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(a) Assume the people are investigated one after another: Let X be the number of people investigated until the organization finds the first person with the disease:...

Question

(a) Assume the people are investigated one after another: Let X be the number of people investigated until the organization finds the first person with the disease: The average and standard deviation of X are about(b) With X as in (a), the chance that the organization has to investigate 2,000,000 people before finding their first person with the disease is(c) In group of 10 people chosen at random by the organization, the probability that at least three people have the disease is equal to(d) In

(a) Assume the people are investigated one after another: Let X be the number of people investigated until the organization finds the first person with the disease: The average and standard deviation of X are about (b) With X as in (a), the chance that the organization has to investigate 2,000,000 people before finding their first person with the disease is (c) In group of 10 people chosen at random by the organization, the probability that at least three people have the disease is equal to (d) In a group of 200,000 people chosen at random by the organization, the number of peo ple in the group with the disease is best modeled as a (e) With the situation as in (d), the number of people in the group with the disease has an expectation and standard deviation respectively equal to (f) With the situation as in (d), the probability that at least two people will have the disease is approximately equal to



Answers

When studying the spread of an epidemic, we assume that th
probability that an infected individual will spread the diseas
an uninfected individual is a function of the distance betwee
them. Consider a circular city of radius 10 mi in which the population is uniformly distributed. For an uninfected individual at a fixed point $A\left(x_{0}, y_{0}\right),$ assume that the probability
function is given by
$$f(P)=\frac{1}{20}[20-d(P, A)]$$
where $d(P, A)$ denotes the distance between $P$ and $A$
(a) Suppose the exposure of a person to the disease is the
sum of the probabilities of catching the disease from all
members of the population. Assume that the infected
people are uniformly distributed throughout the city, with k infected individuals per square mile. Find a double
integral that represents the exposure of a person residing
at $A .$
(b) Evaluate the integral for the case in which $A$ is the center
of the city and for the case in which $A$ is located on the
edge of the city. Where would you prefer to live?

Okay, we are going to answer question number 25 in your textbook water taxi safety. When a taxi, When a water taxi sank in the Baltimore harbor, an investigation revealed that safe passenger load for the water taxi was £3500. It was also noted that the mean weight of a passenger was assumed to be 100 and £40 assume a worst case scenario in which all the passengers are adult men. This could easily occur in a city that hosts conventions in which people the same gender often travel in groups, Assume the weights and men are normally distributed with a mean of £182.. a standard deviation of £40.8. So we have a mean of 182.9 And the standard deviation of 40.8. So you can see there's a lot of variability in this data set. I'm going to plot the distribution part A says, If one man is randomly selected, find the, find the probability that he weighs less than £174. Uh This is a new value suggested by the National Transportation and Safety Board. The probability that he weighs less than £174. I'm looking for the value to the left of 1 74 it's It's 41.3 per 7%. So a pretty small probability um that 174 might need to be re evaluated by the safety board. Part B says with a low limit of £3,500. How many male passengers are allowed. If we assume in me in a week of 140. So all we need to do is take 3500 And divide it by 140. And we could see that 25 passengers would be allowed on the boat. Uh Part C says with a lower limit of £3500. How many male passengers are allowed if we use a new mean of 182.9. So 3500 Divided by 1 82.9. Yeah. Is 19.1361. So approximately 19.1361 passengers would be allowed on the boat. Party says, why is it necessary to periodically review and revise the number of passengers that are allowed to board? Uh Well you can see that, you know, the old standard was 100 and £40. Unfortunately, people's weights have tend to be increasing over time. Uh So they assumed 100 and £40 per person would be good enough and they saw that that was too much. So they should re evaluate it since the um weight of people is constantly changing and they might want to re evaluate this periodically

