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.

Question

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&#039;t have diabetes and is diagnosed as having diabetes ( such diagnoses are called &quot;false positives&quot;).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 &quot;false negatives&quot;).Note: It will be helpful to first draw an appropriate tree diagram modeling the situation.

Raheem Donnelly · Accepted Answer

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)=1−0.0836=0.9164 Also,P(C∣D)=0.945 ⇒P(F∣D)=1−0.945=0.055 And,P(C∣N)=0.02 ⇒P(F∣N)=1−0.02=0.98 Step 3 1. The probability that a randomly selected adult over 40 doesn&#039;t have diabetes and is diagnosed as having diabetes (such diagnoses are called &quot;false positives&quot;) is P(N and C) and is calculated as shown, P(N and C)=P(N)×P(C∣N) =0.9164×0.02 =0.018328 ≈0.018 Step 4 Answer. 1: The probability that a randomly selected adult over 40 doesn&#039;t have diabetes and is diagnosed as having diabetes (such diagnoses are called &quot;false positives&quot;) 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(F∣D)×P(D)+P(F∣N)×P(N) =0.055×0.0836+0.98×0.9164 =0.004598+0.898072 =0.90267≈0.9027 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 &quot;false negatives&quot;) is P(D∣F) and is calculated as shown, P(D∣F)=P(F and D)P(F) =P(F∣D)×P(D)P(F) =0.055×0.08360.9027 =0.00509≈0.0051 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 &quot;false negatives&quot;) is 0.0051.

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

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