Question 1199757
**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