Discussion 1 Statistics 103 Spring 2016 Instructions n i=1 xi = 100, n (2xi + 10) and i=1 4. (Level 2) Suppose n = 300. Find Hi Everyone. Here are the discussion/lab problems. The TAs will go over the solutions in detail during the discussion section (and hopefully have enough extra time to take general questions). There are also extra practice problems for you to do at home. I've tried to assigned each problem a diculty rating: level 1, 2 or 3. This is somewhat subjective but I'd like to think that the levels correspond to easy, moderate and hard. I will try to post these discussion problems a few days in advance of the discussion sections on Tuesdays. Try to work them before the TA goes over them. This will help you think of questions to ask the TAs. After the discussion section I will upload some answers and detailed solutions. I don't have solutions to all the questions but hopefully enough that you can gure out how to do many of the problems on your own. and Answer: 3200; 34608 5. (Level 2) Show that n i=1 (xi 1 n x)2 = 2 + x 1 n n i=1 x2 i At home practice 1. (Level 2) In a particular geographical area, the demand function for each person i for a particular good is d qi = 100 + .2xi 9p d where qi is quantity demanded for person i, xi is person i's income and p is the commodity's price. For rms in this area the supply function for this good is s qj = 50 + 20p Topics covered 1. Practice with n 2 i=1 xi = 152 n 2 i=1 (2xi + 10) . n k=1 2. Linear properties of s where qj is the quantity supplied of the good by rm j. There are 5 rms and 100 consumers. Total income of the consumers is $100, 000. Equilibrium occurs where notation. 1 n n k=1 . 100 5 d qi = i=1 Discussion section problems s qj i=1 (i.e. , market demand equals market supply). Find the equilibrium price. 1. (Level 1) Suppose x = 11 and dene yi = 2xi 5. Find y . Answer: $30.25 Answer: 17 2. (Level 1) Suppose x1 = 2, x2 = 1 and x3 = 0. Compute 2. (Level 1) Suppose that the tax ti paid by individual i is 3 i=1 the following function of his or her income xi : 5xi + 2 and 3 1 i=1 xi +2 Answer: 11; 1.75 ti = 0.1(xi 2000) 3. (Level 2) Suppose yi = 2xi 5. Find (i.e. , there is a at tax of 10% on all income over $2000). You know that total income in a geographical area containing 500 individuals is $5, 000, 000. What will the total tax revenue be? 1 n 1 n n 2 i=1 (xi x) n 2 i=1 (yi y ) . = 1202 and dene Answer: 4808 4. (Level 1) There are three wage earners in a family (Mom, Dad, and one child). Their earnings are $20, 000, $15, 000 and $3000 per year, respectively. Taxes for each person are Answer: 400,000 3. (Level 1) If I stack 100 books on-top of each-other and the ti = 200 + .1xi + .00001x2 . i average width of the books is 1.2 inches, then how high will this stack be in centimeters? Find the total taxes paid by this family per year. Answer: 304.8cm Answer: $ 9540 1 5. (Level 1) If last weeks average temperature was 65o Fahren- 18. (Level 1) Suppose x = 1202 and dene yi = 2xi + 5. Find heit, what was the average temperature in Celsius? y. Answer: 18.33 Answer: 2409 19. (Level 2) Suppose 6. (Level 1) If last months average temperature was 65o 1 n Fahrenheit, what was the average temperature in Celsius? dene yi = axi + b. Answer: 18.33 n 2 i=1 (xi x) = v and n 1 Find n i=1 (yi y )2 . Answer: a2 v n i=1 1 n x2 = 10 and i n 1 2 Dene yi = 2xi + 1 and nd n i=1 yi . 