MATH 2070 Mixed Practice KEY Sections (25) 900(.95 )

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1 1. The demand for board games can be modeled by D( p ) = 9(.9) p thousand games where p is the price in dollars per game. Find the consumers surplus when the market price for the board game is $. per game. Show all of your work using the algebraic method. N p D( p) = lim 9(.9 ) N p 9(.9 ) = lim N ln(.9) ( N ) () 9(.9 ) 9(.9 ) ( N ) 9(.9 ) ().9 ) () 9(.9 ) = lim N ln(.9) = lim lim N ln(.9) N 9( = () ln(.9) = ln(.9) ln(.9) thousand dollars N () 9(.9 ) ln(.9) -. Suppose overall demand for kerosene in the United States is given by D( p) = 1.p + 1. million gallons where p is the price per gallon in dollars. Check point: D()=1.1 Is demand for kerosene elastic, inelastic, or unit elastic when kerosene is sold for $.49 per gallon? Show the mathematical justification for your answer. η = p 4 ( 4. p ) 1.p + 1. and when kerosene is sold for $.49 per gallon. 4 ( ).49 4.(.49) η(.49) =.69 < 1 Demand is inelastic 1.(.49) + 1. For what interval of prices is demand for kerosene elastic? 4 ( 4. p ) p Solving η = = 1 p 1.6 Check η(1) = 1. > 1 1.p + 1. Demand is elastic when < p < 1.6 1

2 . The life expectancy (in years) of a certain brand of clock radio is a continuous random variable if x with probability density function f( x) = ( x + ). Find the probability that a otherwise randomly selected clock radio lasts: a. between and years P( x ) = f ( x) dx =.14 b. at most 6 years 6 P( x 6) = f ( x) dx =.7 c. more than 6 years Px ( > 6) = f( xdx ) =. or Px ( > 6) = 1 Px ( 6) = 1.7 =. d. exactly years P( x = ) = f ( x) dx = 6 4. Determine if each of the following could represent probability density functions. Show supporting work. a. b. m(x) x -1 1 r(t) t 1) m( x) for all x? Yes. 1) r() t for all t? ) m( x) dx = 1? r( t) < when. < t < 1 1 m( x) dx = m( x) dx = bh ( 1.)( 1.) = 1.1 Therefore, r(t )is not a P.D.F.. Therefore, m(x) is not a P.D.F.

3 c. 1+ 4x 4 x if x 1 gx ( ) = otherwise 1) g( x) for all x? Yes. (Include a sketch here in your work) ) g( x) dx = 1? 1 g( x) dx = g( x) dx =. Therefore, g(x) is not a P.D.F.. The demand for the Who Wants to Be A Millionaire computer game sold online can be modeled by D( p) = -.9 p + 1 thousand games where $p per game is the price for each game. a. Label the units on the axes of the above graph. b. What is the consumers willingness and ability to spend on the computer game, when 9, games are demanded? Shade this area on the graph and find this amount. Solve D(p) = Pmax = $4.8 per game Solve D(p) = 9 thousand p = $.8 per game CWAS = p q + Pmax po D( p) 4.8 $.8 = ( 9 thousand games ) + D( p) game.8 = $.8 thousand + $ thousand = $ thousand thousand games.8 p dollars per game The demand for digital cameras can be modeled by d( p) = -.7 p -.9 p+ 1 hundred cameras when the price is $p per camera. The supply of these cameras can be described when p < 6 by s( p) = hundred cameras when the price is $p per camera..6 p -.9 p-.1 when p 6 Sketch both graphs on the axes provided to the right. Label each of the following (with specific values when possible): d(p) d(p) s(p) the shutdown price pmax s(p) the equilibrium price & quantity the producer surplus at equilibrium the consumer surplus at equilibrium PS CS

4 7. Evaluate the improper integral values in the table. N 1 1 dx x 1 1 dx x , , N using the numerical method. Report unrounded Conclusion: 1 1 dx x = - 8. In a certain city, the daily use of water (in hundreds of gallons) per household is a continuous.1x random variable with probability density function =.1e if x f x ( ). otherwise Find and interpret each of the following. Shade the corresponding area on the graph. a. 1 f ( x) dx.49 There is a 4.9% chance that a randomly selected household uses between and 1 gallons of water daily. b. f ( x) dx = 1 f ( x) dx.474 There is a 47.4% chance that a randomly selected household uses more than gallons of water each day. 4

