Chapter 8 Homework Solutions Compiled by Joe Kahlig

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1 homewk problems, B-copyright Joe Kahlig Chapter Solutions, Page Chapter omewk Solutions Compiled by Joe Kahlig You are counting the number of games and there are a limited number of games in a tennis match. Answer: Finite discrete (b) your counting the nubmer of tickets. Answer: Infinite discrete (c) Time is an interval and it doesn t skip values. Answer: Continuous (d) The number may be very large(hopefully), but it is still only a fixed number. Answer: Finite discrete (e) Temperature is an interval and it doesn t skip values. Answer: Continuous. There are = non-heart cards in a deck, so the maximum number of cards you could draw is without drawing a heart. So the wst case scenario is 0 cards drawn. Answer: Finite discrete. Values: X =,,..., There are a total of cards that will be made. Three of them will have a wd with three letters: Get, Its, fun. Answer: letters (b) histogram / / / / / (b) Continuous Values: {x = time in hours 0 X } (c) You could always roll a one, so it might not happen that you roll a six. Answer: Infinite discrete Values: X =,,,,.... The areas of the rectangles must add to one since the rectangles represent. The missing rectangle has an area of 0.. Answer: = = 0.0. Let P(X = ) = J then P(X = ) = J P(X = ) P(X+) = (from the histogram). P(X = )+P(X +) = 0. J +J = 0. and get J = 0. Answer: 0. = P(X = )+P(X = )+P(X = ). Divide the frequency by the total number of students who have waited to get relative frequency( ). students 0 0 (b) histogram. There can be different answers depending where your intervals start. speed(x) freq x < 0 0 x < x < 0 0 x < x < 0 0 x < (b) prob dist. speed(x) prob x < 0 /0 0 x < /0 x < 0 /0 0 x < /0 x < 0 /0 0 x < /0. frequency table

2 homewk problems, B-copyright Joe Kahlig Chapter Solutions, Page grade(x) freq 0 x 0 0 x 0 x 0 x 0 0 x 0 x 0 x (b) prob dist. grade(x) freq 0 x 0/ 0 x / 0 x / 0 x 0/ 0 x / 0 x / 0 x /. Remember that the remainder is what is left over after perfming long division(by hand). F example: divide by has a remainder of since goies into two times(this gives = ) and will be left over. remainder 0 0. The tree shows the experiment. Notice the tree stops on the third level since either a head is tossed the coin has been tossed three times. / / T / / Toss T / Tosses Tosses / T Tosses Use the branches to get the. Answer: tosses. P(X = 0) = C(,0)C(,) C(,) (b) P(X = ) = C(,)C(,) C(,). P(X = ) = C(,)C(,) C(,) = 0 0 (b) P(X ) = C(,0)C(,) C(,) + C(,)C(,) C(,) + C(,)C(,) C(,) = 0 0 P(X ) = P(X = ) = C(,)C(,0) C(,). E(x) = = (b) histogram To calculate P(X = 0) remember that the probabilities must add to. E(X) = =.. Write out the cards and give the sce to each card. Note: the der of the numbers is not imptant. Card Sce Card Sce Card Sce,,, 0,, 0,,,,, Answer: sce (b) E(x) = =.. The probabilities may be computed using a tree combinations. 0 hearts 0 0 (b) E(x) = = 0.. Use a dice chart to find the probabilities. Green Die Red Die 0 hearts (b) E(x) = E(X) =

3 homewk problems, B-copyright Joe Kahlig Chapter Solutions, Page. Note: X is the net winnings. (b). X 00. X = profit on a chip. X Answer: E(x) = 0.+( )0.0 =. 0. X is your net winnings. hearts E(X) = ( )/+( )/+( )/+/ E(X) =. Use a tree to set up the distribution. / / T.0 / / / / net winnings X prob (b) E(x) =. so the game is not fair.. Use a tree combinations to find the probabilities. X is your net winnings and A be the cost of the game. red red 0 red X -A A-A 0-A prob 0 0 If the game is fair then E(x) = 0 0 = 0 ( A)+ 0 (A)+ ( A) 0 = 0( A)+A 0A A = 0 A = 0 =. So to make it fair( as fair as possible) charge $... X is the your net winnings. X. (b) No, the expected winnings are negative. F this problem the game favs the person running the game. (c) Let A =Price of the game, then solve the following equaiton, X A A A 0 = ( A)/+( A)/+( A)/ 0 = ( A)+( A)+( AA) A =. Answer: $.0. X is the revenue at each location. Location A X Location B X Expected value of each location: Location A: =$ Location B: = $. (b) Total revenue at location B is 00. = 0 me than 0 = 0 people. + =. + =. simplify P(J) P(J C ) = = Answer: to. P(A C ) = + =. P(E) = 0 and P(F) =. Since E and F are independent, P(E F) = P(E)P(F) P(E F) = 0 = 0 0. P(E) = 0. prob of th card is a heart given the infmation is Answer: to. Mean =. Median = Mode =. Mean =. Median = 0. Mode = and. The fifth sce is less than equal to since is the median and there are sces that are above this number.

