Question

not all visitors to a certain companys website are customers or potential customers. in fact, the companys executives estimate that about 12% 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.

(if necessary, consult a list of formulas.)

(a) estimate the number of visitors in the sample who actually are looking for the companys 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.

Explanation:

Identify the distribution parameters

We model the number of visitors looking for other websites as a binomial random variable \(X \sim \text{Binomial}(n, p)\).
The sample size is \(n = 50\).
The probability that a visitor is looking for other websites is \(p = 12\% = 0.12\).
Therefore, the probability that a visitor is actually looking for the company's website is:

$$q = 1 - p = 1 - 0.12 = 0.88$$

Let \(Y\) be the number of visitors looking for the company's website, where \(Y \sim \text{Binomial}(n, q)\) with \(n = 50\) and \(q = 0.88\).

Calculate the expected value

Using the Binomial Distribution and Sample Mean concepts, the expectation of \(Y\) is:

$$E[Y] = n \cdot q$$

$$E[Y] = 50 \cdot 0.88 = 44$$

Calculate the standard deviation

Using the Sample Standard Deviation concept, the standard deviation of a binomial distribution is:

$$\sigma_Y = \sqrt{n \cdot q \cdot (1 - q)}$$

$$\sigma_Y = \sqrt{50 \cdot 0.88 \cdot 0.12}$$

$$\sigma_Y = \sqrt{5.28} \approx 2.297825$$

Rounding to at least three decimal places gives \(2.298\).

Answer:

Question (a)

<blank>44</blank>

Question (b)

<blank>2.298</blank>