The image shows a problem about the probability of dogs being spayed or neutered, with a provided percentage. Here’s a breakdown of the problem:

• 88% of owned dogs in the United States are spayed or neutered.
• The goal is to determine the probability of various outcomes when selecting 40 owned dogs randomly.

Let's solve each part:

Binomial Distribution Formula:

Since the probability of success (spayed/neutered) is fixed, this is a binomial probability problem. The formula for binomial probability is:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ where:

• $n = 40$ (number of trials, i.e., dogs),
• $k$ is the number of dogs that are spayed/neutered (varies for each part),
• $p = 0.88$ (probability of success),
• $1 - p = 0.12$ (probability of failure).

Parts to solve:

1. Exactly 26 dogs are spayed or neutered: $P(X = 26)$
2. At most 28 dogs are spayed or neutered: $P(X \leq 28)$
3. At least 28 dogs are spayed or neutered: $P(X \geq 28)$
4. Between 26 and 32 (including both): $P(26 \leq X \leq 32)$

I'll calculate the probabilities for each part using the binomial distribution. Let me do the math for these values.Here are the results for the given problem:

1. Exactly 26 dogs are spayed or neutered: Probability ≈ 0.0001
2. At most 28 dogs are spayed or neutered: Probability ≈ 0.0019
3. At least 28 dogs are spayed or neutered: Probability ≈ 0.9995
4. Between 26 and 32 dogs (including both): Probability ≈ 0.0996

These values differ slightly from what you entered, which may have caused errors in the system. Would you like any further explanation on how these were calculated?

Related Questions:

1. What is the probability of exactly 30 dogs being spayed or neutered?
2. How would the probabilities change if the percentage of spayed dogs was 85% instead of 88%?
3. What is the expected number of spayed dogs out of 40?
4. How does the variance affect the spread of probabilities in this binomial distribution?
5. How would this problem change if it was a normal approximation?

Tip:

When dealing with large binomial distributions, you can use the normal approximation to simplify calculations.