In a certain community, 8% of all adults over 50 have diabetes. If a health service in this community correctly diagnoses 95% of all persons with diabetes as having the disease and incorrectly diagnoses 2% of all persons without diabetes as having the disease, find the probabilities that
a) the community health service will diagnose an adult over 50 as having diabetes.
b) a person over 50 diagnosed by the health service as having diabetes actually has the disease.

Answers

Answer 1

Answer and Step-by-step explanation:

Solution:

Given:

Let A is the event of a person over 50 diagnosed by health service.

B1 is the event that an adult over 50 actually has diabetes.

B2 is the event that an adult over 50 does not have diabetes.

P (B1) = 8% = 0.08

And P (B2) = 1 – P (B1)

                  = 1 – 0.08

                  = 0.92

Correctly diagnose of all persons with diabetes= 95%

P (A/B1) = 0.95

Incorrectly diagnose of all persons without diabetes=   2%

P (A/B2) = 0.02

(a) ) the community health service will diagnose an adult over 50 as having diabetes.

P (B1) P (A/B1) = (0.08) (0.95) = 0.076

P (B2) p (A/B2) = (0.92) (0.02) =0.0184

P (B1) P (A/B1) + P (B2) p (A/B2) =  0.076 + 0.0184

                                                     = 0.0944

(b)  A person over 50 diagnosed by the health service as having diabetes actually has the disease.

By using formula:

P (B1/A) = P (B1) P (A/B1) / P (B1) P(A/B1) + P (B2) P (A/B2)

Put all the given values:

= (0.08) (0.95) / (0.08) (0.95) + (0.92) (0.02)

=0.076 / 0.076 + 0.0184

=0.076 / 0.0944

= 0.8050

Answer 2

Using conditional probability, it is found that there is a:

a) 0.0944 = 9.44% probability that the community health service will diagnose an adult over 50 as having diabetes.

b) 0.8051 = 80.51% probability that a person over 50 diagnosed by the health service as having diabetes actually has the disease.

Conditional Probability

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened. [tex]P(A \cap B)[/tex] is the probability of both A and B happening. P(A) is the probability of A happening.

Item a:

The percentages of a positive test is:

95% of 8%(have diabetes).2% of 92%(do not have diabetes).

Hence:

[tex]P(A) = 0.95(0.08) + 0.02(0.92) = 0.0944[/tex]

0.0944 = 9.44% probability that the community health service will diagnose an adult over 50 as having diabetes.

Item b:

Event A: Positive test.Event B: Has the disease.

From item a, [tex]P(A) = 0.0944[/tex].

The probability of both a positive test and having the disease is:

[tex]P(A \cap B) = 0.95(0.08)[/tex]

Hence, the conditional probability is:

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.95(0.08)}{0.0944} = 0.8051[/tex]

0.8051 = 80.51% probability that a person over 50 diagnosed by the health service as having diabetes actually has the disease.

A similar problem is given at https://brainly.com/question/14398287


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Step-by-step explanation:

10. (6x - 7)° + (3x - 29)° = 180°

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Answers

Answer:

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Step-by-step explanation:

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Answer:

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Answers

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Step-by-step explanation:

3x+4<22

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

There are 34,500 ft2 of land shared between Mabel, Adeline, and Charlie.

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Alabama 114
Arizona 261
Maryland, Delaware, DC 319
Connecticut 183
Florida 654
Georgia 313
Hawaii 58
Maine 98
Minnesota, Dakotas 335
Missouri, Kansas 324
Mississippi 52
Nebraska 125
North Carolina 342
Nevada 66
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Ohio 798
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Arkansas 74
Illinois 632
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Kindly check explanation

Step-by-step explanation:

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Hi there !

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