SOLUTION: 8% of all Americans live in poverty. If 37 Americans are randomly selected, find the probability that a. Exactly 3 of them live in poverty. b. At most 5 of them live in po

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Question 1199757: 8% of all Americans live in poverty. If 37 Americans are randomly selected, find the probability that
a. Exactly 3 of them live in poverty.
b. At most 5 of them live in poverty.
c. At least 4 of them live in poverty.
d. Between 2 and 6 (including 2 and 6) of them live in poverty.

Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website!
**a) Probability of exactly 3 of them live in poverty.**
* This is a binomial probability problem.
* n = 37 (sample size)
* p = 0.08 (probability of success - living in poverty)
* q = 1 - p = 0.92 (probability of failure - not living in poverty)
* x = 3 (number of successes)
* P(X = 3) = (37C3) * (0.08)^3 * (0.92)^(37-3)
* where (37C3) is the number of combinations of 37 things taken 3 at a time.
* Using a binomial probability calculator or statistical software:
P(X = 3) ≈ 0.2079
**b) Probability of at most 5 of them live in poverty.**
* P(X ≤ 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
* Use a cumulative binomial probability calculator or statistical software:
P(X ≤ 5) ≈ 0.9589
**c) Probability of at least 4 of them live in poverty.**
* P(X ≥ 4) = P(X = 4) + P(X = 5) + ... + P(X = 37)
* Alternatively:
P(X ≥ 4) = 1 - P(X ≤ 3)
P(X ≥ 4) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))
P(X ≥ 4) ≈ 1 - 0.7847 = 0.2153
**d) Probability of between 2 and 6 of them live in poverty (inclusive):**
* P(2 ≤ X ≤ 6) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)
* Use a cumulative binomial probability calculator or statistical software to find:
P(2 ≤ X ≤ 6) ≈ 0.9983
**In summary:**
* a) P(X = 3) ≈ 0.2079
* b) P(X ≤ 5) ≈ 0.9589
* c) P(X ≥ 4) ≈ 0.2153
* d) P(2 ≤ X ≤ 6) ≈ 0.9983