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Krissy wanted to understand whether grade level had any relationship to their opinion on extending the school day. She surveyed some students and displayed the results in the table below:


In favor Opposed Undecided
Grade 9 6 2 7
Grade 10 5 11 8
Grade 11 12 15 11
Grade 12 17 5 13


Compare P(Grade 11 | opposed) with P(opposed | Grade 11).

Respuesta :

Answer:

c

Step-by-step explanation:

Answer:

P(Grade 11 | opposed) =0.4545

P(opposed | Grade 11)=0.3947

Step-by-step explanation:

The table displaying the data is given below as:

                  In favor Opposed Undecided     Total

Grade 9          6                    2                  7                  15

Grade 10         5                   11                  8                  24

Grade 11          12                 15                  11                 38

Grade 12         17                  5                   13                 35

Total             40                 33                 39              112

Now we are asked to compare the conditional probability i.e. we are asked to compare  P(Grade 11 | opposed) with P(opposed | Grade 11).

let A denote the event that the student is in grade 11.

and B denote the event of opposing the decision.

Then A∩B denote the event of grade 11 students who opposed.

Hence, we are asked to compare:

P(A|B) and P(B|A)

We know that:

[tex]P(A|B)=\dfrac{P(A\bigcap B)}{P(B)}\\\\and\\\\P(B|A)=\dfrac{P(A\bigcap B)}{P(A)}[/tex]

Now from the table we have:

P(A)=38/112

P(B)=33/112

P(A∩B)=15/112

Hence,

[tex]P(A|B)=\dfrac{\dfrac{15}{112}}{\dfrac{33}{112}}\\\\\\P(A|B)=\dfrac{15}{33}=0.4545[/tex]

Similarly:

[tex]P(B|A)=\dfrac{\dfrac{15}{112}}{\dfrac{38}{112}}\\\\\\P(B\A)=\dfrac{15}{38}=0.3947[/tex]

Hence,

P(Grade 11 | opposed) =0.4545

P(opposed | Grade 11)=0.3947

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