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 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 twice of the corresponding population variance for measurements on lead. Assume S to be independent of $2. Assume that In (LC50) is Normally distributed. Find the value of a such that P (a ≤ $) 0.95. Note: Please round your answer to 4 decimal places.

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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 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 twice of the corresponding population variance for
measurements on lead. Assume S2 to be independent of $2.
Assume that In (LC50) is Normally distributed.
Find the value of a such that P (a ≤ $) = 0.95.
Note: Please round your answer to 4 decimal places.
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 S 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 twice of the corresponding population variance for measurements on lead. Assume S2 to be independent of $2. Assume that In (LC50) is Normally distributed. Find the value of a such that P (a ≤ $) = 0.95. Note: Please round your answer to 4 decimal places.
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