Joint and Marginal Probability Distributions
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Person takes bus & subway to work. Bus runs every 20 min (X) & subway every 4 min (Y). Assume timing of the bus and subway are independent and uniform.
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Please help find the joint and marginal distributions for this problem. Since they are independent I know that f(x,y) = fx(x) * fy(y) but do not understand how to take that and the data given in the problem to find the marginal and joint distribution. I also have to find the Expected value of X & Y.
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TA : The question is not well-defined. What are X and Y? The waiting time?
X is how often the bus runs 20 (waiting time 0 < X < 20)
Y is how often the subway runs 4 (waiting time 0 < Y < 4).
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Solution Summary
Joint and marginal probability distributions are investigated. The solution is detailed and well presented.
Education
- BSc , Wuhan Univ. China
- MA, Shandong Univ.
Recent Feedback
- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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