A School of Public Health completed a study on alcohol consumption on college campuses. They concluded that 20.1?% of women attending? all-women colleges abstained from? alcohol, compared to 17.1?% of women attending coeducational colleges. Approximately 5.2?% of women college students attend? all-women schools. Complete parts? (a) and? (b) below.

Complete question:

A School of Public Health completed a study on alcohol consumption on college campuses. They concluded that 20.1?% of women attending? all-women colleges abstained from? alcohol, compared to 17.1?% of women attending coeducational colleges. Approximately 5.2?% of women college students attend? all-women schools. Complete parts? (a) and? (b) below.

a) what is the probability that a randomly selected female student abstains from alcohol

b) If a randomly selected female student abstains from alcohol, what is the probability she attends a coeducational college?

a) 0.173

b) 0.94

Step-by-step explanation:

Given:

Probability of women attending all women colleges, probabilty of abstained from alcohol P(Ab|AW) = 0.201

Probability of women attending coed colleges, probabilty of abstained from alcohol P(Ab|CE) =0.171

Let probability of women attending all women college, P(AW) =0.052

Let probability of women attending coed, P(CE) = 1 - 0.052 =0.948

a) To find the probability that a randomly selected female student abstains from alcohol, use the formula below:

P(Ab) =P(Ab|AW)*P(AW)+P(Ab|CE)*P(CE)

=(0.201 * 0.052) + (0.171 * 0.948) = 0.17256

Probability that a randomly selected female student abstains from alcohol is 0.17256 ≈ 0.173

b) If randomly selected female student abstains from alcohol, probability she attends a coeducational college. Use the formula below:

P(CE|Ab) =P(Ab|CE)*P(CE)/P(Ab) =

≈ 0.94

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

step-by-step explanation: