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Let Xand Y be random variables with E(X) 20, E(Y) = 8, Sd(Xi 7, SD(Y) 2, Corr(x,Yi-0,6 Find E(2x-Tv) and SD(?x-Zv) Let X and be random varlables with SD(X) = 5, SD(...

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Let Xand Y be random variables with E(X) 20, E(Y) = 8, Sd(Xi 7, SD(Y) 2, Corr(x,Yi-0,6 Find E(2x-Tv) and SD(?x-Zv) Let X and be random varlables with SD(X) = 5, SD(Y) = 7, SD(2x+Sv)-27. Find Cou(x,YJ:A college sophomore is taking courses in a semester He is willing to assume that he MI Bet an A with probability of 0.1, B with probability of 0.8,and a € with probability of 0.1, He fines X1, X2 and X3 to be number of A's, 8*$ and C' $ respectively; that he gets in this semester What Is

Let Xand Y be random variables with E(X) 20, E(Y) = 8, Sd(Xi 7, SD(Y) 2, Corr(x,Yi-0,6 Find E(2x-Tv) and SD(?x-Zv) Let X and be random varlables with SD(X) = 5, SD(Y) = 7, SD(2x+Sv)-27. Find Cou(x,YJ: A college sophomore is taking courses in a semester He is willing to assume that he MI Bet an A with probability of 0.1, B with probability of 0.8,and a € with probability of 0.1, He fines X1, X2 and X3 to be number of A's, 8*$ and C' $ respectively; that he gets in this semester What Is probabillty that he gets 1 A Given that he Bot . 2C's



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Exercise 28 introduced random variables $X$ and $Y,$ the number of cars and buses, respectively,
carried by a ferry on a single trip. The joint pmf of $X$ and $Y$ is given in the table in Exercise $9 .$ It is
readily verified that $X$ and $Y$ are indepent.
(a) Compute the expected value, variance, and standard deviation of the total number of vehicles on a single trip.
(b) If each car is charged $\$ 3$ and each bus $\$ 10,$ compute the expected value, variance, and standard deviation of the revenue resulting from a single trip.

In this problem. We're looking at an instructor who has a class of students and is grading papers from the information that the problem provides. We know that these stand simple size is 40 students and that the mean grading time for each paper is six minutes, and the standard deviation for the grading time is also six minutes. Because the sample size 40 is greater than 30 we know that the central limit theorem also applies, and that means that the total time spent creating the papers, let's call it Big T is an approximately normal distribution. And because of that, we can then calculate the mean and the standard deviation for the total time distribution, which will be, well, Call me a T, and if I t. This is based on the sample size mean and standard deviation, so the equations will be and times the sample mean and the square root of n times the sample standard deviation, which, when you plug in the numbers, comes out to be 240 minutes, along with approximately a standard deviation of 37.9 five. So for the first part of the problem, we are trying to determine the probability that in grading all these papers and starting at 6 50 um, he will be able to watch his show, which will be at 11 o'clock if we transferred this or Trans, convert this to numbers or convert this two minutes. We are essentially trying to find the probability that the total time spent grading t is going to be less than 250 minutes, which is the time between 6:50:11 p.m. Which is when he would be grading and before the TV show starts. And using the central limit theorem and Z a standard normal random variable. We can then approximate this probability to be the probability that Z is going to be less than T minus the total meat divided by the total standard deviation. And once we plug in the numbers, these numbers, um, we have the probability of Z being less than 250 minus 2 40 which is the mean, um, time spent creating total time spent grading and 30 divided by 37.9 five. This comes out to be zero 0.2 six cool, and if we refer to the Z score table that the probability that we get from there is zero point 60 to 6, or approximately 60 0.3%. This means that the probability that the lecture or the instructor is going to be able to finish grading before his show starts is approximately 60.3% for the second problem. We are looking at the probability that he will miss time or essentially miss part of his show, and from the second part tells us that the show starts at 11 10 PM rather than 11 p.m. And based on that and converted to numbers, we are trying to find the probability that T will be greater than 260 minutes, which is the time between 6 50 11 10 PM, which then approximates to the probability that Z a random, a standard normal random variable will be greater than T minus mut divided by by T and plugged in that we get probability that Z is less sorry. Greater than 260 minus 2 40 divided by 37 0.95 Since we're looking at greater than we are looking at the right side of the normal distribution, which means that with respect to disease score, we're going to have one minus the probability for that Z score which calculated out, which is this value is going to be zero point five two seven and looking to the Z score table and looking at 0.53 on the Z score table, we find that this equates to one minus 0.7019 which comes out to be 0.2 98 one, which means that there is approximately a 29.8% chance that the instructor will miss part of his show while grading all of his papers.

