This problem involves calculating probabilities based on a binomial distribution. The binomial distribution applies because we have two possible outcomes for each dog: either they are spayed/neutered or they are not. The probability of success (spayed/neutered) is $p = 0.60$, and the number of trials is $n = 30$.

For a binomial distribution, the probability of exactly $k$ successes in $n$ trials is given by the formula:

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

where:

• $\binom{n}{k}$ is the binomial coefficient, representing the number of ways to choose $k$ successes out of $n$ trials,
• $p$ is the probability of success (0.60 in this case),
• $(1 - p)$ is the probability of failure (0.40 in this case).

We will use this formula and its variations to calculate the probabilities for each part.

a. Exactly 19 of them are spayed or neutered

Here, $k = 19$, $n = 30$, and $p = 0.60$.

$P(X = 19) = \binom{30}{19} (0.60)^{19} (0.40)^{30-19}$

Using a binomial probability calculator or formula, we get:

$P(X = 19) \approx 0.1303$

b. At most 20 of them are spayed or neutered

This means $X \leq 20$, so we sum the probabilities for all values from 0 to 20:

$P(X \leq 20) = \sum_{k=0}^{20} P(X = k)$

Using a binomial distribution table or calculator, we find:

$P(X \leq 20) \approx 0.8348$

c. At least 16 of them are spayed or neutered

This means $X \geq 16$, so we sum the probabilities for all values from 16 to 30:

$P(X \geq 16) = 1 - P(X \leq 15)$

Using a binomial distribution table or calculator, we find:

$P(X \geq 16) \approx 1 - 0.1915 = 0.8085$

d. Between 16 and 20 (including 16 and 20) of them are spayed or neutered

This means $16 \leq X \leq 20$, so we sum the probabilities for $X = 16, 17, 18, 19, 20$:

$P(16 \leq X \leq 20) = P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

Using a binomial distribution table or calculator, we find:

$P(16 \leq X \leq 20) \approx 0.7073$

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Let me know if you'd like any further details or explanations!

Here are some follow-up questions to consider:

1. What is the probability of exactly 15 dogs being spayed or neutered?
2. What is the expected number of dogs to be spayed or neutered in this group of 30 dogs?
3. How would the probabilities change if the probability of spaying/neutering was 70% instead of 60%?
4. What is the standard deviation of the number of spayed or neutered dogs in the sample of 30?
5. How does the binomial distribution differ from the normal distribution for large values of $n$?

Tip: When dealing with binomial distributions, if $n$ is large and $p$ is not too close to 0 or 1, you can approximate the binomial distribution using a normal distribution.