A common measure of toxicity for any pollutant is the concentration of the pollutant that will kill half of the test species in a given amount of time (usually 96 hours for fish species). This measure is called LC50. Let S denote the sample variance for a random sample of eight In (LC50) values for copper and S2 denote the sample variance for a random sample of six In (LC50) values for lead, both samples using the same species fish and are assumed to be independent. The population variance for measurements on copper is assumed to be half of the corresponding population variance for measurements on lead. Find the value of c such that P (S≤ c) = 0.95. Note: Please round your answer to 4 decimal places.
A common measure of toxicity for any pollutant is the concentration of the pollutant that will kill half of the test species in a given amount of time (usually 96 hours for fish species). This measure is called LC50. Let S denote the sample variance for a random sample of eight In (LC50) values for copper and S2 denote the sample variance for a random sample of six In (LC50) values for lead, both samples using the same species fish and are assumed to be independent. The population variance for measurements on copper is assumed to be half of the corresponding population variance for measurements on lead. Find the value of c such that P (S≤ c) = 0.95. Note: Please round your answer to 4 decimal places.
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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Transcribed Image Text:A common measure of toxicity for any pollutant is the concentration of the pollutant that will kill half of the
test species in a given amount of time (usually 96 hours for fish species). This measure is called LC50. Let S²
denote the sample variance for a random sample of eight In (LC50) values for copper and S denote the
sample variance for a random sample of six In (LC50) values for lead, both samples using the same species of
fish and are assumed to be independent. The population variance for measurements on copper is assumed to
be half of the corresponding population variance for measurements on lead. Find the value of c such that
P(Sc): = 0.95.
Note: Please round your answer to 4 decimal places.
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