ec41s10_hw3_sol

# ec41s10_hw3_sol - Kata Bognar [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ Economics 41...

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Kata Bognar [email protected] Economics 41 Statistics for Economists UCLA Spring 2010 Homework Assignment 3. - solutions by Minji Kang - NOTE: Please show your calculations for Questions 5-6,9-10,13-16. The exercises are from the textbook (Weiss: Introductory Statistics). Random Variables Basics. 1. The probability distribution table of a discrete random variable lists: (a) the bottom half of the values that the random variable can assume and their corre- sponding probabilities (b) all the values that the random variable can assume and their correspond- ing probabilities (c) all the values that the random variable can assume and their corresponding frequen- cies (d) the top half of the values that the random variable can assume and their corre- sponding probabilities 2. A continuous random variable is a random variable that can: (a) assume only countable values (b) assume any value in one or more intervals (c) have no random sample (d) assume no continuous random frequency 3. The probability that a continuous random variable X assumes a single value is always: (a) less than 1 (b) greater than zero (c) equal to zero (d) between zero and 1 4. The following table lists the relative frequency distribution of the number of refrigerators owned by all families in a city. x relative frequency 0 0.02 1 0.65 2 0.26 3 0.07 Find the probability that a randomly selected family owns exactly one refrigerators! P ( X = 1) = 0 . 65 1

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x P(X = x) 2 0.15 3 0.32 4 0.24 5 0.13 6 0.08 7 0.06 8 0.02 Find the mean of the random variable X ! E [ X ] = 2 · 0 . 15 + 3 · 0 . 32 + 4 · 0 . 24 + 5 · 0 . 13 + 6 · 0 . 08 + 7 · 0 . 06 + 8 · 0 . 02 = 3 . 93 6. Solve Exercise 5.11 Answer: (a) { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } (b) { Y = 7 } (c) { Y = 7 } = { (1 , 6) , (2 , 5) , (3 , 4) , (4 , 3) , (5 , 2) , (6 , 1) } P ( Y = 7) =
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## This note was uploaded on 05/31/2010 for the course ECON 41 taught by Professor Guggenberger during the Spring '07 term at UCLA.

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ec41s10_hw3_sol - Kata Bognar [email protected] Economics 41...

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