# Retailing and Large Online Retailers Essay

Problem set 3
Crystal Yu

1. Do problem 4.8 and 4.9 page 149 (2 points) You need to show your work to get credit.

4.8
Does it take more time to be removed from an email list than it used to take? A study of 100 large online retailers revealed the following:

Need Three Or More Clicks to Be Removed

Year Yes No Total
2009 39 61 100
2008 7 93 100
46 154 200

a. Give an example of a simple event
Need less than 3 more clicks to be removed.
b. Give an example of a joint event
Need three or more clicks to be removed in 2008
(2 events : needs 3 or more clicks and happened in 2008)
c. What is the complement of “Needs three or more clicks to be removed from an email list”? Need less than 3 more clicks.
d.Why is “Needs three or more clicks to be removed from an email list in 2009” a joint event? It neeeds to fit both criteria, 3 or more clicks and occur in the year 2009

4.9

Referring to the contingency table in Problem 4.8, if a large online retailer is selected at random, what is the probability that

a. you needed three or more clicks to be removed from an email list? P

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= 46/200 = 23%
b. you needed three or more clicks o be removed from an email list in 2009? P = 39/200 = 19.5%
c. you needed three or more clicks to be removed from an email list or were a large online retailer surveyed in 2009?

Let say event A = “needed three or more clicks to be removed from an email list” –> P(A) = 23% Let say event B = “were a large online retailer surveyed in 2009” —> P(B) = 100/200 = 50% So P(A or B) = P(A) + P(B) – P(A and B) = 23% + 50% – 19.5% = 53.5%

d. Explain the difference in the results in (b) and (c).

2. Do problem 4.21 page 157 (2 points) You need to show your work to get credit. Table
Year Yes No Total
2009 39 61 100
2008 7 93 100
46 154 200

a. Given that three or more clicks are needed to be removed from an email list, what is the probability that this occurred in 2009? There were 46 retailers surveyed in two years that need three or more clicks to be removed, 39 of them happened in 2009 –> So P = 39/46 = 84.78%

b. Given that the year 2009 is involved, what is the probability that three or more clicks are needed to be removed from an email list? Of 100 retailers surveyed in 2009, 39 need there or more clicks to be removed —> So P = 39/100 = 39%

c. Explain the difference in the results in (a) and (b).
Results in (a) and (b) are different because the conditions are different. Condition of (a) means for those retailers that “needed three or more clicks to be removed” how many were surveyed in 2009. While condition of (b) means
the opposite way, for those surveyed in 2009, how many retailers need 3 or more clicks.

d. Are needing three or more clicks to be removed from an email list and the year independent? They are not independent. Refer to Equation 4.5 in textbook, p.154 shows that two events A and B are independent if and only if P(A?B) = P(A) Calculate the probabilities of the events :

P(need three or more clicks to be removed) = 46/200 = 23%
P(need three or more clicks to be removed given condition that occurred in 2009) = 39%

Chapter 6 Individual work

Using a Normal Distribution to find probabilities: You may use the tables in your text or PhStat – Probability Distributions – Normal. Regardless of which method you use, either draw a sketch and indicate what probabilities (formulas such as P(X10 for example) for what is required for each part of the problem. To make sketches that you can copy and paste into Word for the areas under the bell-curve you have several tools to offer: (1) you can use the NormalDistribution powerpoint slide attached to this assignment, (2) the NormalDistributionTool.xls Excel tool, (3) do it yourself in Excel by following the instructions at http://www.youtube.com/watch?v=FvY2LwacSg0, (4) draw it by hand then scan it or take a picture and paste it into your file, (4) Use an online graphing tools such as

3. Do problem 6.9 Page 214 (2 points) Remember, you need sketches to get credit.

Mean: \$21
Standard deviation: \$5
a.Find probability that a randomly selected person will spend more than \$25 Z
= (\$25 – \$21)/\$5 = 0.8
Table E.2 shows probability of Z = 0.8 is 0.7881 meaning 78.81% a selected person will spend \$25 or less. So the probability that a randomly selected person will spend more than \$25 is 21.19% (100% – 78.81% = 21.19%)

b. Find probability that a randomly selected person will spend between \$10 and \$20.
Z(\$10) = (\$10 – \$21)/\$5 = – 2.2 ? P(X

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