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SOA Exam P Question 114 | Conditional Variance of Discrete Distribution
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A diagnostic test for the presence of a disease has two possible outcomes: 1 for
disease present and 0 for disease not present. Let X denote the disease state (0 or 1) of a
patient, and let Y denote the outcome of the diagnostic test. The joint probability function
of X and Y is given by:
P[X = 0, Y = 0] = 0.800
P[X = 1, Y = 0] = 0.050
P[X = 0, Y = 1] = 0.025
P[X = 1, Y = 1] = 0.125
Calculate Var( 1) Y X = .
(A) 0.13
(B) 0.15
(C) 0.20
(D) 0.51
(E) 0.71
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disease present and 0 for disease not present. Let X denote the disease state (0 or 1) of a
patient, and let Y denote the outcome of the diagnostic test. The joint probability function
of X and Y is given by:
P[X = 0, Y = 0] = 0.800
P[X = 1, Y = 0] = 0.050
P[X = 0, Y = 1] = 0.025
P[X = 1, Y = 1] = 0.125
Calculate Var( 1) Y X = .
(A) 0.13
(B) 0.15
(C) 0.20
(D) 0.51
(E) 0.71
You can find the link to the questions below:
The link to the answers below:
Check out my other channel Rumi's Life: