According to the American Veterinary Medical Association (AVMA),  of households in the United States own a dog as a pet 33.9%. Suppose that a company that sells dog food would like to establish a focus group to gather input on a new dog food marketing campaign. The company plans to contact 25 randomly selected households to invite people to join the focus group.
a. Compute the probability that 10 of these 25 households own a dog as a pet. (to 4 decimals)

b. Compute the probability that 2 or fewer of these 25 households own a dog as a pet. (to 5 decimals)

c. For the sample of 25 households, compute the expected number of households who own a dog as a pet. (to 1 decimal)

d. For the sample of 25 households, compute standard deviation of households who own a dog as a pet. (to 4 decimals)

This problem can be solved using the **binomial distribution** since it involves determining the probability of a certain number of successes (households owning a dog) in a fixed number of independent trials (households sampled), where each trial has two possible outcomes: either the household owns a dog, or it does not.

The probability mass function (PMF) of a binomial distribution is given by:

\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n-k}
\]

Where:
- \( n \) = 25 (number of trials, i.e., the households contacted)
- \( k \) = the number of successes (households owning a dog)
- \( p = 0.339 \) (probability of success, i.e., a household owning a dog)
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) is the binomial coefficient.

### a. Compute the probability that 10 of these 25 households own a dog as a pet.
We need to compute \( P(X = 10) \).

\[
P(X = 10) = \binom{25}{10} \cdot (0.339)^{10} \cdot (1 - 0.339)^{25 - 10}
\]

### b. Compute the probability that 2 or fewer of these 25 households own a dog as a pet.
This is the cumulative probability \( P(X \leq 2) \), which is the sum of the probabilities for \( X = 0 \), \( X = 1 \), and \( X = 2 \).

\[
P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)
\]

### c. Compute the expected number of households who own a dog as a pet.
The expected value \( E(X) \) for a binomial distribution is given by:

\[
E(X) = n \cdot p
\]

### d. Compute the standard deviation of the number of households who own a dog as a pet.
The standard deviation \( \sigma \) for a binomial distribution is:

\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]

Let's calculate each part.Here are the results:

### a. The probability that 10 of the 25 households own a dog as a pet is:
\[
P(X = 10) = 0.1316
\]

### b. The probability that 2 or fewer of the 25 households own a dog as a pet is:
\[
P(X \leq 2) = 0.0030
\]

### c. The expected number of households who own a dog as a pet is:
\[
E(X) = 8.5
\]

### d. The standard deviation of the number of households who own a dog as a pet is:
\[
\sigma = 2.3668
\]

If you would like further details or have any questions, feel free to ask!

Here are some related questions to consider:
1. How does changing the sample size affect the expected value and standard deviation?
2. What happens to the probabilities if the proportion of dog-owning households changes?
3. How would the results change if you sampled 50 households instead of 25?
4. Can you use a normal approximation for this binomial distribution?
5. What is the probability that more than 15 households own a dog?

**Tip:** When dealing with binomial distributions, if \(n\) is large and \(p\) is not too close to 0 or 1, the normal approximation can often be used for easier computation.

According to the American Veterinary Medical Association (AVMA), 33.9% of households in the United States own a dog as a pet. Suppose that a company that sells dog food would like to establish a focus group to gather input on a new dog food marketing campaign. The company plans to contact 25 randomly selected households to invite people to join the focus group. a. Compute the probability that 10 of these 25 households own a dog as a pet. b. Compute the probability that 2 or fewer of these 25 households own a dog as a pet. c. For the sample of 25 households, compute the expected number of households who own a dog as a pet. d. For the sample of 25 households, compute standard deviation of households who own a dog as a pet.

Explore the binomial distribution to calculate the probability of dog ownership in a random sample of 25 households. Learn how to compute the probability for exactly 10 dog-owning households, the probability for 2 or fewer, the expected number, and standard deviation. Suitable for students in Grades 10-12.

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