Question

O 7 days 30 45 days D CHF 5 days 4 days CHF 8P<90 15 -0 5 days 65 O BP<90 70 BP 90 30 Previous MI Previous MI 25 4 days 3 days 0.825 3 days 4 days 0 BP 90 BP 90 85 45 CHF O 70 4 days 70 0 5 days CHF 2 days 3 days BP 90 25 BP-90 65 Dr. Smith's Practice Patterns Dr. Jone's Practice Patterns 4 days 2 days

(Please make sure that the previous answer is wrong), do not copy it and paste, please.)

1. What is the expected length of stay for each of the clinicians?

2. What is the expected length of stay for Dr. Smith if he were to take care of patients of Dr. Jones?

3. What is the expected length of stay for Dr. Jones if he were to take of patients of Dr. Smith?

Answer 1

Sol:

1. ANS: Expected length for each clinician: Dr. Smith expected length of stay is 3.60875 days. Dr. Jones expected length of stay is 4.67275 days. 2. ANS: Expected Length of stay for Dr. Smith: According to this case decision tree will be changed but the time of the particular situation will not be changed. So in the first case Dr. Smith will have decision tree of Dr. Jones. E'[Dr. smith New] =0.85 * 0.65 * 0.30 * 4.5+0.85 * 0.65 * 0.70 * 4.5 +0.85 * 0.35 * 0.30 * 5+ 0.85 * 0.35 * 0.70 * 4+ 0.15 * 0.70 * 0.45 * 3+ 0.15 * 0.70 * 0.55 * 4+ 0.15 * 0.30 * 0.65 * 2 +0.15 * 0.30 * 0.35 *4 = 4.25975 days 3. ANS: Expected Length of stay for Dr. Jones: According to this case decision tree will be changed but the time of the particular situation will not be changed. So in the first case Dr. Jones will have decision tree of Dr. Smith. E' [Dr. Jones New]=0.25 * 0.15 * 6+0.25 * 0.85 * 0.70 * 4 +0.25 * 0.85 * 0.30 * 3+0.75 * 0.70 * 0.85 * 4+0.75 * 0.70 * 0.15 * 5+ 0.75 * 0.30* 0.25 * 3+0.75 * 0.30 * 0.75 * 2 = 3.69625 days.