This preview shows page 1. Sign up to view the full content.
Unformatted text preview:  S 1) = . 1; p ( R 1  S 1) = . 9 . Moreover, we know p ( S 0) = 2 3 and p ( S 1) = 1 3 . Thus p ( S 1  R 1) = p ( R 1  S 1) p ( S 1) p ( R 1  S 1) p ( S 1) + p ( R 1  S 0) p ( S 0) = . 9( 1 3 ) . 9( 1 3 ) + . 2( 2 3 ) ≈ . 69 Problem 2 (#6.4.????b) . A factory produces cans of soda. The expected number of cans is 10,000; the variance is 1000. Use Chebyshev’s inequality to write a lower bound on the probability that the number of cans produced is between 9000 and 11000. Solution. Chebyshev’s inequality states that P (  NE [ N ]  ≥ r ) ≤ V [ N ] /r 2 . In this case we are interested in r = 1000 (properly speaking, 1001, but let’s make the math simple, so P (  NE [ N ]  ≥ r ) ≤ 1 / 1000. Thus the probability of the number of cans produced being between 9000 and 11000 is at least 0.999. 1...
View
Full
Document
 Spring '11
 MichealAlain
 Math

Click to edit the document details