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A disease is known to be present in 2% of a population - Edexcel - A-Level Maths Statistics - Question 1 - 2008 - Paper 2

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A disease is known to be present in 2% of a population. A test is developed to help determine whether or not someone has the disease. Given that a person has the di... show full transcript

Worked Solution & Example Answer:A disease is known to be present in 2% of a population - Edexcel - A-Level Maths Statistics - Question 1 - 2008 - Paper 2

Step 1

Draw a tree diagram to represent this information.

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Answer

To represent the situation as a tree diagram:

  1. Start with a point representing the total population.
  2. Branch out to two paths: one for having the disease (2% or 0.02) and one for not having the disease (98% or 0.98).
  3. From the 'Disease' branch, create two further branches: one for a positive test (95% or 0.95) and one for a negative test (5% or 0.05).
  4. From the 'No Disease' branch, create two branches: one for a positive test (3% or 0.03) and one for a negative test (97% or 0.97).
  5. The final tree diagram visualizes the probabilities as follows:
                      +--- Disease (0.02)
           +---
           |          +--- Positive Test (0.95)
           |--- +---
                      +--- Negative Test (0.05)
           |          
           |          +--- No Disease (0.98)
           +---
                      +--- Positive Test (0.03)
           |          
                      +--- Negative Test (0.97)

This diagram accurately reflects the disease prevalence and the test outcomes.

Step 2

Find the probability that the test is positive.

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Answer

To find the probability that the test is positive, we can use the law of total probability:

P(PositiveTest)=P(Disease)imesP(PositiveDisease)+P(NoDisease)imesP(PositiveNoDisease)P(Positive Test) = P(Disease) imes P(Positive | Disease) + P(No Disease) imes P(Positive | No Disease)

Substituting in the given values:

P(PositiveTest)=(0.02imes0.95)+(0.98imes0.03)P(Positive Test) = (0.02 imes 0.95) + (0.98 imes 0.03)

Calculating this gives:

P(PositiveTest)=0.019+0.0294=0.0484P(Positive Test) = 0.019 + 0.0294 = 0.0484

Thus, the probability that the test is positive is approximately 0.0484.

Step 3

Find the probability that he does not have the disease.

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Answer

Given that the test is positive, we use Bayes' theorem to find the probability that the person does not have the disease:

P(NoDiseasePositiveTest)=P(PositiveTestNoDisease)imesP(NoDisease)P(PositiveTest)P(No Disease | Positive Test) = \frac{P(Positive Test | No Disease) imes P(No Disease)}{P(Positive Test)}

We already calculated P(PositiveTest)P(Positive Test) and have:

  • P(PositiveNoDisease)=0.03P(Positive | No Disease) = 0.03
  • P(NoDisease)=0.98P(No Disease) = 0.98

Now substituting in:

P(NoDiseasePositiveTest)=0.03×0.980.0484P(No Disease | Positive Test) = \frac{0.03 \times 0.98}{0.0484}

This simplifies to:

P(NoDiseasePositiveTest)=0.02940.04840.607P(No Disease | Positive Test) = \frac{0.0294}{0.0484} \approx 0.607

Therefore, the probability that the person does not have the disease given that the test is positive is approximately 0.607.

Step 4

Comment on the usefulness of this test.

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Answer

The usefulness of the test can be evaluated based on its probability of correctly identifying individuals who do not have the disease despite a positive test result.

Given that approximately 60.7% of individuals who test positive do not have the disease, this indicates that the test has a relatively high false positive rate. This suggests that the test is not very useful for determining the presence of the disease in the population, as a significant proportion of positive results do not correlate with actual disease presence.

In summary, while the test has high sensitivity (95%) for those with the disease, its low specificity (3% positive test for those without the disease) makes it less reliable, leading to potential misdiagnoses.

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