It is estimated that aproximately 8.36% Americans are afflicted with Diabetes . Suppose that a ceratin diagnostic evaluation for diabetes will correct

Ramsey 2021-03-08 Answered
It is estimated that aproximately 8.36% Americans are afflicted with Diabetes .
Suppose that a ceratin diagnostic evaluation for diabetes will correctly diagnose 94.5% of all adults over 40 with diabetes as having the disease and incorrectly diagnoses 2% of all adults over 40 without diabetes as having the disease .
1) Find the probability that a randamly selected adult over 40 doesn't have diabetes and is diagnosed as having diabetes ( such diagnoses are called "false positives").
2) Find the probability that a randomly selected adult of 40 is diagnosed as not having diabetes.
3) Find the probability that a randomly selected adult over 40 actually has diabetes , given that he/she is diagnosed as not having diabetes (such diagnoses are called "false negatives").
Note: It will be helpful to first draw an appropriate tree diagram modeling the situation.
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Expert Answer

Raheem Donnelly
Answered 2021-03-09 Author has 75 answers

Step 1
For the given problem let us assume ,
D = event that people have diabetes
N = event that people have no diabetes
C = event that people diagnosed having diabetes
F = event that people diagnosed not having diabetes
Step 2
Now from the given information in the problem, we calculate the probabilities as shown,
P(D)=0.0836
P(N)=10.0836=0.9164
Also,P(CD)=0.945
P(FD)=10.945=0.055
And,P(CN)=0.02
P(FN)=10.02=0.98
Step 3
1. The probability that a randomly selected adult over 40 doesn't have diabetes and is diagnosed as having diabetes (such diagnoses are called "false positives") is P(N and C) and is calculated as shown,
P(N and C)=P(N)×P(CN)
=0.9164×0.02
=0.018328
0.018
Step 4
Answer. 1: The probability that a randomly selected adult over 40 doesn't have diabetes and is diagnosed as having diabetes (such diagnoses are called "false positives") is 0.018
Step 5
2. The probability that a randomly selected adult of 40 is diagnosed as not having diabetes is P(F) and is calculated as shown,
P(F)=P(F and D)+P(F and N)
=P(FD)×P(D)+P(FN)×P(N)
=0.055×0.0836+0.98×0.9164
=0.004598+0.898072
=0.902670.9027
Step 6
Answer.2: The probability that a randomly selected adult of 40 is diagnosed as not having diabetes is 0.9027
Step 7
3. The probability that a randomly selected adult over 40 actually has diabetes, given that he/she is diagnosed as not having diabetes (such diagnoses are called "false negatives") is P(DF) and is calculated as shown,
P(DF)=P(F and D)P(F)
=P(FD)×P(D)P(F)
=0.055×0.08360.9027
=0.005090.0051
Step 8
Answer. 3: The probability that a randomly selected adult over 40 actually has diabetes, given that he/she is diagnosed as not having diabetes (such diagnoses are called "false negatives") is 0.0051.

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