7. (Level 2) Suppose 1 n n i=1 20. (Level 2) Suppose x = w and dene yi = axi + b. Find y . xi = 2. Answer: aw + b Answer: 49 n i=1 xi = n (xi x)2 . i=1 8. (Level 2) If what is 100, n i=1 x2 = 152 and n = 400 i Answer: 127 9. (Level 3) True or False: does there exist a list of numbers n i=1 x1 , . . . , xn such that n = 40. n i=1 xi = 100, Answer: False (hint: should tive?) n i=1 (xi x2 = 152 and i x)2 always be posi 10. (Level 2) For any list of numbers x1 , x2 , . . . , xn what is the n i=1 (xi value of x)? Answer: 0 1 n Find 1 n n i=1 x2 = 11 and dene yi = 2xi . i n i=1 xi = 10 and dene yi = 2xi . 1 n 2 yi . n i=1 1 n 11. (Level 1) Suppose n i=1 x2 = 10 and dene yi = 2xi . i Answer: 44 12. (Level 1) Suppose Find y . Answer: 20 13. (Level 1) Suppose Find 1 n n i=1 2 yi . Answer: 40 14. (Level 2) Suppose Find 1 n n i=1 (yi 1 n 2 n 2 i=1 (xi x) = 10 and dene yi = 2xi . y) . Answer: 40 15. (Level 2) Suppose yi = 2xi + 5. Find 1 n 1 n n 2 i=1 (xi x) n 2 i=1 (yi y ) . 1 n n 2 i=1 (xi x) = n 2 i=1 (yi y ) . = 1202 and dene Answer: 4808 16. (Level 2) Suppose 10xi 5000. Find 1 n 45 and dene yi = Answer: 4500 17. (Level 1) Suppose Find 1 n n i=1 2 yi . 1 n n i=1 x2 = 120 and dene yi = 2xi . i Answer: 480 2 1. , Solution Person i d qi =100+ 0.2 x i9 p The supply s q j=50+20 p There are 5 firms and 100 customers and the total income of consumers is 100000 Therefore 100 xi=100000 (to get the value of person i's income) x i=1000 Now equating the two equation for the firm and consumers, we have 100 ( 100+0.2 (1000 )9 p )=5 (50+20 p ) 10000+20000900 p=250+ 100 p p=30.25 2. Solution x 1=2, x 2=1, x 3=0 3 5 x i+ 2=( (5 2)+2 ) +( (5 1)+ 2 )+( ( 5 0 )+ 2) i=1 11 3 1 1 1 x 1 = 2+2 + 1+2 + 0+2 = 7 +2 4 i=1 i 1.75 3. Solution n 1 x ( xi )2=1202 n i=1 y i=2 xi 5 n Substituting n y i=2 xi 5 in n 1 ( y y )2 we get; n i=1 i 1 x ( y y )2= 1 ( 2 x i5( 2 5 )) 2= n i=1 i n i=1 n n 1 x ( 2 x i2 x )2 = 1 4 ( xi ) 2 n i=1 n i=1 Taking out the constant n 4 1 x ( x )2 n i=1 i 4808 4. t i =200+ 0.1 x i +0.00001 x 2 i Therefore, t=200+0.1 20000+0.00001 ( 20000 )2 200+0.1 15000+0.00001 ( 15000 )2200+ 0.13000+ 0.0000 9540 5. Solution Fahrenheit to Celsius conversion Week's average in Fahrenheit is 65o The relationship between Temperature in Celsius and temperature in Fahrenheit is given by T ( f 32) 5 9 T c = ( 6532 ) 5 9 18.33 6. Solution Fahrenheit to Celsius conversion Month's average in Fahrenheit is 65o The relationship between Temperature in Celsius and temperature in Fahrenheit is given by T ( f 32) 5 9 T c = ( 6532 ) 18.33 