SOLVE QUESTION 3 only don't do 1 or 2 I already finished with that. If you can't leave it at least I won't loose my question. I SAID PLEASE DO WUESTION 3 number 3 only and expla # Question## 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: \[2.54, 1.64, 0.87, 3.10, 2.78, 1.91, 2.36, 3.17, 2.45, 3.72\]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?

Answered: 1. A data set is Gaussian with an…

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?

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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SOLVE QUESTION 3 only don't do 1 or 2 I already finished with that. If you can't leave it at least I won't loose my question.

I SAID PLEASE DO WUESTION 3 number 3 only and expla

# Question

## 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:
\[2.54, 1.64, 0.87, 3.10, 2.78, 1.91, 2.36, 3.17, 2.45, 3.72\]

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?

Expert Solution

Step 1

In the problem 1, let us assume that lognormal distribution instead of normal.

$M e a n = 3 S \tan d a r d d e v i a t i o n = 2$

For log normal distribution,

$M e a n = e^{μ + \frac{σ^{2}}{2}} 3 = e^{μ + \frac{σ^{2}}{2}} \ln 3 = μ + \frac{σ^{2}}{2} 2 \ln 3 = 2 μ + σ^{2}$

$V a r i a n c e = (e^{σ^{2}} - 1) e^{2 μ + σ^{2}} 4 = (e^{σ^{2}} - 1) e^{2 \ln 3} 4 = (e^{σ^{2}} - 1) e^{\ln 9} 4 = (e^{σ^{2}} - 1) 9 \frac{4}{9} = e^{σ^{2}} - 1 e^{σ^{2}} = \frac{13}{9} σ^{2} = 0.3677$

Therefore,

$2 \ln 3 = 2 μ + σ^{2} 2.1972 = 2 μ + 0.3677 1.8295 = 2 μ μ = 0.9148$

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