4 homewk problems, B-copyright Joe Kahlig Chapter Solutions, Page. Answers will vary. I used the middle of each interval =.0. Answers will vary. used the middle of each interval. Estimated Mean: 0.. Enter the x-value in list and the frequency in list. use the command: -Var Stats L,L mean: x =. median = mode = standard deviation: σ x =. variance: (σ x ) =. (b) mean: x =. median = mode = and standard deviation: σ x =. variance: (σ x ) =.. Enter the x-value in list and the frequency in list. use the command: -Var Stats L,L mean: x =.0 (b) median =. (c) mode = 0 (d) standard deviation: S x =. (e) variance: S x = 00. (f) Q = At least % of the people surveyed drink fewer Dr. Peppers during the semester. Q = median =. At least 0% of the people surveyed drink. fewer Dr. Peppers during the semester. Q = 0 At least % of the people surveyed drink 0 fewer Dr. Peppers during the semester.. Answers will vary. I used the middle of each interval. mean =. (b) standard deviation: σ x =. (c) Enter the age in list and the frequency in list. use the command: -Var Stats L,L Mean =. Median = Mode = (b) Q = At least % of the cars are years younger. Q =median = At least 0% of the cars are years younger. Q = At least % of the cars are years younger. (c) Sample since there are me than 000 cars on campus. (d) S x =. (e) mean +S x =. mean S x = 0. Between 0. years and. years (f) mean +.S x =.0 mean.s x = 0.0 Between 0.0 years and.0 years. Create a distribution from the histogram. Enter the x-values in list and the in list. use the command: -Var Stats L,L E(x) = x =. (b) σ x =.0 (c) varience = (σ x ) = Use Chebychev s inequality. µ+kσ =. 0+k. =. k = P(. X.) =. Use Chebychev s inequality. µ+kσ =. +k. =. k = 0. P(. X.) 0. =. Note: Chebyshev s inequality doesn t really give useful infmation f this problem.. Use Chebychev s inequality. µ+kσ = = 0+k k = P( X ) Answer:. = (b) Want to compute: P(X < )+P(X > ) notice that: P(X < )+P(X > ) = P( X ) µ+kσ = = 0+k k = 0 P( X ) 0 = 0. Answer: 0.0. Use Chebychev s inequality. µ+kσ = 0 00+k. = 0 k = P( X 0) (/) = 0. We would expect at least at least boxes to have between and 0 paperclips.

