Question

A random sample of a 132 students were active they live on campus or off campus the following contingency table gives the 2 way classification of the response

Answer 1

From the given table, we know tha the grand total is 132, then,

`P(\text{Female }\cap Off\text{ campus)=}(24)/(132)=(12)/(66)=(4)/(22)=(2)/(11)`

Similarly,

`P(\text{Male }\cap On\text{ campus)=}(52)/(132)=(26)/(66)=(13)/(33)`

On the other hand, we have

`P(\text{offcampus}\cup Male)=P(offcampus)+P(male)-P(\text{offcampus}\cap Male)`

from the given table, it yields

`P(\text{offcampus}\cup Male)=(13+24)/(132)+(52+13)/(132)-(13)/(132)`

which gives

`\begin{gathered} P(\text{offcampus}\cup Male)=(13+24+52+13-13)/(132) \\ P(\text{offcampus}\cup Male)=(89)/(132) \end{gathered}`

And finally,

`P(on\text{campus}\cup female)=P(oncampus)+P(female)-P(\text{oncampus}\cap female)`

which gives

`\begin{gathered} P(on\text{campus}\cup female)=(52+43)/(132)+(43+24)/(132)-(43)/(132) \\ P(on\text{campus}\cup female)=(52+43+43+24-43)/(132) \\ P(on\text{campus}\cup female)=(119)/(132) \end{gathered}`

Therefore, the answers are:

`\begin{gathered} P(\text{Female }\cap Off\text{ campus)}=0.182 \\ P(\text{Male }\cap On\text{ campus)=}0.394 \\ P(\text{offcampus}\cup Male)=0.674 \\ P(on\text{campus}\cup female)=0.902 \end{gathered}`