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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State/Region Membership
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
New Jersey, Bermuda 321
Alaska, Idaho, Montana, Oregon, Washington 893
New York 690
Ohio 798
Oklahoma 201
Arkansas 74
Illinois 632
Indiana 216
Iowa 76
Kentucky 155
Louisiana 161
Michigan 304
Massachusetts 530
California 1,110
New Mexico 93
Pennsylvania 631
Rhode Island 75
Colorado 344
South Carolina 264
Texas 1,120
Tennessee 122
Utah 45
Virginia 304
Vermont, New Hampshire 192
Wisconsin 316 West Virginia 63
Use statistical software to answer the following questions.
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Mean?
Median?
Standard deviation?
b-1. Find the coefficient of skewness, using the software method? (Round your answer to 2 decimal places.)
Coefficient of skewness?
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Answers

Answer:

Kindly check explanation

Step-by-step explanation:

Given the data :

114, 261, 319, 183,654,313,58,98,335,324,52,125,342,66,321,893,690,798,201,74,632,216,76,155,161,304,530,1110,93,631,75,344,264,1120,122,45,304,192,316,63

USING CALCULATOR :

The mean(m) of the data: = ΣX/n

n = sample size = 40

m = 12974/40

m = 324.4

Median = ((n + 1)/2) th term = 41/2 = (20 + 21)th term / 2

= (261 + 264) / 2 = 262.5

Standard deviation (s) = sqrt(Σ(x - m)²/n))

s = 281. 74

B.) Coefficient of skewness (Ks) :

3(mean - median) / standard deviation

3(324.4 - 262.5) / 281.74

= 0.66

Mild positive skewness Given a positive skew Coefficient value.

(d-1) Are there any outliers?

Yes

Four outliers

(d-2) What are the limits for outliers? (Round your answers to 1 decimal place. Negative amounts should be indicated by a minus sign.)

Lower bound = Q1 - (1.5 * IQR)

Upper bound = Q3 + (1.5 * IQR)

IQR = Q3 - Q1

Q3 = upper quartile = 343 ; Q1 = 106 ; Q2 = 262.5

IQR = 343 - 106 = 237

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Upper bound = 343 + (1.5 * 237) = 698.5

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Answers

Answer:

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Step-by-step explanation:

A) (3,-2)

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

Hi there !

3x + 2 = 11

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Answers

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3x+4<22

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Answers

Answer:

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Answers

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Answers

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