5 9 7. Solution n n 1 x 2=10 , 1 x i=2 n i=1 i n i=1 y i=2 xi +1 Find n 1 y2 n i=1 i Substituting y i=2 xi +1 in this equation, we get n 1 ( 2 x i+1 ) 2 n i=1 Expanding ( 2 xi +1 ) 2 2 We get 4 x i +4 xi +1 n 1 4 x2 + 4 x i+1 n i=1 i Summing up each part we have (4 10)+ ( 4 2 )+ 1 49 8. Solution n n i=1 i=1 x i=100 , x 2=152 ,n=400 i n x ( xi )2=? i=1 x = 100 100 1 = = n 400 4 n ( i=1 x i n 1 2 1 = x 2 xi (when expanded) i 4 4 i=1 ) Now taking the summation of each part; 1 152 100=127 4 9. Solution False, no list of numbers exists. n x ( xi )2 Should always be positive i=1 x2 Expanding ( x i ) we get; x 22 x i + x 2 x i n x x x 22 x i + 2=152500+2.52 i i=1 This gives a negative value which implies that no list of numbers exists. 10. Solution If we pick any random numbers to prove this e.g. 1, 2, 3, 4, 5 to represent x1 , x2 , x3 , x4 ... ... ... . Then x =3 5 Now ( x i x ) =( 13 ) +( 23 )+ (33 ) +( 43 )+ ( 53 ) i=1 The answer is 0 11. Solution and define =0 2 2 2 y i =( 2 x i) =4 x i (substituting and opening the brackets) n 1 y2 n i=1 i Now taking out the constant, we have n 4 1 x2 n i=1 i 4 11 44 12. Solution y =2 (i.e. the relation between y and x remains the same) x y =2 10 20 13. Solution is? Now substituting y i=2 xi n we get; n 1 ( 2 x i )2=4 1 x 2 n i=1 n i=1 i 4 10 40 14. Solution is? Substitute y i=2 xi ( y i ) 2 y ( 2 xi2 )2=4 ( x i )2 x x n 1 x2 = 4 n ( x i ) i=1 4 10 40 15. Solution Substitute We get 2 y i=2 xi +5 y in ( y i ) and expanding 2 x x2 x2 ( 2 xi +5(2 +5)) =( 2 xi 2 ) =4 ( x i ) n n 1 x ( y i y )2=4 1 ( x i )2 n i=1 n i=1 4 1202 4808 16. Solution Substitute y i=10 x i5000 2 y in ( y i ) 2 and expanding 2 x x x We get (10 x i5000(10 5000) ) =(10 xi +10 ) =100 ( x i ) n n 1 ( y i y )2=100 1 ( x i x )2 n i=1 n i=1 100 45 4500 2 17. Solution Substitute y i=2 x i in ( y i )2 and expanding 2 2 We get (2 x i ) =4 ( x i ) n n 1 ( y )2=4 1 ( x i )2 n i=1 i n i=1 4 120 480 18. Solution y =2 + 5 (since the relation of x and y remains the same) x y =2 1202+5 2409 19. Solution Substitute y i=a x i +b in ( y i )2 y and expanding We get 2 2 2 x x x ( a x i +b(a +b) ) =( a xi a ) =a ( x i ) n 2 n 1 x ( y y )2=a2 1 ( xi ) 2 n i=1 i n i =1 a2 v 20. Solution y =a x +b (since the relationship between y and x remains the same) but =w x Therefore, y =aw+b 1. , Solution Person i d qi =100+ 0.2 x i9 p The supply s q j=50+20 p There are 5 firms and 100 customers and the total income of consumers is 100000 Therefore 100 xi=100000 (to get the value of person i's income) x i=1000 Now equating the two equation for the firm and consumers, we have 100 ( 100+0.2 (1000 )9 p )=5 (50+20 p ) 10000+20000900 p=250+ 100 p p=30.25 2. Solution x 1=2, x 2=1, x 3=0 3 5 x i+ 2=( (5 2)+2 ) +( (5 1)+ 2 )+( ( 5 0 )+ 2) i=1 11 3 1 1 1 x 1 = 2+2 + 1+2 + 0+2 = 7 +2 4 i=1 i 1.75 3. Solution n 1 x ( xi )2=1202 n i=1 y i=2 xi 5 n Substituting n y i=2 xi 5 in n 1 ( y y )2 we get; n i=1 i 1 x ( y y )2= 1 ( 2 x i5( 2 5 )) 2= n i=1 i n