Applied Statistics in Business and Economics
Applied Statistics in Business and Economics
5th Edition
ISBN: 9781259329050
Author: DOANE
Publisher: MCG
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Chapter 5.3, Problem 21SE

a.

To determine

Find the probability P(S).

a.

Expert Solution
Check Mark

Answer to Problem 21SE

The probability P(S) is 0.754.

Explanation of Solution

Calculation:

The given information is that the event S represent the randomly chosen female aged 18-24 is a smoker and the event C represents that the randomly chosen female aged 18-24 is Caucasian. Also, P(S)=0.246, P(C)=0.830 and P(SC)=0.232.

The formula for finding probability P(S) is,

P(S)=1P(S)

Substitute P(S)=0.246

P(S)=10.246=0.754

Thus, the probability P(S) is 0.754.

Hence, it can be concluded that there is a 75.4% (=0.754×100%) chance that the female aged 18-24 is a non-smoker.

b.

To determine

Find the probability P(SC).

b.

Expert Solution
Check Mark

Answer to Problem 21SE

The probability P(SC) is 0.844.

Explanation of Solution

Calculation:

The formula for finding probability P(SC) is,

P(SC)=P(S)+P(C)P(SC)

Substitute P(S)=0.246, P(C)=0.830 and P(SC)=0.232

P(SC)=0.246+0.8300.232=0.844

Thus, the probability P(SC) is 0.844.

Hence, it can be concluded that there is an 84.4% (=0.844×100%) chance that the female aged 18-24 is a smoker or Caucasian.

c.

To determine

Find the probability P(S|C).

c.

Expert Solution
Check Mark

Answer to Problem 21SE

The probability P(S|C) is 0.2795.

Explanation of Solution

Calculation:

The formula for finding probability P(S|C) is,

P(S|C)=P(SC)P(C)

Substitute P(C)=0.830 and P(SC)=0.232

P(S|C)=0.2320.830=0.2795

Thus, the probability P(S|C) is 0.2795.

Hence, it can be concluded that there is an 27.95% (=0.2795×100%) chance that the female aged 18-24 is smoker given that female aged 18-24 is a Caucasian.

d.

To determine

Find the probability P(S|C).

d.

Expert Solution
Check Mark

Answer to Problem 21SE

The probability P(S|C) is 0.0824.

Explanation of Solution

Calculation:

The formula for finding probability P(S|C) is,

P(S|C)=P(SC)P(C)=P(S)P(SC)1P(C)          P(SC)+P(SC)=P(S)

Substitute P(S)=0.246, P(C)=0.830 and P(SC)=0.232

P(S|C)=0.2460.23210.830=0.0140.17=0.0824

Thus, the probability P(S|C) is 0.0824.

Hence, it can be concluded that there is an 8.24% (=0.0824×100%) chance that the female aged 18-24 is smoker given that female aged 18-24 is a not Caucasian.

