Not all visitors to a certain company's website are customers or potential customers. In fact, the company's executives estimate that about 8% of all visitors to the website are looking for *other* websites. Assume that this estimate is correct and that a random sample of 40 visitors to the website is taken. (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 random 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 at least three decimal places.

This problem involves a binomial distribution, where you are given the probability of success (a visitor looking for the company's website) and the number of trials (visitors sampled). Let's solve each part step by step.

### (a) Finding the Mean of the Relevant Distribution

The mean (or expectation) \( E(X) \) for a binomial distribution is given by the formula:
\[
E(X) = n \cdot p
\]
where:
- \( n = 40 \) is the number of trials (visitors sampled),
- \( p = 0.92 \) is the probability of success (since 8% are looking for other websites, 92% are looking for the company's website).

Now, calculate the mean:
\[
E(X) = 40 \cdot 0.92 = 36.8
\]
So, the mean number of visitors in the sample who are actually looking for the company's website is **36.8**.

### (b) Finding the Standard Deviation of the Distribution

The standard deviation \( \sigma \) for a binomial distribution is calculated using the formula:
\[
\sigma = \sqrt{n \cdot p \cdot (1 - p)}
\]
where:
- \( n = 40 \) is the number of trials,
- \( p = 0.92 \) is the probability of success,
- \( 1 - p = 0.08 \) is the probability of failure (visitors looking for other websites).

Now, calculate the standard deviation:
\[
\sigma = \sqrt{40 \cdot 0.92 \cdot 0.08} = \sqrt{40 \cdot 0.0736} = \sqrt{2.944} \approx 1.715
\]
So, the standard deviation of the number of visitors who are actually looking for the company's website is approximately **1.715**.

### Summary
- (a) The mean number of visitors looking for the company's website: **36.8**.
- (b) The standard deviation of the distribution: **1.715**.

Would you like to explore any further aspects of this problem, or do you have any additional questions?

### Related Questions:
1. What is the probability that exactly 36 visitors are looking for the company's website?
2. How would the mean change if the sample size increased to 100 visitors?
3. What is the significance of the standard deviation in this context?
4. Can you calculate the probability that more than 38 visitors are looking for the company’s website?
5. How does changing the probability \( p \) affect the mean and standard deviation?
6. What is the probability that at most 35 visitors are looking for the company’s website?
7. How would the variance of the distribution be calculated from the standard deviation?
8. How do binomial distributions approximate to normal distributions with larger sample sizes?

**Tip:** In binomial distributions, as the sample size increases, the distribution becomes more symmetric and can often be approximated by a normal distribution, which simplifies calculations for large samples.

Learn how to estimate the number of visitors interested in a company's website using probability and the binomial distribution. Calculate the mean and standard deviation based on a given probability of interest. Suitable for high school students studying probability.

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