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. A data set is Gaussian with an average of 3 and a standard deviation of 2. a. What percentage of the data set is negative? b. What percentage of data falls between 2 and 5?

. A data set is Gaussian with an average of 3 and a standard deviation of 2. a. What percentage of the data set is negative? b. What percentage of data falls between 2 and 5?

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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HWK 5: Gaussian (Normal) Distribution 1. A data set is Gaussian with an average of 3 and a standard deviation of 2. a. What percentage of the data set is negative? b. What percentage of data falls between 2 and 5? 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?
Transcribed Image Text:HWK 5: Gaussian (Normal) Distribution 1. A data set is Gaussian with an average of 3 and a standard deviation of 2. a. What percentage of the data set is negative? b. What percentage of data falls between 2 and 5? 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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