Chapter 8 Homework Solutions Compiled by Joe Kahlig. speed(x) freq 25 x < x < x < x < x < x < 55 5

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1 H homework problems, C-copyright Joe Kahlig Chapter Solutions, Page Chapter Homework Solutions Compiled by Joe Kahlig. (a) finite discrete (b) infinite discrete (c) continuous (d) finite discrete (e) continuous. (a) Finite discrete.,,..., (b) continuous {t = time in hours 0 t } (c) infinite discrete,,,,.... Let P(X = 6) = J then P(X = ) = J P(X = ) + P(X + 6) =. (from the histogram). solve J + J = 0. and get J = 0. Answer: (a) students 0 6 (b) histogram speed(x) freq x < x < 5 5 x < x < 5 5 x < x < 55 5 (b) prob dist. speed(x) prob x < 0 6/0 0 x < 5 /0 5 x < 0 9/0 0 x < 5 /0 5 x < 50 5/0 50 x < 55 5/0. (a) frequency table grade(x) freq 90 x 99 0 x 9 0 x 9 60 x x 59 0 x 9 0 x 9 (b) prob dist. 5 6 grade(x) freq 90 x 99 /56 6. (a) letters 5 6 (b) histogram 5/ / / 0 x 9 /56 0 x 9 /56 60 x 69 /56 50 x 59 /56 0 x 9 /56 / / 0 x 9 / (a) There can be different answers depending where you intervals start. 9. remainder 0

2 H homework problems, C-copyright Joe Kahlig Chapter Solutions, Page. tosses 9. (a) C(,0)C(,) C(5,) (b) C(,)C(,) C(5,). (a) P(X = ) = C(5,) C(,) C(,) = 0 0 (b) P(X ) =. (a) C(5,0) C(,) C(,) 9 + C(5,) C(,) C(,) + C(5,) C(,) C(,) = 0 or P(X ) = P(X = ) = C(5,) C(,0) C(,) (b) histogram E(X) = (a) 6. (a). (a) (b)..5 (b) score hearts 0 9 X E(x) = ( ) 0.05 = E(X) = $ 0. (a) x prob (b) E(x) =. so the game is not fair. 5. Let X = your net winnings and A be the cost of the game. red red 0 red X -A A-A 0-A prob If the game is fair then E(x) = 0 0 = ( A) + 6 (A) + 6 ( A) 0 = 0( A) + A A A = 0 A = 0 =. so to make it fair(or as fair as possible) charge $... (a) -.5 (b) no (c) $.50. (a) Location A: $5 Location B: $5.5.. (b) more than to 9. P(A C ) =. P(E F) = 9 9 = to. Mean =.9 Median = 5 Mode = 6. Mean =. Median = 0.5 Mode = 9 and. The fifth score is less than or equal to.. Answers will vary. I used the midpoint of each interval = Answers will vary. used the median of each interval. Estimated Mean: (a). (b) 6.56 (c) -0. (a) mean =.5 median = mode = standard deviation =. variance =.56 (b) mean =. median = mode = and 5 standard deviation = variance = 9.5

3 H homework problems, C-copyright Joe Kahlig Chapter Solutions, Page. (a) mean =.0 (b) median =.5 (c) mode = 90 (d) standard deviation = S x =.69 (e) variance = S x = 0. (f) Q = At least % of the people surveyed drink or fewer Dr. Peppers during the semester. Q = median =.5 At least 50% of the people surveyed drink.5 or fewer Dr. Peppers during the semester. Q = 90 At least 5% of the people surveyed drink 90 or fewer Dr. Peppers during the semester. 9. (a) Mean =.6 Median = Mode = (b) Q = At least % of the cars are years or younger. Q =median = At least 50% of the cars are years or younger. Q = At least 5% of the cars are years or younger. (c) sample since there are more than 000 cars on campus. (d).665 (e) between 0.99 years and.6 years (f) between 0.06 years and 5.0 years 0. (a).5 (b).606 (c) P(. X.) = 9. P(. X.) 0.6 =. note: Chebyshev s inequality doesn t really give useful information for this problem.. (a) P(9 X ) k = 05 + k to get k = Answer:.95 = 5 6 (b) solve = 05 + k to get k =. P(5 X ) = 0.99 Answer: 0.0. solve for k 0 + k. = 6 k = 5 P(9 X 6) (5/) = 0. we would expect at least or at least boxes to have between 9 and 6 paperclips. 5. n = ; p = /5; r = 6,, (a) (c) (a). (b) (a) ( ( 6) ) 6 (c) = (d) E(X) = 6 =.666 expected grade is * E(X) = (a) 0. (b) (a) n=, p= 6, r= 6 Answer: (b) n=, p = 6, r=0,,, Answer: E(x) = (a) n = 0, p = 0., r = Answer: 0.6 (b) n=0, p = 0., r =,, 9, 0 Answer: 0. (c) = n = ; p = /; r =,,, 5, 6, Answer: (a) ( ) 0 ( 5 ) 5 (b) n = 6, p = 0 5 and r = Answer: (a) (c) E(x) =.069 (b) normalcdf(-,.5, 0, ) = 0.5 (c) normalcdf(-0.5, E99, 0, ) = 0. (d) normalcdf(-e99,.5 0, ) = 0.99 (e) 0. (f) A = invnorm(0.6,0,) = 0. (g) J=invNorm(-.9,0,) = A = (a) 0.6

4 H homework problems, C-copyright Joe Kahlig Chapter Solutions, Page.6 (c) 0.95 (d) (a) (c) 0. (d) (a) 6.655% (b).5% 6. A = A = (a) (c) (a). (b).005 (c) about 9 6. (a) normalcdf(,e99,0,5) = 0.05 (c) 6 or. 6. invnorm(0.,,.5) =.05 minutes 69. (a) normalcdf(9.,e99,.,.) = (a) minimum.99 and maximum.005 (b).5% (c) so approximately 55.. (a). (c).5. (a).695 (b).9 (c) 6.66%. (a).65 (b) 9.9 so about 9. (a).066.% 5..0 so approximately years. A =.06 B =.65 C = There are two different methods to approximate a binomial distribution with a normal curve. Your answers will depend on which style your instructor used in class. Method I: uses the 0.5 adjustment to account for converting from a discrete random variable to a continuous random variable. use the numbers directly without any adjustment.. Method I: normalcdf(.5,e99,0*., 0..6) = 0.05 normalcdf(,e99,0*., 0..6) = N = 000, p = 0., q = 0.. µ = NP = 600, σ = (a) Method I: normalcdf(e99, 69.5, 600, 000..) Answer: normalcdf(e99, 69, 600, 000..) Answer: 0.96 (b) Method I: normalcdf(5.5, 69.5, 600, 000..) Answer: 0. normalcdf(55, 69, 600, 000..) Answer: Method I: normalcdf(69.5, 9.5, 00 0., 00..9) Answer: 0.0 normalcdf(0, 9, 00 0., 00..9) Answer: 0.. (a) Method I: (b) Method I: (c) Method I: (a) Method I:

5 H homework problems, C-copyright Joe Kahlig Chapter Solutions, Page 5 (b) Method I: