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

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,
${\displaystyle P\left(D\right)=0.0836}$
${\displaystyle \Rightarrow P\left(N\right)=1-0.0836=0.9164}$
${\displaystyle \text{Also},P(C\mid D)=0.945}$
${\displaystyle \Rightarrow P(F\mid D)=1-0.945=0.055}$
${\displaystyle \text{And},P(C\mid N)=0.02}$
${\displaystyle \Rightarrow P(F\mid N)=1-0.02=0.98}$
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,
${\displaystyle P\left(N\text{and}C\right)=P\left(N\right)\times P(C\mid N)}$
${\displaystyle =0.9164\times 0.02}$
${\displaystyle =0.018328}$
${\displaystyle \approx 0.018}$
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,
${\displaystyle P\left(F\right)=P\left(F\text{and}D\right)+P\left(F\text{and}N\right)}$
${\displaystyle =P(F\mid D)\times P\left(D\right)+P(F\mid N)\times P\left(N\right)}$
${\displaystyle =0.055\times 0.0836+0.98\times 0.9164}$
${\displaystyle =0.004598+0.898072}$
${\displaystyle =0.90267\approx 0.9027}$
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 "false negatives") is ${\displaystyle P(D\mid F)}$ and is calculated as shown,
${\displaystyle P(D\mid F)=\frac{P\left(F\text{and}D\right)}{P\left(F\right)}}$
${\displaystyle =\frac{P(F\mid D)\times P\left(D\right)}{P\left(F\right)}}$
${\displaystyle =\frac{0.055\times 0.0836}{0.9027}}$
${\displaystyle =0.00509\approx 0.0051}$
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 "false negatives") is 0.0051.