61% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 46 owned dogs are randomly selected, find the probability that

a. Exactly 29 of them are spayed or neutered.

b. At most 29 of them are spayed or neutered.

c. At least 28 of them are spayed or neutered.

d. Between 28 and 32 (including 28 and 32) of them are spayed or neutered.

A good calculator is found at: Stattrek Binomial Calculator Round answers to at least 4 decimal places.

A binomial distribution, with a success rate of p on each trial, gives us the probability of x number of success in n number of trials, using the formula:

P(X = x) nCx.pˣ.qⁿ⁻ˣ, where q = 1 - p.

In the question, we are informed that 61% of owned dogs in the United States are spayed or neutered, and are given that 46 owned dogs are randomly selected.

This can be seen as a binomial probability distribution, with n = 46, and p = 61% = 0.61, q = 1 - p = 1 - 0.61 = 0.39.

(a) We are asked for the probability of exactly 29 of them being spayed or neutered.

Thus, x = 29, and we need to find P(X = 29).

Using the given calculator, P(X = 29) = 0.1163.

(b) We are asked for the probability of at most 29 of them being spayed or neutered.

Thus, we need to find P(X ≤ 29).

Using the given calculator, P(X ≤ 29) = 0.6648.

(c) We are asked for the probability of at least 28 of them being spayed or neutered.

Thus, we need to find P(X ≥ 28).

Using the given calculator, P(X ≥ 28) = 0.5714.

(d) We are asked for the probability between 28 and 32 of them are spayed or neutered.

Thus, we need to find P(28 ≤ X ≤ 32), which can be shown as;

P(28 ≤ X ≤ 32) = P(X ≤ 32) - P(X < 28).

Using the given calculator, P(X ≤ 32) = 0.91209.

Using the given calculator, P(X < 28) = 0.42864.

Thus, P(28 ≤ X ≤ 32) = P(X ≤ 32) - P(X < 28) = 0.91209 - 0.42864 = 0.48345.

Thus, using the binomial distribution, the probability that:

(a) Exactly 29 of them are spayed or neutered, that is, P(X = 29) = 0.1163.

(b) At most 29 of them are spayed or neutered, that is, P(X ≤ 29) = 0.6648.

(c) At least 28 of them are spayed or neutered, that is, P(X ≥ 28) = 0.5714.

(d) Between 28 and 32 (including 28 and 32) of them are spayed or neutered, that is, P(28 ≤ X ≤ 32) = 0.48345.

Learn more about the binomial distribution at

https://brainly.com/question/24756209

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