1.In a certain year, 88% of all Caucasians in the U.S., 77% of all African-Ameri
ID: 3150287 • Letter: 1
Question
1.In a certain year, 88% of all Caucasians in the U.S., 77% of all African-Americans, 77% of all Hispanics, and 77% of residents not classified into one of these groups used the Internet for e-mail. At that time, the U.S. population was 69% Caucasian, 12% African-American, and 14% Hispanic. What percentage of U.S. residents who used the Internet for e-mail were Hispanic? (Round your answer to the nearest whole percent.)
2.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 fifth as likely to develop diabetes compared with men who were sedentary. Assume that one-fourth of all middle-aged men are 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.)
Explanation / Answer
1.Percentage of Us residents =(0.69*0.88+ 0.12*0.77 + 0.14*0.77 + 0.05*0.77 )*100
=84.59% =85%
2.sample size= 5990
n(diabetes) =202
P(active men to develop diabetes) =5*P(sedentary men to develop diabetes)
fraction of active men =P(active men) =1/4 => fraction of sedentary men =3/4
let *P(sedentary men to develop diabetes) = q => P(active men to develop diabetes)=5q
(5q)*(1/4) *5590 + q*(3/4)*5590 =202
q=0.01806
n(active and with diabetes) =5q*(1/4) *5590 =75.75 =76
P(active/diabetes) =P(active and with diabetes)/P(diabetes) =76/202
=0.376
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