Questions: In a certain area, 39% of people own a dog. Complete parts a and b below. a. Find the probability that exactly 6 out of 20 randomly selected people in the area own a dog. The probability that exactly 6 out of 20 randomly selected people in the area own a dog is (Type an integer or decimal rounded to three decimal places as needed.)

Solution Steps

To solve this problem, we can use the binomial probability formula. The scenario involves a fixed number of trials (20 people), each with two possible outcomes (owning a dog or not), and a constant probability of success (39% or 0.39). We need to find the probability of exactly 6 successes (people owning a dog) out of 20 trials.

1. Define the number of trials \( n = 20 \).
2. Define the probability of success \( p = 0.39 \).
3. Define the number of successes \( k = 6 \).
4. Use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(\binom{n}{k}\) is the binomial coefficient.

We are given a binomial distribution problem where:

• The number of trials \( n = 20 \).
• The probability of success (owning a dog) \( p = 0.39 \).
• The number of successes we are interested in \( k = 6 \).

The probability of exactly \( k \) successes in \( n \) trials is given by the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Substituting the given values: \[ P(X = 6) = \binom{20}{6} (0.39)^6 (1-0.39)^{20-6} \]

The binomial coefficient \(\binom{20}{6}\) is calculated as: \[ \binom{20}{6} = \frac{20!}{6!(20-6)!} = 38760 \]

Substitute the values into the formula: \[ P(X = 6) = 38760 \times (0.39)^6 \times (0.61)^{14} \] \[ P(X = 6) \approx 0.1347 \]

Final Answer