Question

4. 165/3 points 1/3 Submissions Used My Notes Ask Your T A professor has noticed that students that attend class regularty, miss no more than two classes per term, generally get better grades. For the class, the overall percent of students who attend regularly is 6。6. Of those who come to class on a regular basis, 45% receive A's. Of those who don't attend regularly, only 6% get A's. Draw a tree diagram like the one in the image, where R "attends class regulary , R does not attend class regularly, A earned an A", A did not eam an A (a) Based on your tree diagram fill in the appropriate matching probabilities, enter your answer as a proportion with three decimal places А.orP(R)-|66 B. or P(R)34 С.orRAIR) F.or P(AIR)-96 (bjAmong aW students what proportion earn an A and don't attend class regularly?34 (c)what is the chance a randomly chosen student will earn an A in the class? 331 (d)Given a student earned an A what is the chance they attend class regularly? 897 x Hint Use the total law of probabilly. Mint: PRIA Show My Work (optiona

Answer 1

a) P(R) = 0.66

P(R') = 0.34

P(A|R) = 0.45

P(A'|R) = 0.55

P(A|R') = 0.06

P(A'|R') = 0.94

b) Proportion of students who earn an A and dont attend class regulary = 0.06*0.34 = 0.0204

c) Proportion of students who will earn an A in the class, P(A) = 0.06*0.34 + 0.45*0.66

= 0.0204 + 0.297 = 0.3174

d) Proportion of students who earned an A, given that they attend class regularly =

$P(R|A) = \frac{P(A|R)*P(R)}{P(A)} = \frac{0.45*0.66}{0.3174}$

= 0.9357