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, 2,205 japanese americans, and 3,970 whites.
in words, explain what it means to pick one person from the study who is \japanese american or smokes 21 to 30 cigarettes per day.\
○ the person must be both japanese american and smoke 21 to 30 cigarettes per day, and the sample space includes everyone in the study.
○ the person must be either japanese american or smoke 21 to 30 cigarettes per day, and the sample space is reduced to those who smoke 21 to 30 cigarettes per day.
○ the person must be both japanese american and smoke 21 to 30 cigarettes per day, and the sample space is reduced to those who are either japanese american or smoke 21 to 30 cigarettes per day.
○ the person must be either japanese american or smoke 21 to 30 cigarettes per day, and the sample space includes everyone in the study.
○ the person must be either japanese american or smoke 21 to 30 cigarettes per day, and the sample space is reduced to those who are japanese american.
find the probability of this. (enter your probability as a fraction.)

Explanation:

Part 1: Interpreting "Japanese American OR smokes 21 to 30 cigarettes per day"

• "OR" in probability/set theory means the person satisfies at least one of the conditions (being Japanese American, smoking 21 - 30 cigs/day, or both).
• The sample space for any probability in the study is all people in the study (since we’re picking from the entire study group).
• Let's analyze each option:
• First option: "both" is "AND", not "OR" – incorrect.
• Second option: Sample space is not reduced to 21 - 30 smokers (sample space is all in study) – incorrect.
• Third option: "both" is "AND", and sample space reduction is wrong – incorrect.
• Fourth option: Correctly defines "OR" (either condition) and sample space as all in the study.
• Fifth option: Sample space reduced to Japanese Americans – incorrect.

Step1: Find total number of people in the study

First, calculate the number of people in each smoking category:

• At most 10 cigs/day: \( 9886 + 2745 + 12831 + 8378 + 7650 \)

\( = 9886 + 2745 = 12631 \); \( 12631 + 12831 = 25462 \); \( 25462 + 8378 = 33840 \); \( 33840 + 7650 = 41490 \)

• 11 - 20 cigs/day: \( 6514 + 3062 + 4932 + 10680 + 9877 \)

\( 6514 + 3062 = 9576 \); \( 9576 + 4932 = 14508 \); \( 14508 + 10680 = 25188 \); \( 25188 + 9877 = 35065 \)

• 21 - 30 cigs/day: \( 1671 + 1419 + 1406 + 4715 + 6062 \)

\( 1671 + 1419 = 3090 \); \( 3090 + 1406 = 4496 \); \( 4496 + 4715 = 9211 \); \( 9211 + 6062 = 15273 \)

• At least 31 cigs/day: \( 759 + 788 + 800 + 2305 + 3970 \)

\( 759 + 788 = 1547 \); \( 1547 + 800 = 2347 \); \( 2347 + 2305 = 4652 \); \( 4652 + 3970 = 8622 \)

Total people: \( 41490 + 35065 + 15273 + 8622 \)
\( 41490 + 35065 = 76555 \); \( 76555 + 15273 = 91828 \); \( 91828 + 8622 = 100450 \)

Step2: Find number of people who are Japanese American OR smoke 21 - 30 cigs/day

Using the principle of inclusion - exclusion: \( n(A \cup B) = n(A) + n(B) - n(A \cap B) \), where:

• \( A \): Japanese Americans (sum across all smoking categories)
• \( B \): People who smoke 21 - 30 cigs/day
• \( A \cap B \): Japanese Americans who smoke 21 - 30 cigs/day

Calculate \( n(A) \):
Japanese Americans in each category: \( 8378 + 10680 + 4715 + 2305 \)
\( 8378 + 10680 = 19058 \); \( 19058 + 4715 = 23773 \); \( 23773 + 2305 = 26078 \)

Calculate \( n(B) \): Already found as \( 15273 \) (from Step1: 21 - 30 cigs/day total)

Calculate \( n(A \cap B) \): Japanese Americans who smoke 21 - 30 cigs/day = \( 4715 \)

Thus, \( n(A \cup B) = 26078 + 15273 - 4715 \)
\( 26078 + 15273 = 41351 \); \( 41351 - 4715 = 36636 \)

Step3: Calculate the probability

Probability \( P = \frac{n(A \cup B)}{\text{Total number of people}} = \frac{36636}{100450} \)
Simplify the fraction: Divide numerator and denominator by 2: \( \frac{18318}{50225} \) (check if further simplification is possible: GCD of 18318 and 50225. Let's check GCD(50225, 18318):
50225 ÷ 18318 = 2 with remainder 13589
18318 ÷ 13589 = 1 with remainder 4729
13589 ÷ 4729 = 2 with remainder 4131
4729 ÷ 4131 = 1 with remainder 598
4131 ÷ 598 = 6 with remainder 543
598 ÷ 543 = 1 with remainder 55
543 ÷ 55 = 9 with remainder 48
55 ÷ 48 = 1 with remainder 7
48 ÷ 7 = 6 with remainder 6
7 ÷ 6 = 1 with remainder 1
6 ÷ 1 = 6 with remainder 0. So GCD is 1. Thus, simplified fraction is \( \frac{18318}{50225} \) (or check if we made a calculation error in total people: Let's re - calculate total people:

At most 10: 9886 + 2745 = 12631; 12631 + 12831 = 25462; 25462 + 8378 = 33840; 33840 + 7650 = 41490 ✔️

11 - 20: 6514 + 3062 = 9576; 9576 + 4932 = 14508; 14508 + 10680 = 25188; 25188 + 9877 = 35065 ✔️

21 - 30: 1671 + 1419 = 3090; 3090 + 1406 = 4496; 4496 + 4715 = 9211; 9211 + 6062 = 15273 ✔️

At least 31: 759 + 788 = 1547; 1547 + 800 = 2347; 2347 + 2305 = 4652; 4652 + 3970 = 8622 ✔️

Total: 41490 + 35065 = 76555; 76555 + 15273 = 91828; 91828 + 8622 = 100450 ✔️

Japanese Americans: 8378 (≤10) + 10680 (11 - 20) + 4715 (21 - 30) + 2305 (≥31) = 8378 + 10680 = 19058; 19058 + 4715 = 23773; 23773 + 2305 = 26078 ✔️

\( n(A \cup B) = 26078 + 15273 - 4715 = 36636 \) ✔️

So probability is \( \frac{36636}{100450} = \frac{18318}{50225} \) (or check if we miscalculated \( n(A \cup B) \):

Wait, let's re - check \( n(A) \):

Wait, at most 10 cigs/day: Japanese Americans = 8378

11 - 20 cigs/day: 10680

21 - 30 cigs/day: 4715

At least 31 cigs/day: 2305

Sum: 8378 + 10680 = 19058; 19058 + 4715 = 23773; 23773 + 2305 =…

Answer:

D. The person must be either Japanese American or smoke 21 to 30 cigarettes per day, and the sample space includes everyone in the study.

Part 2: Calculating the Probability