As a marketing manager, you are tasked with selecting a website to place your advertisement. The following sampled data shows the number of user visits per month over the last 3 three years: Website 1: 10357, 10537, 10767, 10561, 10544, 10581, 10602, 10665, 10335, 10419, 10737, 10410, 10485, 10601, 10458, 10472, 10435, 10375, 10436, 10510, 10345, 10559, 10520, 10425, 10351, 10465, 10491, 10671, 10366, 10440, 10618, 10606, 10406, 10538, 10449, 10462
Website 2: 11067, 11029, 10888, 10789, 10914, 10663, 10787, 11140, 11042, 11074, 10868, 10853, 10900, 11088, 10991, 10928, 10959, 11126, 11033, 11114, 11150, 11155, 11027, 10900, 11015, 11123, 10953, 11181, 10855, 10731, 10971, 10770, 11070, 11122, 11018, 10903 Since the behavior of internet users can be considered a natural process, consider the number of views to follow normal distribution. In addition, please assume no autocorrelation or time-series nature of the data. Based on the data above, provide the answers to the following question:
A. What is the average and standard deviation of viewership of each website?
B. Is viewership different between these two websites? If yes, which website provides more views?
C. Suppose that your manager requires at least 12000 views per month. What is the probability of 12000 views happening on each website?
D. Which website provides more consistent view? How would you measure it?
E. Which website would you recommend for your advertisement?

Kindly check explanation

Step-by-step explanation:

Given the data :

Website 1 : 10357, 10537, 10767, 10561, 10544, 10581, 10602, 10665, 10335, 10419, 10737, 10410, 10485, 10601, 10458, 10472, 10435, 10375, 10436, 10510, 10345, 10559, 10520, 10425, 10351, 10465, 10491, 10671, 10366, 10440, 10618, 10606, 10406, 10538, 10449, 10462

Mean, xbar = ΣX/ n ; n = sample size = 36

Xbar = 377999 / 36 = 10499.9722

Standard deviation, s = √[(x - xbar)² / (n-1]

Using calculator :

Standard deviation (Website 1 :), s = 110.239865

Website 2 : 11067, 11029, 10888, 10789, 10914, 10663, 10787, 11140, 11042, 11074, 10868, 10853, 10900, 11088, 10991, 10928, 10959, 11126, 11033, 11114, 11150, 11155, 11027, 10900, 11015, 11123, 10953, 11181, 10855, 10731, 10971, 10770, 11070, 11122, 11018, 10903

Mean, xbar = ΣX/ n ; n = sample size = 36

Xbar = 395197 / 36 = 10977.6944

Standard deviation, s = √[(x - xbar)² / (n-1]

Using calculator :

Standard deviation (Website 2), s = 132.617995

2.)

Yes, the viewership between the two websites are different with the second website has a higher mean viewership with a mean of 10977.6944.

3.)

The probability of 12000 views per month on each website :

Probability = Mean viewership per month / required viewership

Website 1 :

P(12000) = 10499.9722 / 12000 = 0.8749

Website 2 :

P(12000) = 10977.6944 / 12000 = 0.9148

4.)

More consistent website :

We use the standard deviation value, the higher the standard deviation, the higher the variability :

Website 1 should be more consistent has it has a Lower standard deviation score, hence, should show lower variability than website 2.

5.)

Website suitable for advertisement should be one with higher viewership per month in other to reach a larger audience. Hence, website 2 should be recommended for advertisement.

the answer is C,  <NPQ