In this problem. We are looking at an instructor who has a class of students and it's creating papers from the problem. We know that the sample size is 40 students so and equals 40. And we also know that the mean grading time spent on each paper or mu is six minutes. Finally, we know that the standard deviation for the mean creating time is also six minutes. Because the sample size 40 is larger than a 30. We know that the central limit theorem is applicable, which means that the total time spent creating let's call it Big T is has an approximately normal distribution. And from that we also can find the mean and the standard deviation for that for big T, which is, um, you t and fi t these are not. These are directly related to the sample means and standard sample standard deviations, and we find them using the equations for the mean total time spent grading being the samples, times the sample mean and for the standard deviation. We have the square root of these number of samples times the sample standard deviation, and when we calculate these out, he's come out to be 240 minutes for the sample. Ah, for the mean total Times bank rating and we also have approximately 37.95 for the standard deviation of the time spent total time spent reading in the first part of the problem. We're trying to determine the probability that creating 40 papers the instructor will be done by 11 PM for the news and translating this two minutes in actual numbers were essentially trying to figure out the probability that the total time spent grading or big T is going to be less than 250 minutes, which is the time before 6 50 between 6, 50 and 11 o'clock, which is the time that ah, he starts grading and the time that he hopes to be done grading and using the central Limit theorem. And with Z being a standard normal random variable, we can approximate this probability to the probability that Z is less than the target time where the probability of time that we're examining, which is 2 50 minus the total time spent ratings mean divided by the standard deviation and using the numbers that we have, we get the probability that Z it's less than yeah, zero point 264 or which then we can use the Z score to figure out the corresponding probability and from the Z score table at the end of the book, we can find out that this is the score of 0.264 is correlates with a probability of 0.602 six, which means that there is approximately a 60.3% chance that the instructor will be able to finish grading before his show for the second part of the problem. Similarly, we're trying to determine the probability that he will actually miss part of the show, and we find out that the specific segment of the show he wants to watch starts at 11 10. So translated to numbers, we are looking at the probability that big T, the total time spent grading is going to be greater than 260 minutes and similarly to the previous part. We use the same equation with Z being an approximately normal random variable, but we replace the number that we were looking at 2 50 with 260 minutes instead which is the probability that the is going to be greater than 0.5 to 7. Since we're looking at the right side of the distribution, we have to subtract the probability from one so trans this translates to one minus the probability associated with disease score of zero point five 27 When we refer to disease Score table, we then find out that this becomes one minus 0.7 oh 19 which is 0.2981 or approximately 29 point heat percent, which means that the instructor has a 29.8% chance of missing his at least some time of his part of the show that he wants to watch.

In this problem. We are looking at an instructor who has a class of students and it's creating papers from the problem. We know that B sample size is 40 students so and equals 40 and we also know that the mean grading time spent on each paper or mu is six minutes. Finally, we know that the standard deviation for the mean creating time is also six minutes. Because the sample size 40 is larger than a 30. We know that the central limit theorem is applicable, which means that the total time spent creating let's call it Big T is has an approximately normal distribution. And from that we also can find the mean and the standard deviation for that for big T, which is, um, you t and fi t these are not. These are directly related to the sample means and standard sample standard deviations, and we find them using the equations for the mean total time spent grading being the samples, times the sample mean and for the standard deviation. We have the square root of these number of samples times the sample standard deviation, and when we calculate these out, he's come out to be 240 minutes for the sample. Ah, for the mean total Times bank rating and we also have approximately 37.95 for the standard deviation of the time spent total time spent reading in the first part of the problem. We're trying to determine the probability that creating 40 papers the instructor will be done by 11 PM for the news and translating this two minutes in actual numbers were essentially trying to figure out the probability that the total time spent grading or big T is going to be less than 250 minutes, which is the time before 6 50 between 6, 50 and 11 o'clock, which is the time that ah, he starts grading and the time that he hopes to be done grading and using the central Limit theorem. And with Z being a standard normal random variable, we can approximate this probability to the probability that Z is less than the target time where the probability of time that we're examining, which is 2 50 minus the total time spent ratings mean divided by the standard deviation and using the numbers that we have, we get the probability that Z it's less than yeah, zero point 264 or which then we can use the Z score to figure out the corresponding probability and from the Z score table at the end of the book, we can find out that this is the score of 0.264 is correlates with a probability of 0.602 six, which means that there is approximately a 60.3% chance that the instructor will be able to finish grading before his show for the second part of the problem. Similarly, we're trying to determine the probability that he will actually miss part of the show, and we find out that the specific segment of the show he wants to watch starts at 11 10. So translated to numbers, we are looking at the probability that big T, the total time spent grading is going to be greater than 260 minutes and similarly to the previous part. We use the same equation with Z being an approximately normal random variable, but we replace the number that we were looking at 2 50 with 260 minutes instead which is the probability that the is going to be greater than 0.5 to 7. Since we're looking at the right side of the distribution, we have to subtract the probability from one so trans this translates to one minus the probability associated with disease score of zero point five 27 When we refer to disease Score table, we then find out that this becomes one minus 0.7 oh 19 which is 0.2981 or approximately 29 point heat percent, which means that the instructor has a 29.8% chance of missing his at least some time of his part of the show that he wants to watch.


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