Question

Consider the midterm and final for a statistics class. Suppose 13% of students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final. You randomly pick up a final exam and notice the student received an A. What is the probability that this student earned an A on the midterm?The end-goal is to find P(midterm = A|final = A).

Answer 1

Answer:

There is a 38.97% probability that this student earned an A on the midterm.

Step-by-step explanation:

The first step is that we have to find the percentage of students who got an A on the final exam.

Suppose 13% students earned an A on the midterm. Of those students who earned an A on the midterm, 47% received an A on the final, and 11% of the students who earned lower than an A on the midterm received an A on the final.

This means that

Of the 13% of students who earned an A on the midterm, 47% received an A on the final. Also, of the 87% who did not earn an A on the midterm, 11% received an A on the final.

So, the percentage of students who got an A on the final exam is

$P_{A} = 0.13(0.47) + 0.87(0.11) = 0.1568$

To find the probability that this student earned an A on the final test also earned on the midterm, we divide the percentage of students who got an A on both tests by the percentage of students who got an A on the final test.

The percentage of students who got an A on both tests is:

$P_{AA} = 0.13(0.47) = 0.0611$

The probability that the student also earned an A on the midterm is

$P = \frac{P_{AA}}{P_{A}} = \frac{0.0611}{0.1568} = 0.3897$

There is a 38.97% probability that this student earned an A on the midterm.