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The aricle "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants' (Water Research [984: [169-_[174) suggests the uniform distributio...

Question

The aricle "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants' (Water Research [984: [169-_[174) suggests the uniform distribution the interval (7,5, 20) as model for depth (cm) of the bioturbation layer in sediment in certain region. What arc the mean and variance of dcpth? b. What is the cdf of depth?What the probability that obsenved depth is At mostIO? Between [0 and I5? What is the probability that the obscrved depth is within[ standard deviation of the me

The aricle "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants' (Water Research [984: [169-_[174) suggests the uniform distribution the interval (7,5, 20) as model for depth (cm) of the bioturbation layer in sediment in certain region. What arc the mean and variance of dcpth? b. What is the cdf of depth? What the probability that obsenved depth is At mostIO? Between [0 and I5? What is the probability that the obscrved depth is within[ standard deviation of the mean value? Within standard deviations?



Answers

The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Res. $1984 : 1169-1174$ ) suggests the uniform distribution on the interval $[7.5,20]$ as a model for depth $(\mathrm{cm})$ of the bioturbation layer in sediment in a certain region.
(a) What are the mean and variance of depth?
(b) What is the cdf of depth?
(c) What is the probability that observed depth is at most 10$?$ Between 10 and 15$?$
(d) What is the probability that the observed depth is within 1 standard deviation of the mean value?

This question covers drawing normal distribution plots. So in order to draw a normal distribution plot, you must first know how much entry CSR there's 22 entries and we find the corresponding table three for 22 entries which begins with negative 1.91 from here. We can construct a table with these values sorted in order. So I would and put these in a calculator or some application that can sort numbers from highest, lowest. So the lowest point 17 and the highest I believe is 7.6. So you would sort this and find the corresponding um table three values. So the lowest value of table three corresponds to the lowest value from this dataset and so is the highest corresponds to the highest. And in between they also have the same pattern. Yeah, so once we have that we can figure out, we can plot it in an X. Y axis. Okay. Yeah. Once we plotted in an X Y axes we can determine two overall pattern of where the majority of the data is. And once we determine the overall pattern, it is pretty clear what data points are. Outliers. Yeah. So in this case there are two outliers right here and there are outliers because they don't follow the general trend, Right? So the outliers are 6.7 and 7.6 um Mimos per square meter per day of effusive oxygen uptake. Um And after this we can determine whether it is normally distributed or not. Yeah. So nose we normal since the majority of the data um follows this trend. If we just eliminated the outliers, then we know that it's mostly normal, so it's normal they distributed.

In this question we want to answer. The variable is normally distributed with me and musical 30.9. And standard deviation Sigma 25 specifically want to answer the three questions at the bottom. A thirsty. This question is challenging understanding uh normally distributed variables to solve. We need to utilize the following information on the left. The fact that the Z scores and probabilities can be mapped onto one another. The A Z table for a standard distribution are invariable. If we call it X is not a standard normal because it has a non zero mean and non standard deviation on one standard deviation. Thus it is an ex but the convertible standard normal, we simply use the conversion D equals x minus mu over sigma on the right With the information left and the right. We have all, we need to answer a three c. So a what is probably except for the 100 Xena is 100 minus two point number. 25 equals 1.96. So the equivalent problem is probably that Z is greater than not which is 0.25 from a Z table for me what is probably the X is less than 25 Xena at this time is negative. 1.36 from a Z table probably is less than 0.1501. Finally and see what economic seated about 5% of the population. We want the X amount for which the area of the normal curve to the right is 50.5. Probably the greater than 0.5 gives, you know, equal 1.64. Thus inverting the equation in the upper right gives economical Xena signal plus new equals 91.9.

In this exercise were given a log. Normal distribution with parameters mu equals 1.9 and Sigma equals 0.9 For part A were asked for the mean and standard deviation of the distribution. So from our textbook, we have the formulas given here. So we simply have to plug our parameter values into these formulas to solve for the mean and variance. So first to mean we have e to the 1.9 plus 0.9 squared over two, and this comes out to 10.24 for the variance E to the two times 1.9 plus 0.9 squared times e to the 0.9 squared minus one, and this comes out to 125 0.395 and that gives us a standard deviation. Have 11.198 So that answer is part A. Now, for part B. We want to find the probability that X is at most 10 and is between five and 10, so we'll start with exes at most 10. This is equal to the probability that the natural log of X, his lessner equal to the natural log of 10. So here I'm just subtracting the parameter mu and then dividing by the parameter sigma to standardize it. This comes out to 0.673 So the probability of being 10 or less in this distribution is 100.673 And now for the probability that X is between five and 10 can be expressed like this in standardized form, this is 0.673 minus 0.373 and it comes out to 0.299

The scenario for this exercise is that organisms are present in water being discharged from a ship according to a Muslim process at a rate of 10 organisms per meter cubed. So this is a possum process where the number of events, the number of organisms is not a function of time but is a function of water volume. So we are given a rate of penn organisms per meter cubed. So they put some parameter is equal to they're rate times the volume of water in meters cube. So the number of organisms present is a personal random variable with parameter lambda times volume for part A were asked to calculate the probability that one cubic meter of discharge contains at least eight organisms. So here they put some parameter is equal to 10. So what is the probability that we have at least eight organisms? This is equal to one, minus the probability that we have, at most seven. For a calculation like this, you probably want to use software. So I'm going to use our So we're looking at one minus the cumulative probability of seven, uh, events. So that is P. P. O. S seven and the plus on parameter is 10 and we get 0.78 And so the probability of having at least eight organisms in 1 m cubed of discharge is 0.78 Next for B were asked the probability that the number of organisms in 1.5 m cubed of discharge exceeds its mean value by more than one standard deviation. So first, let's calculate mu. So that is it's gonna be the rate times 1.5 m cube, which gives 15. So we know that this is the mean number of organisms present in 1.5 m cube for a possible random variable. The variance on the number present is also equal to um you, which is 15, which tells us that the standard deviation on the number is equal to the square root of 15. So we're looking for the probability that the number of organisms present exceeds the mean plus one standard deviation. So we're looking for the probability that the number exceeds the main plus one standard deviation since we can only have an integer number of organisms present. We know that this is equivalent to saying that the probability of X being at least 19 organisms in the 1.5 m cube, and this is also equal to one minus the probability the cumulative probability for 18. Once again, we will use our to solve this equation. So we have one minus a cumulative probability up to 18 when the rate is 15, when the when the personal parameter is 15, that's mu and we get 0.18 So we could say that the probability that the number of organisms in 1.5 m Cuban of discharge exceeds the mean value by more than one standard deviation is 0.18 and next for C were asked what amount of discharge would be required for the probability of a containing at least one organism to be 0.999 So we're looking for the probability that the number of organisms being at least one is 0.999 In other words, one minus. The probability of having exactly zero organisms is 0.999 or the probability of having zero organisms present is 0.1 So if we plug X equals zero into the probability mass function for a Poisson random variable. This is equal to E to the minus. Um, you I'm basically just using the probability mass function except X equals zero. And so we have this equation here. So if we take the natural logarithms of both sides, we have minus mu is equal to the log rhythm of zero point 001 and this gives us mu equal to 6.91 So that tells us what our parameter mu needs to be. Remember, we're looking for the amount that would result in this probability. So mu is equal to They're rate times the volume or the amount. And this means that the volume must equal zero point 691 m cube. So for a volume of discharge of 0.691 m, cubed, the probability of having at least one organism is 0.999


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