To solve this problem, we can use the binomial distribution, which is appropriate since the question involves a fixed number of trials (15 people) with two possible outcomes (owning a dog or not).

Part a:

We need to find the probability that exactly 7 out of 15 randomly selected people own a dog. The formula for the probability of exactly $k$ successes (in this case, owning a dog) in $n$ trials (in this case, selecting 15 people) is given by the binomial probability formula:

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

Where:

• $n = 15$ (the number of trials)
• $k = 7$ (the number of people owning a dog)
• $p = 0.36$ (the probability of a person owning a dog)
• $1 - p = 0.64$ (the probability of a person not owning a dog)

First, we calculate the binomial coefficient $\binom{n}{k}$:

$\binom{15}{7} = \frac{15!}{7!(15-7)!} = 6435$

Then we compute the probability:

$P(X = 7) = 6435 \times (0.36)^7 \times (0.64)^8$

Using a calculator, this gives us:

$P(X = 7) \approx 0.142$

Part b:

Now, we need to find the probability that 7 or fewer out of the 15 people own a dog. This is the cumulative probability for $X \leq 7$, which can be expressed as:

$P(X \leq 7) = P(X = 0) + P(X = 1) + \cdots + P(X = 7)$

To calculate this, we sum the binomial probabilities from $X = 0$ to $X = 7$. We can either use a binomial cumulative distribution table or a calculator that supports binomial cumulative distribution functions.

Using a binomial calculator or statistical software, we find:

$P(X \leq 7) \approx 0.746$

Thus, the probability that 7 or fewer people in a random sample of 15 own a dog is approximately 0.746.

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Summary:

• Part a: The probability that exactly 7 out of 15 randomly selected people own a dog is 0.142.
• Part b: The probability that 7 or fewer out of 15 randomly selected people own a dog is 0.746.

Would you like further clarification or a more detailed breakdown of any of these steps?

Related questions:

1. What is the probability of exactly 3 people owning a dog out of 15?
2. How would the probabilities change if the percentage of people who own a dog increased to 50%?
3. How do you calculate the mean and variance of a binomial distribution?
4. How can you use the binomial distribution to approximate a normal distribution when $n$ is large?
5. What would be the probability of more than 10 people owning a dog in a sample of 15?

Tip:

For large numbers of trials (such as $n > 30$), the binomial distribution can often be approximated by a normal distribution for easier calculations.