Yeah. In this question we're pretending that we're living in the land of an epidemic, in which case the we're assuming that the probability of that an individual who is infected with some disease will spread it to someone else. And we're going to assume that it's a function of the distance between them. We know that we're in a circular city where the radius of 10 miles where the population is uniformly distributed and the population and the probability function that you'll get infected is written down below. And the first part of the question is we're going to suppose that we find a double integral that can represent the probability of the exposure of a person living at any point. And we're going to be assuming that there are K infected individuals in a square mile. So yeah. How do we do this? So let's let's create a rectangle and let's assume that its dimensions are dX and DY, respectively. Yeah. Yes. Mhm. Come on. And we know that the were given already a probability of infection. So we're going to plug in p into whatever point we have here, you know that the probability of any infection is going to be given by that function. Okay. Yeah. Times K. So that's so this is our differential region. So we have the double integral over the region R of D. We have what's called a density function and then we have the actual yeah, rectangle. We have the function times a differential area. And we know that the we know that the distance as a function of two points can be regarded as x minus x not squared plus why minus why not squared? And take the whole route of that. We know that that's the total distance. Mhm. So you can say that the integral is factoring the k out. We have to integrate over the region R 20 minus the distance. Which is this number here. Mm. Damn X minus X. Not squared plus why minus? Why not squared? That will all be divided by 20 D. A. And we have to and we need to remember what the region R. Is And were given that are is a circular city with the radius of 10 miles. So we know that it's just the circle X squared plus Y squared Being less than or equal to 100 because the radius is 10. So that's just the first part and there's a distance from the center of the city. We have our X not and why not being zero? So we want to find the probability that you get infected at the edge. And in the center let's start with the center. Yeah. And because of that that means that we can we can express our exposure probability as the double integral over the region R. And I'm going to simplify, I'm going to rewrite the inte grand a little bit By just dividing the 20th. So we have 1 -1/20 times the square root of X squared plus Y squared integrated with respect to the differential area. And we're going to convert this to polar coordinates. Remember how we have our hi of are equal to the square of x squared plus y squared. And our differential area it's just our time crd photo. And remember our radius will be integrating over the whole polar radius and will be integrating over the whole circle that the circle spain that this whole spans bus Therefore are integral becomes the integral of K times the integral from 0 to 2 pi times the integral from 0 to 10. We'll have one minus We'll have 1 -2. Okay, over 20 times are D R D theta right? We can split this integral. We can split this and thrill into two. One guy. Okay and multiply that together. Yeah. Yeah. And so we'll carry out the integration. The integral of one is just too high and we'll have here are squared over two minus R cubed over 60 Going from 0 to 10 Which results in a probability of of 200 pi over three K. Mhm. So let's now let's consider let's consider the edge case. And just for this case we'll have We'll set X not equal to 10 for instance let's make our life easier. Mhm. That means that are integral will become Okay. Times you had to go from syrup, times you enter all over the region. R Will have 1 -1/20 Times The Square Root of X -10 squared minus y squared integrated with respect to the differential area. And once again we'll convert this to polar coordinates. Mhm. So that means that this will be equal to K. Times integral from 0 to 2 pi integral from 0 to 10. And we'll have 1 -1/20. Mhm. Times yeah. Our times co sign of data minus 10 squared plus R squared times the sine of data. So this will be the entire ensure that we have to integrate. Yes. And I did this on the calculator And you get that this is approximately equal to 1:36 K. Uh huh. Clearly I would like to live on the edge because the chance that I get exposed is a lot less than if I was to live at the center. That's how you do this question.

