Not all visitors to a certain company's website are customers or potential customers. In fact, the company's executives estimate that about 10% of all visitors to the website are looking for other websites. Assume that this estimate is correct and that a random sample of 50 visitors to the website is taken.

Answer the following.

(a) Estimate the number of visitors in the sample who actually are looking for the company's website by giving the mean of the relevant distribution (that is, the expectation of the relevant distribution variable). Do not round your response.

(b) Quantify the uncertainty of your estimate by giving the standard deviation of the distribution, Round your response

To solve this problem, we'll use the binomial distribution. The problem involves estimating the number of visitors in a sample who are actually looking for the company's website.

### Part (a): Estimate the Mean

The mean of a binomial distribution, which represents the expected number of successes (in this case, visitors who are looking for the company's website), is given by:

\[
\mu = n \times p
\]

Where:
- \(n = 50\) is the sample size,
- \(p = 1 - 0.10 = 0.90\) is the probability that a visitor is actually looking for the company's website.

Now, let's calculate the mean:

\[
\mu = 50 \times 0.90 = 45
\]

So, the expected number of visitors in the sample who are looking for the company's website is **45**.

### Part (b): Standard Deviation

The standard deviation of a binomial distribution is given by:

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

Where:
- \(n = 50\),
- \(p = 0.90\) (probability that a visitor is looking for the company's website),
- \(1 - p = 0.10\) (probability that a visitor is looking for another website).

Now, let's calculate the standard deviation:

\[
\sigma = \sqrt{50 \times 0.90 \times 0.10} = \sqrt{50 \times 0.09} = \sqrt{4.5} \approx 2.121
\]

So, the standard deviation of the distribution is approximately **2.121**.

### Summary of Answers:
- **(a)** The mean (expected number of visitors who are actually looking for the company's website) is **45**.
- **(b)** The standard deviation of the distribution is approximately **2.121**.

Would you like any further details or explanations?

### Related Questions:
1. What is the probability that exactly 45 visitors out of 50 are looking for the company's website?
2. How would the mean and standard deviation change if the sample size increased to 100?
3. What is the probability that fewer than 40 visitors are looking for the company's website?
4. How would the standard deviation change if the probability of looking for the website increased to 95%?
5. What assumptions are made when using the binomial distribution in this context?
6. How would you calculate the probability that more than 47 visitors are looking for the company's website?
7. What if the estimate of 10% was not accurate? How would you reassess the probabilities?
8. Can the normal approximation to the binomial distribution be used in this case?

### Tip:
When calculating probabilities for large samples using the binomial distribution, consider using the normal approximation for easier computation, especially when the conditions for the approximation are met.

Explore how to estimate visitor intent on a company's website using the binomial distribution. Calculate the mean and standard deviation for a sample of 50 visitors to determine the expected number of visitors who are interested in the company's website.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation