Question

According to an online survey by Harris Interactive for job site CareerBuilder, 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.
Slept on the Job IT Professional Government Professional
Yes 155 256
No 145 144

1. Convert the contingency table into a joint probability table. (Round the intermediate calculations and final answers to 4 decimal places.)
2. What is the probability that a randomly selected worker is an IT professional? (Round the intermediate calculations and final answers to 4 decimal places.)
3. What is the probability that a randomly selected worker slept on the job? (Round the intermediate calculations and final answers to 4 decimal places.)
4. If a randomly selected worker slept on the job, what is the probability that he/she is an IT professional? (Round the intermediate calculations and final answers to 4 decimal places.)
5. If a randomly selected worker is a government professional, what is the probability that he/she slept on the job? (Round the intermediate calculations and final answers to 4 decimal places.)
6. Are the events "IT Professional and Slept on the Job" independent? Explain.

Answer 1

Answer:

Follows are the solution to this question:

Step-by-step explanation:

$Frequency\ total = 155 + 256 + 144 + 145 = 700;$

In point (1):

Now that it is possible to obtain the joint distribution as:                                                                                                                              

                                 $IT\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Total$

$Yes \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{155}{700} = 0.2214\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{256}{700} = 0.3657 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.5871\\\\No \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{145}{700} = 0.2071\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{144}{700} = 0.2057\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.4128\\\\Total \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.4285 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.5714\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1$

In point (2):

Likely that now It (the IT column total throughout the table above) seems to be the random individual selected = 0.4285;

Here, 0.4285 is the necessary probability.

In point (3):

The way of its worker's sleep at work =0.5871 ( first row total in the above table )

Here 0.5871 was its required probability.

In point (4):

In this as employee slept at work, he is likely to also be from IT to be calculated with:

$=\frac{\text{probability he was just an IT-former but was asleep at work}}{ \text{possibly he worked at work}}\\\\=\frac{0.2214}{0.5871} \\\\ =0.3771 \text{was its probability required here}.$

In point (5):

In an individual is also an official government, its chance of getting sleep at work is measured when:

$= \frac{\text{Probability he is professional in business slept at work}}{\text{or that he is officially in business}} \\\\= \frac{0.3657}{0.5714}\\\\ = 0.6400 \text{is the required probability here}.$

In point (6):

$\to P(IT) \ P(Yes) = 0.4285 \times 0.5871 = 0.2516 \\\\\to P(IT and Yes ) = 0.2214$

Therefore, 2 cases aren't independent, because P(IT) P(Yes) is not the same as P(IT and Yes), which also implies that P(IT|Yes) is not the same as P (IT), therefore "$\text{No because P("IT " | "Yes")} \neq P("IT")$" is correct.