Question

Suppose a university is divided into three campuses: the Main campus, the Downtown campus, and the Bay Area campus. The Main campus has 60% of the university's students, the Downtown campus has 25%, and the Bay Area has the rest. At the Main campus, 12% of the students are statistics majors. At the Downtown campus, 4% of the students are statistics majors. Finally, at the Bay Area campus, 1% of the students are statistics majors.
Required:
a. What proportion of students at the university are statistics majors?
b. Of all statistics majors at the university, what proportion are at the Main campus?

Answer 1

Answer:

a) 0.0835 = 8.35% of students at the university are statistics majors

b) 0.8623 = 86.23% are at the Main campus

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

This is used for item b.

a. What proportion of students at the university are statistics majors?

12% of 60%(At the main campus).

4% of 25%(Downtown campus).

The rest is 100 - 60 + 25) = 15%.

1% of 15%(Bay Area Campus). Si

$p = 0.12*0.6 + 0.04*0.25 + 0.01*0.15 = 0.0835$

0.0835 = 8.35% of students at the university are statistics majors.

b. Of all statistics majors at the university, what proportion are at the Main campus?

Using conditional probability to find the percentage.

Event A: Statistics majors, so $P(A) = 0.0835$

Event B: Main Campus

Intersection of A and B:

12% of 60%. So

$P(A \cap B) = 0.12*0.6 = 0.072$

Percentage:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.072}{0.0835} = 0.8623$

0.8623 = 86.23% are at the Main campus