# 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 helpf

Question
Modeling data distributions
It is estimated that aproximately $$\displaystyle{8.36}\%$$ Americans are afflicted with Diabetes .
Suppose that a ceratin diagnostic evaluation for diabetes will correctly diagnose $$\displaystyle{94.5}\%$$ of all adults over 40 with diabetes as having the disease and incorrectly diagnoses $$\displaystyle{2}\%$$ 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.

2021-03-09

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

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