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#1




ASM 1st Edition (Exam P, Weishaus) Practice Exam 2 Q14, 16,24
Hi,
I'm preparing for the Jan Exam P (2nd attempt) and I was wondering if anyone could help me shed some light on these questions as I've been staring at the answer keys and yet I'm not understanding it clearly. 14) A machine fails when two of its components fail. The failure times of the first component are uniformly distributed on [3,6]. The failure time of the second component, given that the first component failed at time x, is uniformly distributed on [2x,3x]. Calculate the probability that the machine does not fail before time 10. A) 0.52 B) 0.55 C) 0.59 D) 0.62 E)0.65 My thoughts: I draw a graph for y=2x and y=3x and use [3,6] for my x limit and y>10 for my y limit. Get the joint density by 1/(63) * 1/(3x2x) = 1/3x. But I don't know how to use my limits for my integration to find the probability. 16) A charity receives 12 donations in a day. The size of each donation is normally distributed with mean 100 and variance 6400. The charity makes 3 grants each day. Each grant is normally distributed with mean 350 and variance 10,000. The donations and grants are mutually independent. Calculate the probability that the sum of the donations minus the sum of the grants for one day is positive. A) 0.69 B) 0.7 C) 0.71 D) 0.72 E) 0.73 My thoughts: Let x be the sum of all donations and let y be sum of all grants. I found the E[XY] = 150 and Var[xy] = 106800. So then I have to find P[xy>0]. Set z= xy for simplicity. P[z>150/326.8] => p[z<0.46]. But this is the wrong answer. 24) The length of a table is measured. The measurement error is uniformly distributed over [0.05, 0.05]. To get a more accurate measurement, the measurement is repeated n times, and the n measurements are then average. Each measurement is independent of each other measurement. Calculate the minimum number of measurements necessary so that the measurement error is no more than 0.005 in absolute value with probability 95%. A) 90 B) 100 C) 115 D) 122 E) 128 My thoughts: I can set up P[0.005<x<0005]=0.95 where x is measurement error but I don't know how to move forward from here... Thank you in advance. 
#2




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It gets more complicated because the limits (with nonzero density) for the second failure depend on when the first failure was. You can't use 10 to infinity for the inside limits, since there may be 0 density at (x,10). So you need to combine (evaluate separately and add) two double integrals, to make sure that (for example) (4,9) and (4,25) are not included in either but (4,11) is included in one (and not both); similarly for (5.5,10.25), (5.5, 30) and (5.5,12). The limits for the inside integral will be different in the two double integrals. (The limits for the outside limits will probably be different, with no overlapping, but could be the same if you choose the limits on the inside integral to avoid overlapping.) Quote:
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#3




I only have time to help with #14 right now.
Double integrals are not necessary. Let and We want There are 3 cases: Case 1  in this case, X can not exceed 10 so the conditional probability is 0. Note: Case 2  in this case, X could exceed 10 and its interval contains 10 so the conditional probability can be found with the continuous version of the law of total probability. Keep in mind that so our integral in this case will be reduced to that uniform PDF for U. Note: Case 3  in this case, X is certain to exceed 10 since its lower bound does hence the conditional probability is 1. Note: SOLUTION That integral uses because that's the conditional probability that X is at least 10 given the value of U and the other factor is the PDF of U (both conditional on case 2). Does that make sense to you? 
#4




You are correct for #16. As Gandalf suggested: errata
You'll find the quote "On page 480, in the solution to question 16, change Φ(0.459)=0.7057 to Φ(0.459)=0.6768. None of the five answer choices is correct."  For #24 the mean of the average is computed as 0. The variance of the average is computed as ... so the standard deviation is So you want: or look up 0.975 on the normal table and you see it corresponds to z=1.96 so Last edited by Z3ta; 12212017 at 06:21 PM.. 
#5




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I'm still having trouble understanding the limits. I have attached an image that demonstrates how I'm seeing the question. But I don't know how to set the limits near the ? so I can find the probability. Let me know if the link doesn't work or if you don't understand my question. Thank you for identifying the errata. The answer key according to the manual was (E). 
#7




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The answers to the second and third question are good. I really like the idea of not doing the double integrals for the first question. However, I'm don't understand how you broke it up in 3 pieces and how you got those critical points. Also when you say, U E (3, 10/3). I'm not too sure what you mean by that notation. Are you saying U ranging from 3 ~ 10/3? I'm really sorry but could you elaborate a little bit more on why/how you broke this case into 3? I haven't been successful at understanding your explanation. Thank you very much Zeta, and wish you happy holidays! 
#8




OK, that picture is the right way to start, but I'm not sure how the picture works. The line y=2x, y=3x, x=3, x=6 and y=10 are all good lines to have.
It looks like the x=3 line intersects the y=10 line between where it intersects the y=2x line and the y=3x line (even though the intersection with the y=3x line is not shown). Since you have a point labeled (10/3, 10) (a good point to know), I suggest you add the line x=10/3 to your picture. Since you have a point labeled (5, 10) (a good point to know), I suggest you add the line x=5 to your picture, and label all points of intersection among the various lines. Then think about three double integrals. Maybe not the most efficient thing to do, but might help you get to the right result. First, write the double integral that includes all points you want to include between x=5 and x=6. Second, write the double integral that includes all points you want to include between x=10/3 and x=5. Then, write the double integral that includes all points you want to include between x=3 and x=10/3. 
#9




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I worked backwards to determine the numbers 10/3 and 5. I knew that there would be certainty of if since 2U is its lower bound. Similarly, there is certainty that if since 3U is its upper bound. The problem becomes a weighted sum of the conditional probabilities in each of those cases (weighted by the unconditional probability of each case) using the law of total probability. That's the discrete version. The integral itself is an application of the continuous version of the law of total probability (you integrate times the PDF of over the possible values of ). Note: here is really since we're within a specific case of our first use of the law of total probability... but I thought it might be confusing if I wrote that. Does that make sense? Last edited by Z3ta; 12272017 at 09:12 PM.. 
#10




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Thank you very much for your explanation! Happy holidays and a happy new year! 
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