SOLUTION: 84% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 36 owned dogs are randomly selected, find the probability that a. Ex

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Question 1199760: 84% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 36 owned dogs are randomly selected, find the probability that
a. Exactly 31 of them are spayed or neutered.
b. At most 29 of them are spayed or neutered.
c. At least 30 of them are spayed or neutered.
d. Between 29 and 35 (including 29 and 35) of them are spayed or neutered.

Answer by CPhill(1959) (Show Source):
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**a) Probability of exactly 31 spayed/neutered dogs:**
* This is a binomial probability problem.
* n = 36 (sample size)
* p = 0.84 (probability of success - spayed/neutered)
* q = 1 - p = 0.16 (probability of failure - not spayed/neutered)
* x = 31 (number of successes)
* P(X = 31) = (36C31) * (0.84)^31 * (0.16)^(36-31)
* where (36C31) is the number of combinations of 36 things taken 31 at a time.
* Using a binomial probability calculator or statistical software:
P(X = 31) ≈ 0.1223
**b) Probability of at most 29 spayed/neutered dogs:**
* P(X ≤ 29) = P(X = 0) + P(X = 1) + ... + P(X = 29)
* Use a cumulative binomial probability calculator or statistical software:
P(X ≤ 29) ≈ 0.0419
**c) Probability of at least 30 spayed/neutered dogs:**
* P(X ≥ 30) = P(X = 30) + P(X = 31) + ... + P(X = 36)
* Alternatively:
P(X ≥ 30) = 1 - P(X ≤ 29)
P(X ≥ 30) = 1 - 0.0419 = 0.9581
**d) Probability of between 29 and 35 spayed/neutered dogs (inclusive):**
* P(29 ≤ X ≤ 35) = P(X = 29) + P(X = 30) + ... + P(X = 35)
* Use a cumulative binomial probability calculator or statistical software to find:
P(29 ≤ X ≤ 35) ≈ 0.9954
**In summary:**
* a) P(X = 31) ≈ 0.1223
* b) P(X ≤ 29) ≈ 0.0419
* c) P(X ≥ 30) ≈ 0.9581
* d) P(29 ≤ X ≤ 35) ≈ 0.9954