HWKsolutions3lastsemester

16667foralld

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: or 1 – Pr (Y &lt; 1) = 1 ‐ .343 = 0.657 o Find Pr {Y &lt; 3}. Pr (Y &lt;3 ) = 1 – Pr (Y &gt;=3) = 1‐ 0.027 = .973 o Calculate the mean of Y. 0(.343) + 1(.441) + 2(.189) + 3(.027) = 0.9 3 http://www.stat.ucla.edu/%7Edinov/courses_students.dir/12/Winter/STAT13.1.dir/HWs.dir/HW03.html Y (No. Black) 0 0.343 1 0.441 2 0.189 3 0.027 Total Probability 1.00 (HW3.5) Go to the Coin Die Experiment. o Simulate event independence between the outcome of the die (event B) and the outcome of the coin (event A), by setting the probabilities of both dice to be identical. Run 100 experiments and argue that the observed data implies independence between the events A={Coin=Head} and B={Die=3}, i.e., P(A⋂B) = P(A) P(B), approximately. Using the 1‐6 flat with p=0.5, we have the above situation. We have chosen the setting of X == A == Coin to be P(X=1) = 0.5 and P(X=0) = 0.5, Further, we have chosen the setting of Y == B == Die to be P(Die = d) = 0.16667 for all d. Since we have the empirical distributi...
View Full Document

This note was uploaded on 01/22/2014 for the course STATISTICS 13 taught by Professor Markhansen during the Spring '11 term at UCLA.

Ask a homework question - tutors are online