1. Given the table below from a survey of CU students: What is the probability o
ID: 3335868 • Letter: 1
Question
1. Given the table below from a survey of CU students: What is the probability of Cannabis use?
Alcohol Use
No Alcohol
Cannabis
400
50
No Cannabis
518
232
a) .80
b) .625
c) .375
d) None
2. In a sample of 1000 adults abusing pain medication, 450 are males, and 350 have a history of chronic pain. The probability of being female with chronic pain history is .15. If an adult is picked at random from this sample, find the probability of picking a female or a person with a chronic pain history.
a).50
b) 1.10
c) .90
d) .75
e) None
3.4% of all persons have an allelle that makes them a candidate for an experimental drug. If a research team randomly selects 4 subjects for their study, what is the likelihood that at least 1 will have the allele?
4, Out of 109 subjects, 58 are normal and 51 have neurologic problems. All were subjected to computed tomographic scans (brain scans) that were then interpreted by a neurologist who was blind to their true condition. The neurologist rated the scans on a scale of 1 (most normal in appearance) to 5 (most abnormal in appearance). The results are in the table below:
CT rating by Neurologist
1
2
3
4
5
Total
Normal Subject
33
6
6
11
2
58
Neurological Problems
3
2
2
11
33
51
Total
36
8
8
22
35
109
Assuming a rating of 4 or higher means a subject is diagnosed with neurologic problems, what is the sensitivity of this test?
5. What is the probability of obtaining exactly 3 girls in a random sample of 3 children drawn from a population with 55% girls?
Alcohol Use
No Alcohol
Cannabis
400
50
No Cannabis
518
232
Explanation / Answer
Ans:
1)P(cannabis use)=(400+50)/(400+50+518+232)=450/1200=0.375
Option c is correct.
2)P(males)=0.45
P(female)=1-0.45=0.55
P(chronic pain)=0.35
P(female and chronic pain)=0.15
P(female or chronic pain)=P(female)+P(chronic pain)-P(female and chronic pain)
=0.55+0.35-0.15
=0.75
Option d is correct.
3)Binomial distribution with n=4,p=0.04
P(atleast 1)=1-P(none)=1-4C0*0.040*(1-0.04)4=1-0.964=1-0.8493=0.1507
5)probabilty of being girl,p=0.55
n=3
P(3 girls)=3C3*0.553*(1-0.55)0=0.553=0.1664
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