Question 1209734: 84% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 50 owned dogs are randomly selected, find the probability that
a. Exactly 44 of them are spayed or neutered.
b. At most 43 of them are spayed or neutered.
c. At least 43 of them are spayed or neutered.
d. Between 37 and 41 (including 37 and 41) of them are spayed or neutered.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! This is a binomial probability problem. Here's how to solve it:
* **n** (number of trials) = 50
* **p** (probability of success - spayed/neutered) = 0.84
* **q** (probability of failure - not spayed/neutered) = 1 - p = 0.16
The binomial probability formula is: P(x) = (nCx) * p^x * q^(n-x)
Where nCx represents "n choose x" (the binomial coefficient).
**(a) Exactly 44 are spayed/neutered:**
P(x = 44) = (50C44) * (0.84)^44 * (0.16)^6
P(x = 44) = 1,478,745,000 * 0.002011 * 0.00001678
P(x = 44) ≈ 0.0049
**(b) At most 43 are spayed/neutered:**
This means 0 to 43 are spayed/neutered. It's a cumulative probability. We can use a binomial cumulative distribution function (CDF) calculator or statistical software for this. It's the sum of probabilities from x=0 to x=43.
P(x ≤ 43) ≈ 0.8878 (using a calculator or software)
**(c) At least 43 are spayed/neutered:**
This means 43 to 50 are spayed/neutered. We can use the complement rule:
P(x ≥ 43) = 1 - P(x < 43) = 1 - P(x ≤ 42)
Use a binomial CDF calculator:
P(x ≥ 43) = 1 - 0.8284
P(x ≥ 43) ≈ 0.1716
**(d) Between 37 and 41 (inclusive):**
This means 37, 38, 39, 40, and 41 are spayed/neutered. We can use the CDF:
P(37 ≤ x ≤ 41) = P(x ≤ 41) - P(x ≤ 36)
Use a binomial CDF calculator:
P(37 ≤ x ≤ 41) = 0.4072 - 0.0150
P(37 ≤ x ≤ 41) ≈ 0.3922
**Summary of Answers:**
* (a) P(x = 44) ≈ 0.0049
* (b) P(x ≤ 43) ≈ 0.8878
* (c) P(x ≥ 43) ≈ 0.1716
* (d) P(37 ≤ x ≤ 41) ≈ 0.3922
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