Questions: 80% of owned dogs in the United States are spayed or neutered. Round your answers to four decimal places. If 39 owned dogs are randomly selected, find the probability that a. Exactly 30 of them are spayed or neutered. b. At most 32 of them are spayed or neutered. c. At least 33 of them are spayed or neutered. d. Between 30 and 35 (including 30 and 35) of them are spayed or neutered.

Solution Steps

To find the probability that exactly 30 out of 39 owned dogs are spayed or neutered, we use the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

where:

• \( n = 39 \) (number of trials),
• \( x = 30 \) (number of successes),
• \( p = 0.8 \) (probability of success),
• \( q = 1 - p = 0.2 \) (probability of failure).

Calculating this gives:

\[ P(X = 30) = \binom{39}{30} \cdot (0.8)^{30} \cdot (0.2)^{9} \approx 0.1343 \]

To find the probability that at most 32 dogs are spayed or neutered, we sum the probabilities from 0 to 32:

\[ P(X \leq 32) = \sum_{x=0}^{32} P(X = x) \]

Calculating this, we find:

\[ P(X \leq 32) \approx 0.0 \]

To find the probability that at least 33 dogs are spayed or neutered, we can use the complement rule:

\[ P(X \geq 33) = 1 - P(X \leq 32) \]

Since we found \( P(X \leq 32) \approx 0.0 \), we have:

\[ P(X \geq 33) \approx 1 - 0.0 = 1.0 \]

Final Answer