2. A sample of pond water is evaluated for bacteria by counting the number of bacterial FRNA genes per milliliter. The counts (in billions/ml) are [2.54, 1.64, 0.87, 3.10, 2.78, 1.92, 2.36, 3.17, 2.45, 3.72]. You want to make estimates of the water for the entire pond. What fraction of the pond water would you estimate has above 3 billion/ml bacterial gene count?
2. A sample of pond water is evaluated for bacteria by counting the number of bacterial FRNA genes per milliliter. The counts (in billions/ml) are [2.54, 1.64, 0.87, 3.10, 2.78, 1.92, 2.36, 3.17, 2.45, 3.72]. You want to make estimates of the water for the entire pond. What fraction of the pond water would you estimate has above 3 billion/ml bacterial gene count?
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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![2. A sample of pond water is evaluated for bacteria by counting the number of bacterial rRNA genes per milliliter. The counts (in billions/ml) are:
\[ [2.54, 1.64, 0.87, 3.10, 2.78, 1.92, 2.36, 3.17, 2.45, 3.72]. \]
You want to make estimates of the water for the entire pond. What fraction of the pond water would you estimate has above 3 billion/ml bacterial gene count?
3. Redo the above calculation, but assume it is lognormal instead of normal. Is there a reason to prefer the lognormal to the normal distribution for this case?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24a60763-065d-4deb-b582-ca07220062f2%2Fc16b602e-6b0d-49bf-ad60-2cff4024ac89%2F79amceb_processed.png&w=3840&q=75)
Transcribed Image Text:2. A sample of pond water is evaluated for bacteria by counting the number of bacterial rRNA genes per milliliter. The counts (in billions/ml) are:
\[ [2.54, 1.64, 0.87, 3.10, 2.78, 1.92, 2.36, 3.17, 2.45, 3.72]. \]
You want to make estimates of the water for the entire pond. What fraction of the pond water would you estimate has above 3 billion/ml bacterial gene count?
3. Redo the above calculation, but assume it is lognormal instead of normal. Is there a reason to prefer the lognormal to the normal distribution for this case?
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