A professor has noticed that, even though attendance is not a component of the f

Question

A professor has noticed that, even though attendance is not a component of the final grade for the class, students that attend regularly generally get better grades. In fact, 48% of those who come to class on a regular basis receive A's. Only 4% who do not attend regularly get A's. Overall, 60% of students attend regularly. Based on this class profile, suppose we are randomly selecting a single student from this class, and answer the questions below.Hint #1: pretend that there are 1000 students in the class and use the values given in the problem to construct the appropriate contingency table. Round cell frequencies to the nearest integer Hint #2: No joke, you really need to use hint #1. Hint #3: The first step to using hint #1 is to calculate the totals for those who attend regularly and do not attend regularly. A) P(receives A's | attends regularly) = Use 2-decimal precision in your answer B) P(receives A's | does not attend regularly) = Use 2-decimal precision in your answer C) P(receives A's) = Use 3-decimal precision in your answer D) P(attends regularly | receives A's) Use 3-decimal precision in your answer E) P(does not attend regularly | does not receive A's) =

Explanation / Answer

here from above:

A) P(receives A's | attends regularly) =0.48

B) P(receives A's | does not attend regularly) = 0.04

c)P(receives A's) = P( attend regularly)*P(receives A's | attends regularly)+P(does not attend regularly)*P(receives A's | does not attend regularly) =0.6*0.48+(1-0.6)*0.04=0.304

d) P(attends regularly | receives A's) =P( attend regularly)*P(receives A's | attends regularly)/P(receives A's)

=0.6*0.48/0.304 =0.947

e)P (does not receive A's) =1-0.304 =0.696

P(does not attend regularly | does not receive A's)

=P(does not attend regularly)*P(does not receive A's|does not attend regularly)/P(does not receive A's)

=0.4*(1-0.04)/0.696 =0.552

please revert for any clarification required

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