Question

'Question 10. In 2002, a group of medical researchers reported that on average, 30 out of every 10,000 people have colorectal cancer: Of these 30 people with colorectal cancer; 18 will have a positive hemoccult test: Of the remaining 9970 people without colorectal cancer; 400 will have a positive test What is the probability that a person that tested negative is free of colorectal cancer? 0.9987 (b) 0.2 (c) 0.00125'

Name: question 10 in 2002 a group of medical researchers reported that on average 30 out of every 10000 people have colorectal cancer of these 30 people with colorectal cancer 18 will have a posit 41728
Uploaded: 2022-07-03T13:56:33-08:00
Duration: 1 min 41 s
Channel: Nicole Smina
Description: question 10 in 2002 a group of medical researchers reported that on average 30 out of every 10000 people have colorectal cancer of these 30 people with colorectal cancer 18 will have a posit 41728

          'Question 10. In 2002, a group of medical researchers reported that on average, 30 out of every 10,000 people have colorectal cancer: Of these 30 people with colorectal cancer; 18 will have a positive hemoccult test: Of the remaining 9970 people without colorectal cancer; 400 will have a positive test  What is the probability that a person that tested negative is free of colorectal cancer?
0.9987
(b) 0.2
(c) 0.00125'

Added by Elizabeth D.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Nicole Smina

07/03/2022

Step 1

- 12 people with cancer tested negative - 9570 people without cancer tested negative Total = 12 + 9570 = 9582 Show more…

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'Question 10. In 2002, a group of medical researchers reported that on average, 30 out of every 10,000 people have colorectal cancer: Of these 30 people with colorectal cancer; 18 will have a positive hemoccult test: Of the remaining 9970 people without colorectal cancer; 400 will have a positive test What is the probability that a person that tested negative is free of colorectal cancer? 0.9987 (b) 0.2 (c) 0.00125'