5 9. Suppose that h, the total number of hours a student spends each day on Twitter is distributed.h he when h 4 according to the density function Th ( ) =. elsewhere a. Find the average amount of time a student spends on Twitter in a day. 4 µ = h T ( h ) dh = 1.1 hours The average amount of time a student spends on Twitter in a day is 1. hours. b. Find and interpret ( ) Ph>. 4 P( h > ) = T ( h) dh =.1 ; There is a 1.% chance that a randomly selected student spends more than hours on Twitter in a day. c. Find the standard deviation of the total number of hours a student spends on Twitter each day. 4 ( h ) σ = T ( h) dh =.4967 hours σ =.4967 =.61 hours 1. years ago, Colleen started investing into a retirement account at a rate of dollars per year. If the retirement account earned.% annual interest, compounded continuously over the entire -year period, what is the current account balance? Assume that Colleen invested the same amount each year and that the income stream was continuous. a. Explain why this question is asking for the -year future value, NOT the present value. This question is asking for the accumulated value of Colleen s investment and the compounded interest at the end of a -year period. It is not asking for the amount of a one-time investment that would be needed to grow to some future value (i.e. the present value). b. Find the -year future value..( t) $86,89.9 FV = e dt = 11. To save for her child s college education, Raquel would like to invest % of her salary each year into a savings account which earns.% interest, compounded continuously. Write the function R(t) that shows the continuous stream of money flowing into the savings account, assuming Raquel s salary last year was $8, and i. her salary remains constant over the next 18 years. R(t) =.(8) = 19 dollars per year ii. her salary increases by.9% per year over the next 18 years. R(t) =.(8(1.9 t )) = 19(1.9 t ) dollars per year iii. her salary increases by $ per year over the next 18 years. R(t) =.(8 + t) = 19 + t dollars per year

6 1. A Clemson alumnus wants to establish a fund to provide the Math Sciences department with a continuous income stream of R(t) = $, per year to use for scholarships. The fund will earn interest at a rate of 4%, compounded continuously. The amount of the endowment he needs to make now to support the scholarship for T years is equal to the T-year present value of the income stream. Find the amount of the endowment he needs to make now to support the scholarship for 1 years. Round to the nearest whole dollar. 1.4t $6, PV = e dt = 1. For the year ending March 1, 9, sales for Toshiba Corporation were $7.1 billion. Assume Toshiba invests 8.% of their sales amount each year, beginning April 1, 9, and that those investments can earn an APR of 4.8% compounded continuously. Assuming that Toshiba s sales are projected to decrease by 1.% per year for the next five years, find the following: a. Rt (), the function that describes the flow of the company s investments over the next five years. t t ( ( ) ) ( ) Rt ( ) = = billion dollars per year b. the amount invested by Toshiba over the next five years t ( ) dt = 8.74 billion dollars c. the five-year future value t.48( t) ( ) e dt =.1 billion dollars d. the five-year present value.48t e dt =.9 billion dollars t ( ) e. the amount of interest earned over the next five years t.48( t) t ( ) ( ) Interest = FV-Investment = e dt = =.791 billion dollars 6

7 14. The supply for Clemson car flags can be modeled by the following: when p < S( p) = p.19(1.99) + 1 when p flag. hundred flags, where $p per flag is the price of a GRAPH A GRAPH B GRAPH C a. What is the market price when 1, flags are sold? (be careful units are in hundreds!) Solve S(p) = 1 hundred p = $1.87 per flag b. What is the supply of Clemson car flags when the price of one flag is $1? S(1) = 76 hundred flags c. What is the producer revenue when the price of a flag is $1? Shade this region on graph A. PR $1 = ( 76 hundred flags ) = $988 hundred flag d. What the minimum price for which producers will supply Clemson car flags? $ per flag e. What is the producers surplus when the price of a flag is $11? Shade this region on graph B. 11 PS = S( p) = $178. hundred or $17,8.4 f. What is the producer s willingness and ability to receive when the price of a flag is $1? Shade this region on graph C. 1 $1 PWAR = ( S(1) hundred flags ) S( p) flag = $ hundred $ hundred =$17.7 hundred 1 $1 = 11.4 hundred flags S( p) flag = $181.7 hundred $ hundred PWAR ( ) =$17. hundred 7