i=1 n n 1 x ( 2 x i2 x )2 = 1 4 ( xi ) 2 n i=1 n i=1 Taking out the constant n 4 1 x ( x )2 n i=1 i 4808 4. t i =200+ 0.1 x i +0.00001 x 2 i Therefore, t=200+0.1 20000+0.00001 ( 20000 )2 200+0.1 15000+0.00001 ( 15000 )2200+ 0.13000+ 0.0000 9540 5. Solution Fahrenheit to Celsius conversion Week's average in Fahrenheit is 65o The relationship between Temperature in Celsius and temperature in Fahrenheit is given by T ( f 32) 5 9 T c = ( 6532 ) 5 9 18.33 6. Solution Fahrenheit to Celsius conversion Month's average in Fahrenheit is 65o The relationship between Temperature in Celsius and temperature in Fahrenheit is given by T ( f 32) 5 9 T c = ( 6532 ) 18.33 5 9 7. Solution n n 1 x 2=10 , 1 x i=2 n i=1 i n i=1 y i=2 xi +1 Find n 1 y2 n i=1 i Substituting y i=2 xi +1 in this equation, we get n 1 ( 2 x i+1 ) 2 n i=1 Expanding ( 2 xi +1 ) 2 2 We get 4 x i +4 xi +1 n 1 4 x2 + 4 x i+1 n i=1 i Summing up each part we have (4 10)+ ( 4 2 )+ 1 49 8. Solution n n i=1 i=1 x i=100 , x 2=152 ,n=400 i n x ( xi )2=? i=1 x = 100 100 1 = = n 400 4 n ( i=1 x i n 1 2 1 = x 2 xi (when expanded) i 4 4 i=1 ) Now taking the summation of each part; 1 152 100=127 4 9. Solution False, no list of numbers exists. n x ( xi )2 Should always be positive i=1 x2 Expanding ( x i ) we get; x 22 x i + x 2 x i n x x x 22 x i + 2=152500+2.52 i i=1 This gives a negative value which implies that no list of numbers exists. 10. Solution If we pick any random numbers to prove this e.g. 1, 2, 3, 4, 5 to represent x1 , x2 , x3 , x4 ... ... ... . Then x =3 5 Now ( x i x ) =( 13 ) +( 23 )+ (33 ) +( 43 )+ ( 53 ) i=1 The answer is 0 11. Solution and define =0 2 2 2 y i =( 2 x i) =4 x i (substituting and opening the brackets) n 1 y2 n i=1 i Now taking out the constant, we have n 4 1 x2 n i=1 i 4 11 44 12. Solution y =2 (i.e. the relation between y and x remains the same) x y =2 10 20 13. Solution is? Now substituting y i=2 xi n we get; n 1 ( 2 x i )2=4 1 x 2 n i=1 n i=1 i 4 10 40 14. Solution is? Substitute y i=2 xi ( y i ) 2 y ( 2 xi2 )2=4 ( x i )2 x x n 1 x2 = 4 n ( x i ) i=1 4 10 40 15. Solution Substitute We get 2 y i=2 xi +5 y in ( y i ) and expanding 2 x x2 x2 ( 2 xi +5(2 +5)) =( 2 xi 2 ) =4 ( x i ) n n 1 x ( y i y )2=4 1 ( x i )2 n i=1 n i=1 4 1202 4808 16. Solution Substitute y i=10 x i5000 2 y in ( y i ) 2 and expanding 2 x x x We get (10 x i5000(10 5000) ) =(10 xi +10 ) =100 ( x i ) n n 1 ( y i y )2=100 1 ( x i x )2 n i=1 n i=1 100 45 4500 2 17. Solution Substitute y i=2 x i in ( y i )2 and expanding 2 2 We get (2 x i ) =4 ( x i ) n n 1 ( y )2=4 1 ( x i )2 n i=1 i n i=1 4 120 480 18. Solution y =2 + 5 (since the relation of x and y remains the same) x y =2 1202+5 2409 19. Solution Substitute y i=a x i +b in ( y i )2 y and expanding We get 2 2 2 x x x ( a x i +b(a +b) ) =( a xi a ) =a ( x i ) n 2 n 1 x ( y y )2=a2 1 ( xi ) 2 n i=1 i n i =1 a2 v 20. Solution y =a x +b (since the relationship between y and x remains the same) but =w x Therefore, y =aw+b