0.18 × 0.18 ≈ 0.0072, or 0.72%
So, the probability that exactly 4 of the 30 Americans live in poverty is approximately 0.72%.
b. At most 4 of them live in poverty:
To find the probability that at most 4 of the 30 Americans live in poverty, we need to calculate the probability of 4 specific events:
P(at most 4 in poverty) = P(0 people in poverty) + P(1 person in poverty) + P(2 people in poverty) + P(3 people in poverty) + P(4 people in poverty)
Using the same probabilities as before, we get:
P(0 people in poverty) = (1 - 0.18)^30 ≈ 0.52
P(1 person in poverty) = 0.18 × (1 - 0.18)^29 ≈ 0.23
P(2 people in poverty) = 0.18 × 0.18 × (1 - 0.18)^28 ≈ 0.09
P(3 people in poverty) = 0.18 × 0.18 × 0.18 × (1 - 0.18)^27 ≈ 0.06
P(4 people in poverty) = 0.18 × 0.18 × 0.18 × 0.18 × (1 - 0.18)^26 ≈ 0.04
Now, we add up these probabilities to get the probability that at most 4 of the 30 Americans live in poverty:
P(at most 4 in poverty) = 0.52 + 0.23 + 0.09 + 0.06 + 0.04 ≈ 0.94
So, the probability that at most 4 of the 30 Americans live in poverty is approximately 0.94.
c. At least 4 of them live in poverty:
To find the probability that at least 4 of the 30 Americans live in poverty, we need to calculate the probability of 4 specific events:
P(at least 4 in poverty) = P(4 people in poverty) + P(5 people in poverty) + P(6 people in poverty) + ... + P(30 people in poverty)
Using the same probabilities as before, we get:
P(4 people in poverty) = 0.18 × 0.18 × 0.18 × 0.18 ≈ 0.072
P(5 people in poverty) = 0.18 × 0.18 × 0.18 × 0.18 × 0.18 ≈ 0.052
P(6 people in poverty) = 0.18 × 0.18 × 0.18 × 0.18 × 0.18 × 0.18 ≈ 0.036
...
P(30 people in poverty) = 0.18 × 0.18 × 0.18 × 0.18 × 0.18 × 0.18 × 0.18 ≈ 0.0012
Now, we add up these probabilities to get the probability that at least 4 of the 30 Americans live in poverty:
P(at least 4 in poverty) = 0.072 + 0.052 + 0.036 + ... + 0.0012 ≈ 0.17
So, the probability that at least 4 of the 30 Americans live in poverty is approximately 0.17.
d. Between 2 and 9 (including 2 and 9) of them live in poverty:
To find the probability that between 2 and 9 (including 2 and 9) of the 30 Americans live in poverty, we need to subtract the probability of exactly 4 people living in poverty from the probability of at most 4 people living in poverty:
P(between 2 and 9) = P(at most 4) - P(exactly 4)
Using the probabilities we calculated earlier, we get:
P(between 2 and 9) = 0.94 - 0.072 ≈ 0.87
So, the probability that between 2 and 9 (including 2 and 9) of the 30 Americans live in poverty is approximately 0.87.
Explanation: