Question

Suppose that there were 226977 visitors to your website over a period of a week, and of those 56330 clicked on the button to get more information about your product. If you wanted to model the probability distribution of the number of visitors who clicked the button, you would typically estimate the mean of the distribution to be equal to the observed number of visitors who clicked, that is the sample mean: 56330.

Required:
Under this setup, what is the variance of this distribution?

Answer 1

The variance of this distribution is 42353.2

Explanation:

For each visitor, there are only two possible outcomes. Either they clicked on the button, or they did not. Visitors are independent of each other. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The variance of the binomial distribution is:

`V(X) = np(1-p)`

In this question:

226977 visitors, so
`n = 226977`

56330 clicked, so
`p = (56330)/(226977) = 0.2482`

Then

`V(X) = 226977*0.2482*(1 - 0.2482) = 42353.2`

The variance of this distribution is 42353.2