According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that men who were very active (burning about 3,500 calories daily) were a third as likely to develop diabetes compared with men who were sedentary. Assume that one in 10 middle-aged men is very active, and the rest are classified as sedentary. What is the probability that a middle-aged man with diabetes is very active? (Round your answer to four decimal places.)

One in 10 middle-aged men is very active, and the rest are classified as sedentary.
Men who were very active (burning about 3,500 calories daily) were a third as likely to develop diabetes.

From point 1) we can say that the relation between very active and sedentary men is 1 to 9. That is, for each very active man there are 9 sedentary men.
From point two we can say that, if we had the same number of sedentary and very active men, then there would be 3 times more sedentary men with diabetes than very active men with diabetes.

But we do not have the same number of sedentary and very active men. There are 9 times more sedentary than very active. Therefore, there would be 9*3=27 times more sedentary men with diabetes than very active men with diabetes.

In conclusion, the probability that a middle-aged man with diabetes is very active is [tex]\frac{1}{28}=0.0357[/tex]

azikennamdi azikennamdi · Answer 2 · 2022-04-04T17:40:26+02:00

The probability (to four decimal points) that a middle-aged man with diabetes is very active is 0.0625.

What is the formula for the above problem?

The formula is given as:

P(a | d) = P (d | a) P(d|a)p(a)/P(d)

a - means the probability that the man is active.

d - is the probability that the patient is diabetic.

P (d) = (202/5990) * P (a) .........1, and

P(d) = 1/4,

But P(a|d) = P(d*a)/P(d)

= P(d*a)/(P(d*a) + P(d*a'))

= P(d*a)/(P(d*a) + 15*P(d*a)) = 1/16

Thus, 1/16= 0.0625

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