Respuesta :
Answer:
Follows are the solution to this question:
Step-by-step explanation:
[tex]Frequency\ total = 155 + 256 + 144 + 145 = 700;[/tex]
In point (1):
Now that it is possible to obtain the joint distribution as:
[tex]IT\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ G \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Total[/tex]
[tex]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[/tex]
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:
[tex]=\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}.[/tex]
In point (5):
In an individual is also an official government, its chance of getting sleep at work is measured when:
[tex]= \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}.[/tex]
In point (6):
[tex]\to P(IT) \ P(Yes) = 0.4285 \times 0.5871 = 0.2516 \\\\\to P(IT and Yes ) = 0.2214 \\\\[/tex]
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 "[tex]\text{No because P("IT " | "Yes")} \neq P("IT")[/tex]" is correct.