According to a study in The New England Journal of Medicine, 85 of a sample of 5,990 middle-aged men had developed heart disease. It also found that men who were very active (burning about 3,500 calories daily) were half as likely to develop heart disease compared with men who were sedentary. Assume that one in nine middle-aged men are very active, and the rest are classified as sedentary. What is the probability that a middle-aged man with heart disease is very active Please, round your answer to three decimal places.

Question

According to a study in The New England Journal of Medicine, 85 of a sample of 5,990 middle-aged men had developed heart disease. It also found that men who were very active (burning about 3,500 calories daily) were half as likely to develop heart disease compared with men who were sedentary. Assume that one in nine middle-aged men are very active, and the rest are classified as sedentary. What is the probability that a middle-aged man with heart disease is very active  Please, round your answer to three decimal places.

Quizplus · Accepted Answer

Out of 5,990 men, 85 had heart disease. Let H be the event of having heart disease and A be the event of being very active. We know P(A) = 1/9 and P(H|A̅) = probability of heart disease given that the man is sedentary = (85 - P(H|A))/[(8/9) × 5,990]. We also know that P(H|A) = (1/2)×P(H|A̅) because men who are very active are half as likely to develop heart disease compared with men who are sedentary. Substituting P(H|A) into the previous equation, we obtain:
P(H|A̅) = 2P(H|A) = 2(85 - P(H|A))/[(8/9) × 5,990],
solving which gives P(H|A) = 29/546 = 0.053 (rounded to three decimal places). Therefore, the probability that a middle-aged man with heart disease is very active is
P(A|H) = P(H|A)P(A)/P(H) = P(H|A)P(A)/[P(H|A)P(A) + P(H|A̅)(1 - P(A))]
= (0.053)×(1/9)/[(0.053)×(1/9) + (85 - 0.053)(8/9)/(8/9) × 5,990]
≈ 0.059. Therefore, the answer is choice C.

According to a Study in the New England Journal of Medicine