11 HW According to an online survey by Harris Interactive for job site CareerBui
ID: 3274578 • Letter: 1
Question
11 HW
According to an online survey by Harris Interactive for job site CareerBuilder.com, more than half of IT (information technology) workers say they have fallen asleep at work (InformationWeek, September 27, 2007). Sixty-four percent of government workers admitted to falling asleep on the job. Consider the following contingency table that is representative of the survey results.
Convert the contingency table into a joint probability table. (Round your intermediate calculations and final answers to 4 decimal places.)
Job Category
What is the probability that a randomly selected worker is an IT professional? (Round your intermediate calculations and final answers to 4 decimal places.)
What is the probability that a randomly selected worker slept on the job? (Round your intermediate calculations and final answers to 4 decimal places.)
If a randomly selected worker slept on the job, what is the probability that he/she is an IT professional? (Round your intermediate calculations and final answers to 4 decimal places.)
If a randomly selected worker is a government professional, what is the probability that he/she slept on the job? (Round your intermediate calculations and final answers to 4 decimal places.)
According to an online survey by Harris Interactive for job site CareerBuilder.com, more than half of IT (information technology) workers say they have fallen asleep at work (InformationWeek, September 27, 2007). Sixty-four percent of government workers admitted to falling asleep on the job. Consider the following contingency table that is representative of the survey results.
Explanation / Answer
Total frequency = 155 + 256 + 144 + 145 = 700
a) Now the joint probability distribution could be obtained as:
b) Probability that a random selected person is from IT ( IT column total in the above table ) = 0.4285
Therefore 0.4285 is the required probability here.
c) Probability that the worker slept on job = 0.5871 ( first row total in the above table )
Therefore 0.5871 is the required probability here.
d) Given that the worker slept on job, probability that he is from IT is computed as:
= Probability that he is form IT and slept on job / Probability that he slept on job
= 0.2214 / 0.5871
= 0.3771
Therefore 0.3771 is the required probability here.
e) Given that a person is a government official, probability that he slept on job is computed as:
= Probability that is is government official and he slept on job / Probability that he is government official
= 0.3657 / 0.5714
= 0.6400
Therefore 0.6400 is the required probability here.
f) P( IT ) P( Yes ) = 0.4285*0.5871 = 0.2516
P(IT and Yes ) = 0.2214
Therefore the 2 events are not independent as P(IT)P(Yes) is not equal to P(IT and Yes ) which also means that P(IT | Yes ) is not equal to P(IT)
Therefore c) is the correct answer here.
IT G Total Yes 155/700 = 0.2214 256/700 = 0.3657 0.5871 No 145/700 = 0.2071 144/700 = 0.2057 0.4128 Total 0.4285 0.5714 1Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.