Question

The web publisher www.exploreiceland.is (Links to an external site.)Links to an external site. provides information on traveling to Iceland. Access to the website is free but revenues are generated by selling ads that are posted on the website. In the following month, the website has committed to displaying ads to 650,000 viewers, i.e., 650,000 impressions. Based on data from previous months the traffic to the website is estimated to be normally distributed with a mean of 850,000 viewers and a standard deviation of 150,000.

How many impressions should the web publisher have taken on, to be able to guarantee a 95% service level?

Answer 1

1096750 impressions

Explanation:

Given that :

Mean = 850,000

Standard deviation = 150,000

If we assume that X should be the numbers of impressions created;

Then ;

`X \approx N (\mu , \sigma^2)`

Now ; representing x as the value for the number of impression needed ; Then ;

`P(X>x) = 0.95`

`P((X- \mu)/(\sigma) > (x -850000)/(150000)) = 0.95`

`P(Z> ( x -850000)/(150000)) = 0.95`

From normal tables:

`P(Z >1.645) = 0.95`

`(x - 850000)/(150000) =1.645`

(x- 850000) = 1.645(150000)

x - 850000 = 246750

x = 246750 + 850000

x = 1096750 impressions