|
Question 548771: Binomial problems: Mean and standard deviation
Not all visitors to a certain company's website are customers or potential customers. In fact, the company's executives estimate that about 5% 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.
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.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Binomial problems: Mean and standard deviation
Not all visitors to a certain company's website are customers or potential customers. In fact, the company's executives estimate that about 5% 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.
----
n = 50 and p(other) = 0.05 ; p(right) = 0.95
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.
mean = np = 50*0.95 = 47.5
------------------------------------
b)Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.
sqrt[50*0.95*0.05] = 1.5411
=================================
Cheers,
Stan H.
=================================
|
|
|
| |
