CLA 2 Paper & PPT - Advanced Statistical Concepts and Business Analytics

2

Advanced Statistical Concepts and Business Analytics

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A. Probability of a person having schizophrenia if the person CAT shows Atrophy.

Let

P(N) show the probability of a normal person with schizophrenics

P(S) shows the likelihood of a person with schizophrenics

P(A) offer the possibility of a person with Atrophy

There are two events when a person is said to have Atrophied.

When a person suffering from schizophrenia and CAT shows he has Atrophy =30%

A normal person scanned and said to have Atrophy =2%

The probability of having schizophrenia in the USA is said to be 1.5%

Therefore, the probability that a person has schizophrenia given brain atrophy is calculated as

P(Si/E) = P (Si) P(E/Si) ÷∑P(Si) P(E/Si)

P(S/A) = (0.015×0.3) ÷ (0.015×0.3) +(0.0985×0.02)

= 0.045 ÷ (0.045+0.0197)

= 0.18595

= 0.186

B. If Johns CAT showed Atrophy, does the answer help or hurt the case?

The probability is hurting the case that Hinckley suffered mental illness because the probability of 0.186 is very small to decide he suffered mental illness. Therefore, it can not be used to defend Hinckley.

C. Suppose we take a 10% chance that Hinckley has schizophrenia. Find the probability that he has schizophrenia if he shows brain atrophy.

P(S/A) = P(S) P(A/S) ÷ ((P(S) P(A/S) + P(N) P(A/N))

= (0.1×0.3) ÷ ((0.1×0.3) + (0.9×0.02))

=0.03÷ (0.03+ 0.016)

=0.625

D. If the probability of Hinckley having schizophrenia is taken to be 10 per cent, does his scan showed Atrophy in the brain help the case that he suffered brain Illness?

When the probability of Hinckley having schizophrenia is 10%, his scan on brain atrophy helps in the case since the possibility of having brain illness is higher than that of randomly picked Americans.62.5% is a considerable probability he might be having brain illness.

E. If the prior probability of Hinckley, who has schizophrenia, is taken to be 0.25, his likelihood of schizophrenia given his scan showed brain atrophy.

P(S)=0.25

P(N)=0.75

P(S/A) = P(S) P(A/S) ÷ (P(S) P(A/S)) + P(N) P(A/N)

= (0.25×0.3) ÷ (0.25×0.3) +(0.75×0.02)

P(S/A) =0.075 ÷ (0.075+0.015)

P(S/A) = 0.833

F. How strong does the case in E above help the case that Hinckley suffered mental illness?

When the prior probability of Hinckley having schizophrenia is taken to be 0.25, his likelihood of having schizophrenia given his scan on brain atrophy strongly support the case that he is having mental illness since the probability is very high, i.e., 0.833 which is equivalent to 83%

References

1.Goncharenko, (2018). The Bayes’ formula in terms of the multi-optional uncertainty conditional optimality doctrine.