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n case you're wondering what the abbreviations stand for: CLT: central limit theorem TTE: Time to the event (i don't think we took this in class though, could be there for no reason) GCVF: Generalized change of variable formula  CBS: Cauchy bunyakowski schwarz.   If there is an abbreviation you don't understand, please ignor

n case you're wondering what the abbreviations stand for: CLT: central limit theorem TTE: Time to the event (i don't think we took this in class though, could be there for no reason) GCVF: Generalized change of variable formula  CBS: Cauchy bunyakowski schwarz.   If there is an abbreviation you don't understand, please ignor

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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In case you're wondering what the abbreviations stand for:

CLT: central limit theorem

TTE: Time to the event (i don't think we took this in class though, could be there for no reason)

GCVF: Generalized change of variable formula 

CBS: Cauchy bunyakowski schwarz.

 

If there is an abbreviation you don't understand, please ignore it because it might not mean anything. 

(1) We select n = 50 real numbers at random and then round them off to the nearest integers, before adding them all up. Find an approximation to the probability that the difference between the sum of the rounded numbers and the sum of the original numbers is no more than 3. Solve this problem by taking the following three steps one by one. Step (i) Let Let Z1 , Z2, for instance 13.49 and will be rounded to 13 and 13.51 will be rounded to 14. You may round up or down the number 13.50, which carries zero probability anyway. Why is it reasonable to assume that the rounding error , Zn be the original numbers and let N1, N2, ·.. , Nn be their rounded off version. That is, iid U; = Z; - N; Uniform(-,), i = 1,2, .. , n? The correct answer is CLT TTE CBS GCVF None of the above N/A (i- Select One) Step (ii) We want to compute the probability P (| E Z¡ – E1 Ni| < 3), which is equal to (ωP ( ΣU < 3 ) . () P ( Σ UI 3). (c) P (E, U; < 3). (d) P (E, Ui 2 3). i=1 =1 i=1 (e) None of the above The correct answer is (a) (b) (c) (d) (e) N/A (ii- Select One) Step (iii) The desired probability of Step (ii) is approximately equal to P(|N(0, 1)| < 3/12/50) = 0.86 due to TTV CBS GCVF CLT None of the above N/A (iii- Select One)
Transcribed Image Text:(1) We select n = 50 real numbers at random and then round them off to the nearest integers, before adding them all up. Find an approximation to the probability that the difference between the sum of the rounded numbers and the sum of the original numbers is no more than 3. Solve this problem by taking the following three steps one by one. Step (i) Let Let Z1 , Z2, for instance 13.49 and will be rounded to 13 and 13.51 will be rounded to 14. You may round up or down the number 13.50, which carries zero probability anyway. Why is it reasonable to assume that the rounding error , Zn be the original numbers and let N1, N2, ·.. , Nn be their rounded off version. That is, iid U; = Z; - N; Uniform(-,), i = 1,2, .. , n? The correct answer is CLT TTE CBS GCVF None of the above N/A (i- Select One) Step (ii) We want to compute the probability P (| E Z¡ – E1 Ni| < 3), which is equal to (ωP ( ΣU < 3 ) . () P ( Σ UI 3). (c) P (E, U; < 3). (d) P (E, Ui 2 3). i=1 =1 i=1 (e) None of the above The correct answer is (a) (b) (c) (d) (e) N/A (ii- Select One) Step (iii) The desired probability of Step (ii) is approximately equal to P(|N(0, 1)| < 3/12/50) = 0.86 due to TTV CBS GCVF CLT None of the above N/A (iii- Select One)
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