Someone is normal, they would have a mean number of white blood cells in a particular volume of blood of 7500 with a standard deviation of 1750. That is quite a big standard deviation. And they talk about the idea that if you have less than 3500 that you are considered to have leukemia. And so we want to know if you have these are these values, these parameters, what's the likelihood you take one sample and it's less than 3500. So we know we want to convert that to a Z value and we're going to take the 3500 minus the 7500, which is negative 4000, and then divided by the standard deviation and so negative 4000, divided by the 17 50 that comes up to be a Z value, uh negative 2.2 around it. Tonight, I'm going to use a table in my book negative 2.29 and looking across that probability is point 0110 So it's not real likely that this is truly your mean, that you're going to run that low in white blood cell count. Now, what happens if you take two samples and find the average? What's the likelihood that that averages less than 3500? Well, now we're going to convert that to the Z value and we're still going to have 3500 minus 7500. Which is going to be that negative 4000 divided by. But now our standard deviation is reduced. Yeah. Yeah. Let me do that. And so let's find out what that Z value becomes. That Z value, the negative 4000 troops can't type negative 4000 divided by. And we have that 17 50 divided by the square root of two. And that comes out to a Z value of negative 3.23 And the likelihood of that happening is only 0.6 Very unlikely if you're mean is actually 7500 to get the average of two readings to be less than 3500. So what happens if we increase that sample size 23 times during the week and we know that that's going to take that Z value which we're still going to have that difference of 4000 in the top. And now our standard deviation is going to get even smaller. Yeah. And let's see what that becomes. Okay. That's the value becomes negative 3.96 And my chart does not go up that. So I'm going to actually use normal C D E. F. To get that answer. And I'm gonna put my lower limit as negative like 1000 and I'll use that value in my calculator that I have. And that comes out to be about 0.1234 zeros and 38 So we can see that as you increase the sample size, you decrease the likelihood of getting that type of value if you're mean is actually 7500. So if someone does have that from a reading of three, we would definitely think that the person has that loop Luca pina. Right? So this you would be very safe to say if someone scored that low on the average of three readings that they had a problem with that that disease.

In this question, we're told that 20% of individuals have an adverse reaction to a particular drug, so we can see that probability is 0.2 and the drug is administered two individuals, one after another until the first adverse reaction. So we're looking at the number of trials until we have exactly one success, and we're as to define an appropriate random variable and use its distribution to answer the questions and the questions. A through E all pertained to how many trials must occur before we have one adverse reaction. So we can define our random variable X as the number of drug administrations until the first adverse reaction. So if you're thinking this is a negative binomial distribution, that is right, it's also a geometric random variable. And that's just a special case of the negative binomial. When R is equal to one so far, a geometric distribution, the probability mass function is simply one minus p to the exponents, X minus one times p and that is for X is a positive integer for part. A were asked the probability that when the experiment terminates that four individuals have not had adverse reactions, so that means we have the first four individuals do not have a reaction, and then the fifth one has reaction, and that's when it terminates. So this can only be the probability that we have five trials. That's four no reactions, followed by one adverse reaction. So it's 0.8, which is one minus p, and this probability comes out to 0.82 in part B. We're looking for the probability that the drug is administered to exactly five individuals. So this is the same question. Is part a just worded in a different way. If you have four non adverse reactions, followed by one adverse reaction that is the same as having the drug administered to a total of five people for part C, we're asked the probability that at most, four individuals do not have an adverse reaction. That is the same as the probability that at most, five drug administrations take place before it ends. So if we have at most five people being administered the drug, we know that at most, four have had no reaction. So this is the summation from X equals 1 to 5 of 0.8 to the exponents X minus one time 0.2 and this probability is 0.672 for part D were asked how many individuals we would expect do not have an adverse reaction And how many individuals would you expect to be given the drug? So the number that we expect to be given the drug will be the number without an adverse reaction, plus one. So the expected value for X This is the expected number of drug administrations in total for a geometric random variable is given by one overpay. So we expect five people to have been administered the drug, and therefore we expect four people to have not had an adverse reaction, and for party were asked for the probability that the number of admit of individuals given the drug is within one standard deviation of what we expect so mathematically, this could be written as what we expect is the mean, so minus one standard deviation. We're saying, What is the probability that X is within one standard deviation of the mean like that? And so we already know the mean That's from part D. We just have to calculate the standard deviation So the variance on X for a geometric random variable is one minus p over P squared, which comes out to 20. So if we take the square root of that, we get the standard deviation. That's 4.47 So we have that. We're looking for the probability that five minus 4.47 is less than or equal to x. Just listen or equal to five plus 4.47 I recall that X must be an integer. So this comes out to the probability that one is less than X lessner equaled X, which is less than or equal to nine. So the number of trials What is the probability that the number of trials is between one and nine inclusive, more than nine would be more than one standard deviation away from the mean and less than one would be more than one standard deviation away from the mean. Of course, we can't actually have less than one, because this is only defined for excess one or greater. You have to have at least one trial so we can write this as a summation for X equals one 3 to 9 for our probability mass function. And this calculation comes out to 0.866


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