8 Now consider that the demand for Clemson car flags can be modeled by D( p) = -9 p+ 1 hundred flags where $p per flag is the price of one flag. GRAPH A GRAPH B GRAPH C g. Is there a price at which there will not be a demand for Clemson car flags? If so, what is it? If not, explain. Solve D(P) = P max = $17. per flag h. For what price will flags be sold at the equilibrium point? What is the quantity at this point? p * = $11. per flag q * =.84 hundred flags or,84 flags i. Find the following at market equilibrium: Consumer Expenditure (shade on graph A) * * 11. $ CE = p q =.84 hundred flags flag = hundred $ (99.81 if unrounded) ( ) Producer Revenue * * 11. $ PR = p q = (.84 hundred flags) flag = hundred $ (99.81 if unrounded) Consumer Surplus Producer Surplus (shade on graph B) CS = D( p) PS = 11. S( p) = $1.17 hundred or $1,17. = $196.9 hundred or $19, 6.91 Consumer Willingness and Ability to Spend Producer Willingness and Ability to Receive CWAS = CE + CS = hundred $ PWAR = PR - PS = 4.7 hundred $ (or hundred $ if unrounded) (or 4.78 hundred $ if unrounded) Total Social Gain (shade on graph C) TSG=$196.9 hundred + $1.17 hundred = $1. hundred (or $1. hundred if unrounded surpluses are used) 8

9 1. The research department of a steel manufacturer believes that one of the company s rolling machines is producing sheets of steel of varying thickness. The thickness is uniformly distributed with values between 1 and millimeters. Any sheets less than 16 millimeters must be scrapped because they are unacceptable to buyers. a = 1 and b = a. Calculate the average thickness of the sheets produced by this machine. b a 1 µ = a + = 1 + = 17 mm b. What is the probability density function? Name the function f( x ). 1 if 1 x f( x) = otherwise c. Sketch f( x ). Show the mean on the horizontal axis. μ d. What percentage of the steel sheets produced by this machine have to be scrapped? 1 Px ( < 16) = bh = 1 =. % e. % of the steel sheets produced are thicker than 187. millimeters.. 1. = ( s) 1. = s s = 187. mm s 9

10 16. At a certain grocery checkout counter, the waiting times (in minutes) follow an exponential.4x.4e if x distribution, f( x) =. k =.4 if x < a. What is the average wait time at the grocery checkout counter?.4x ( ).4. min µ = x f x dx = xe dx = 1 1 (or if using rules for exponential distributions: µ = = =. min) k.4 b. What is the probability of waiting between and 4 minutes? (a sketch may be helpful here) 4 P( < x < 4) = f ( x) dx =.474 c. What is the probability of waiting less than 9 seconds? (a sketch may be helpful here) 1. P( x < 1.) = f ( x) dx =.41 d. What is the probability of waiting more than minutes to checkout? (a sketch may be helpful here) Px ( > ) = 1 Px ( < ) = 1 f( xdx ) = = The distribution of heights for all American woman aged 18 to 4 is normally distributed with a mean height of 6. inches and a standard deviation of. inches. Use the Empirical Rule to answer the following questions. a. What percentage of American women aged 18 to 4 is between 7. inches and 7 inches tall? P(7. < x< 7) = + = % σ σ b. What percentage of American women aged 18 to 4 is less than 6 inches tall?.68 Px< ( 6) =. + =.84 84%. 1σ 1

11 c. What percentage of American women aged 18 to 4 is between inches tall and 6 inches tall? P( < x< 6) = = % σ 1σ d. What percentage of American women aged 18 to 4 is less than inches tall? Px ( < ) = =.1.1% 99.7% between and 7 which means there is 1% 99.7% =.% left for the two tails. The two tails are equal, so.%/ =.1% is in each tail. σ 19. Scores on a 1-point final exam administered to all applied calculus classes at a large university are normally distributed with a mean of 7. and a standard deviation of 8.6. What percentage of students taking the test had: a. Scores between 6 and 8? (a sketch may be helpful here) P(6 < x < 8) = normalcdf (6,8, 7,, 8.6) = % b. Scores of at least 9? (a sketch may be helpful here) P( x 9) = normalcdf (9, E99, 7,, 8.6) = % c. Scores less than 6? (a sketch may be helpful here) P( x 6) = normalcdf ( E 99, 6, 7,, 8.6) =.8.8% d. Scores that were more than one standard deviation away from the mean? (a sketch may be helpful here) µ σ = = 4.6 µ + σ = = 1.9 Px ( < 4.6) + Px ( > 1.9) = = % e. At what score was the rate of change of the probability density function for the scores a maximum? (a sketch may be helpful here) The location of the first inflection point on the normal pdf corresponds to the relative maximum on the rate of change of the pdf. The inflection point occurs at µ σ = =