Business Statistics Chapter 3

Topics: Event, Bond, Elementary event Pages: 4 (952 words) Published: March 18, 2012
5.1 A credit card customer at Border’s can use Visa (V), MasterCard (M), or American Express (A). The merchandise may be books (B), electronic media (E), or other (O). (a) Enumerate the elementary events in the sample space describing a customer’s purchase. (b) Would each elementary event be equally likely? Explain

A. S= ( V,B), (V,E), (V,O), (M,B), (M,E), (M,O), (A,B), (A,E), (A,O) B. Events are not likely. The quantities are not the same.

5.13 Are these characteristics of a student at your university mutually exclusive or not? Explain. A. A=works 20 hours or more, B= majoring in accounting
-not mutually exclusive
B. A= born in United States, B= born in Canada
-mutually exclusive
C. A= owns a Toyota, B= owns a Honda
-Not mutually exclusive

5.21 Let S be the event that a randomly chosen female aged 18-24 is a smoker. Let C be the event that a randomly chosen female aged 18-24 is Caucasian. Given P(S)=.246, P(C) =.830 and P(SC)= .232, find each probability and express the event in words. (Data are from Statistical Abstract of the United States, 2001.)

A. P(S).P(S)=1-.246. There is a 75.4% chance that a female aged 18-24 is a nonsmoker B. P(SC).-P(SC)=.246 + .830-.232=.844. There is an 84.4% chance that a female aged 18-24 is a smoker or is Caucasian C. P(SC).-P(SC)=.232/.830=.2795. Given that the female aged 18-24 is a Caucasian, there is a 27.95% chance that she smokes D. P(SC‘).-P(SC‘)=P(S)-P(SC)=.0824. Given that the female aged 18-24 is not Caucasian, there is an 8.42% chance that she smokes

5.23 Given P(A)= .40, P(B)=.50 and P(AB)=.05. (a). Find (AB). (b) In this problem, are A and B independent? Explain (a) P(AB)= P(AB)/P(B)=.05/.50=.10
(b) No because the multiplication of both is not equal to .5


| D| P| A| SUM|
N| 13| 6| 6| 25|
O| 58| 30| 21| 109|
R| 8| 7| 7| 22|
SUM| 79| 43| 34| 156|

a. P(D)=.5064
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