Question

an article in the new england journal of medicine, reported about a study of smokers in california and hawaii. in one part of the report, the self - reported ethnicity and smoking levels per day were given. of the people smoking at most ten cigarettes per day, there were 9,886 african americans, 2,745 native hawaiians, 12,831 latinos, 8,378 japanese americans, and 7,650 whites. of the people smoking 11 to 20 cigarettes per day, there were 6,514 african americans, 3,062 native hawaiians, 4,932 latinos, 10,680 japanese americans, and 9,877 whites. of the people smoking 21 to 30 cigarettes per day, there were 1,671 african americans, 1,419 native hawaiians, 1,406 latinos, 4,715 japanese americans, and 6,062 whites. of the people smoking at least 31 cigarettes per day, there were 759 african americans, 788 native hawaiians, 800 latinos, 2,305 japanese americans, and 3,970 whites. suppose that one person from the study is randomly selected. find the probability that person smoked 11 to 20 cigarettes per day. (enter your probability as a fraction.)

Explanation:

Step1: Calculate total number of people in each smoking category

• At most 10 cigarettes: \(9886 + 2745 + 12831 + 8378 + 7650 = 41490\)
• 11 - 20 cigarettes: \(6514 + 3062 + 4932 + 10680 + 9877 = 35065\)
• 21 - 30 cigarettes: \(1671 + 1419 + 1406 + 4715 + 6062 = 15273\)
• At least 31 cigarettes: \(759 + 788 + 800 + 2305 + 3970 = 8622\)

Step2: Calculate total number of people in the study

Total = \(41490 + 35065 + 15273 + 8622 = 100450\)

Step3: Calculate probability of smoking 11 - 20 cigarettes

Probability = \(\frac{\text{Number of people smoking 11 - 20 cigarettes}}{\text{Total number of people}}\) = \(\frac{35065}{100450}\)

Step4: Simplify the fraction

Divide numerator and denominator by 5: \(\frac{35065\div5}{100450\div5} = \frac{7013}{20090}\)

Answer:

\(\frac{7013}{20090}\)