5 homewk problems, B-copyright Joe Kahlig Chapter Solutions, Page ). ( ( ) (b) C(,) C(,) C(,) binompdf(,/,) + binompdf(,/,) + binompdf(,/,) binomcdf(,/,) binompdf(,/,) Answer: binompdf(0,0.,)= C(0,)0. 0. Answer: (b) binomcdf(0,0.,) = 0. (c) binomcdf(0,0.,0) binomcdf(0,0.,) Answer: 0. (d) binomcdf(0,0.,0) binomcdf(0,0.,) Answer: 0.. Note: expected value is an average so do not round the answer. E(X) = np = 00. =. (b) E(X) = np = 00. =. ). ( ( ) (b) binomcdf(0, /,) = 0.0 (c) binompdf(0,/,) + binompdf(0,/,) + binompdf(0,/,) C(0,) ( ( ) ( ) +C(0,) ( ) + ) C(0,) ( ( ) ) Answer: 0. (d) expected umber of questions crect is 0 =. expected grade is 0 * E(X) =. 0. binompdf(,0.0,) = C(,) Answer: 0. (b) E(x) = 0.0 =. Note: expected value is an average so do not round the answer.. binompdf(,, ) = C(,)( ( ) Answer: 0.00 (b) binomcdf(,, )= 0.. E(x) = 0 =. = 0. binomcdf(,0.,) binomcdf(,0.,) = 0.. binompdf(0,0., ) = C(0,)0. 0. Answer: 0. (b) binomcdf(0,0.,0) binomcdf(0,0.,) Answer: 0.0 ) (c) binompdf(0,0., 0) + binompdf(0,0., ) + binompdf(0,0., ) + binompdf(0,0., ) + binompdf(0,0., ) Answer: 0.. µ = 0. = σ = 0.. =. (b) within standard deviation means µ σ X µ+σ.0 X. x =,0,,,,, binomcdf(0,0.,) binomcdf(0,0.,) Answer: 0. (c) X =,,,..., binomcdf(0,0.,) binomcdf(0,0.,) Answer: 0.. binomcdf(,, ) binomcdf(,, ) Answer: 0.00 (b) binomcdf(,, ) binomcdf(,, ) Answer: binomcdf(,, ) binomcdf(,, ) Answer: 0.0. ( 0 ) ( ) (b) binompdf(, 0, ) Answer: 0.0 (c) E(x) = 0 =.0. nmalcdf(.,e,0,) = 0.0 (b) nmalcdf(,., 0, ) = 0. (c) nmalcdf( 0., E, 0, ) = 0. (d) nmalcdf( E,. 0, ) = 0. (e) 0, since z is a continuous random variable. (f) nmalcdf( E,,0,) + nmalcdf(.,e,0,) Answer: 0. (g) A = invnm(0.,0,) = 0. (h) J=invNm(.,0,) = area not between A and A is 0. = 0. Area at each end of the graph is 0. = 0. A = invnm(0.+0.,0,) =.. nmalcdf(,,00,0) = 0. (b) nmalcdf(,0,00,0) = 0. (c) nmalcdf(,e,00,0) = 0. (d) A = invnm(0.,00,0) =.. nmalcdf(,,0,) = 0.

6 homewk problems, B-copyright Joe Kahlig Chapter Solutions, Page (b) nmalcdf(0,,0,) = 0. (c) nmalcdf( E,,0,) = 0. (d) zero since X is a ocntinuous random variable (e) B = invnm(.,0,) =.. µ+.σ = +. = µ.σ =. = nmalcdf(,,,) = 0. Answer:.% (b) µ+σ = + = nmalcdf(, E,,) = 0.0 Answer:.%. st. dev = var = = area to the left of X= nmalcdf( E,,,) = 0. Answer: A = invnm(0.+0.,,) = 0.0. area to the left of X=0 nmalcdf( E,0,0,0) = 0. Area to the right of B is = 0.0 Area to the left of A is..0 = 0. Answer: A =invnm(0.,0,0) =.. nmalcdf( E,,0,0) = 0.. nmalcdf(000,e,000,00) = 0.00 (b) nmalcdf(00,000,000,00) = 0. (c) binompdf(,0.,) = σ = = 0 (b) nmalcdf(0.,.00,.00,0.00) = 0. Accept =.% Answer: 00-. =.% (c) =.. nmalcdf(0,e,.,.) = 0. (b) 0, since this is a continous random variable (c) nmalcdf(,,.,.) = 0.. nmalcdf(,e,., 0.) = 0. (b) nmalcdf(.,.,., 0.) = 0.0 (c) µ+.σ =.+.0. =. µ.σ =.0. =. nmalcdf(.,.,., 0.) = 0. Answser:.%. nmalcdf(,e,, ) = 0. (b) namlcdf( E, 0,,) = =. Answer: about. nmalcdf(, E,, ) = 0.0 (b) nmalcdf( E,,,) = 0.00 Answer: 0.%. nmalcdf(., E,., 0.) = =.0 Answer: approximately. invnm(0.0,0, /) =. years. A = invnm( 0.0,,) =.0 B = invnm( ,,) =. C = invnm( ,,) =. nmalcdf(0, E,000,0) = 0. (b) binompdf(, 0.,) = 0.00 (c) 000. =. approximately. nmalcdf(,e,0,) = 0.0 (b) since the random variable is continuous, the that it takes exactly 0 minutes is zero. (c) nmalcdf(,,0,) = =. approximately. 0. invnm(0.,0,.) =.00 minutes. nmalcdf(.,e,.,.) = 0.0 (b) 0, since this is a continous random variable. minimum length = = 0. maximum length = = =.00