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Chapter 5 Solutions

Applied Statistics in Business and Economics

Ch. 5.2 - Prob. 11SECh. 5.2 - Prob. 12SECh. 5.3 - Prob. 13SECh. 5.3 - Are these events collectively exhaustive or not?...Ch. 5.3 - Prob. 15SECh. 5.3 - Given P(A) = .70, P(B) = .30, and P(A B) = .00,...Ch. 5.3 - Prob. 17SECh. 5.3 - Prob. 18SECh. 5.3 - Prob. 19SECh. 5.3 - Prob. 20SECh. 5.3 - Prob. 21SECh. 5.3 - Prob. 22SECh. 5.4 - Given P(J) = .26, P(K) = .48. If A and B are...Ch. 5.4 - Given P(A) = .40, P(B) = .50. If A and B are...Ch. 5.4 - Prob. 25SECh. 5.4 - Prob. 26SECh. 5.4 - Prob. 27SECh. 5.4 - Prob. 28SECh. 5.4 - The probability that a student has a Visa card...Ch. 5.4 - Bob sets two alarm clocks (battery-powered) to be...Ch. 5.4 - A hospitals backup power system has three...Ch. 5.4 - Over 1,000 people try to climb Mt. Everest every...Ch. 5.4 - Suppose 50 percent of the customers at Pizza...Ch. 5.5 - Prob. 34SECh. 5.5 - The contingency table below summarizes a survey of...Ch. 5.5 - Prob. 36SECh. 5.5 - A survey of 156 introductory statistics students...Ch. 5.5 - This contingency table describes 200 business...Ch. 5.5 - Based on the previous problem, is major...Ch. 5.5 - The following contingency table shows average...Ch. 5.6 - Of grocery shoppers who have a shopping cart, 70...Ch. 5.6 - A study showed that 60 percent of The Wall Street...Ch. 5.7 - A drug test for athletes has a 5 percent false...Ch. 5.7 - Half of a set of the parts are manufactured by...Ch. 5.7 - An airport gamma ray luggage scanner coupled with...Ch. 5.8 - (a) Find 8! without a calculator. Show your work....Ch. 5.8 - Prob. 47SECh. 5.8 - In the Minnesota Northstar Cash Drawing, you pick...Ch. 5.8 - Prob. 49SECh. 5.8 - Prob. 50SECh. 5.8 - Until 2005, the UPC bar code had 12 digits (09)....Ch. 5.8 - Bob has to study for four final exams: accounting...Ch. 5.8 - Prob. 53SECh. 5.8 - Prob. 54SECh. 5.8 - Prob. 55SECh. 5.8 - Prob. 56SECh. 5 - Prob. 1CRCh. 5 - Prob. 2CRCh. 5 - Prob. 3CRCh. 5 - Prob. 4CRCh. 5 - Prob. 5CRCh. 5 - Prob. 6CRCh. 5 - Prob. 7CRCh. 5 - In a contingency table, explain the concepts of...Ch. 5 - Prob. 9CRCh. 5 - Prob. 10CRCh. 5 - Prob. 11CRCh. 5 - Prob. 60CECh. 5 - Prob. 61CECh. 5 - A judge concludes that there is a 20 percent...Ch. 5 - Prob. 63CECh. 5 - Prob. 64CECh. 5 - In the first year after its release, 83 percent of...Ch. 5 - Prob. 66CECh. 5 - Prob. 67CECh. 5 - If Punxsutawney Phil sees his shadow on February...Ch. 5 - Prob. 69CECh. 5 - Bob owns two stocks. There is an 80 percent...Ch. 5 - Prob. 71CECh. 5 - A study showed that trained police officers can...Ch. 5 - The probability that a 2011 Audi A8 will be stolen...Ch. 5 - The probability of being struck by lightning is...Ch. 5 - Prob. 75CECh. 5 - A certain model of remote-control Stanley garage...Ch. 5 - (a) In a certain state, license plates consist of...Ch. 5 - Prob. 78CECh. 5 - Prob. 79CECh. 5 - Prob. 80CECh. 5 - Prob. 81CECh. 5 - A certain airplane has two independent alternators...Ch. 5 - Prob. 83CECh. 5 - Prob. 84CECh. 5 - A turboprop aircraft has two attitude gyroscopes,...Ch. 5 - Which are likely to be independent events? For...Ch. 5 - In child-custody cases, about 70 percent of the...Ch. 5 - A web server hosting company advertises 99.999...Ch. 5 - Prob. 89CECh. 5 - The probability is 1 in 4,000,000 that a single...Ch. 5 - Prob. 91CECh. 5 - Prob. 92CECh. 5 - Four students divided the task of surveying the...Ch. 5 - Refer to the contingency table shown below. (a)...Ch. 5 - Prob. 95CECh. 5 - High levels of cockpit noise in an aircraft can...Ch. 5 - Prob. 97CECh. 5 - A biometric security device using fingerprints...Ch. 5 - Dolon Web Security Consultants requires all job...
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