In a study of 200 auto accident claims, it is found that claims are submitted t months after the accident occurs for t = 0,1,2, .. and there is no withdrawal observation. Ŝ(t) is determined by the product-limit method, and its variance is estimated by Greenwood's formula. Find thee number of claims submitted at time t = 10, given the following value $(8) = 0,22 $(9) = 0,16 Est.Var[$(9)] = 0,00672 Est.Var[$(10)] [$(10)]* = 0,04045
In a study of 200 auto accident claims, it is found that claims are submitted t months after the accident occurs for t = 0,1,2, .. and there is no withdrawal observation. Ŝ(t) is determined by the product-limit method, and its variance is estimated by Greenwood's formula. Find thee number of claims submitted at time t = 10, given the following value $(8) = 0,22 $(9) = 0,16 Est.Var[$(9)] = 0,00672 Est.Var[$(10)] [$(10)]* = 0,04045
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![In a study of 200 auto accident claims, it is found that claims are submitted t months
after the accident occurs for t = 0,1,2,.. and there is no withdrawal observation. $(t) is
determined by the product-limit method, and its variance is estimated by Greenwood's
formula. Find thee number of claims submitted at time t = 10, given the following value
$(8) = 0,22
$(9) = 0,16
Est.Var[$(9)] = 0,00672
Est.Var[$(10)]
[$(10)]*
0,04045](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc5db1628-dc8e-462c-8278-dd1e4808b311%2Fb139ceab-2581-4069-9a9b-e1216919ac8f%2Fv4zr1if_processed.jpeg&w=3840&q=75)
Transcribed Image Text:In a study of 200 auto accident claims, it is found that claims are submitted t months
after the accident occurs for t = 0,1,2,.. and there is no withdrawal observation. $(t) is
determined by the product-limit method, and its variance is estimated by Greenwood's
formula. Find thee number of claims submitted at time t = 10, given the following value
$(8) = 0,22
$(9) = 0,16
Est.Var[$(9)] = 0,00672
Est.Var[$(10)]
[$(10)]*
0,04045
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