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Unit 4 Quiz 2 Review Answers 1. a. P(X = 3) = 0.28 b. P(X ≤ 3) = 0.16 + 0.22 + 0.28 = 0.66 c. P(1<X≤4) = 0.22 + 0.28 + 0.20 = 0.70 d. E(X) = 2.94 units SD(X) = 1.2714 units 2. Probability model for the payout of a prize X = payout P(X=x) -$2 1496/1500 $73 1/1500 $148 1/1500 $248 1/1500 $498 1/1500 For (b) and (c), just put X-values in L1, probabilities in L2 and then do 1-var stats L1, L2: b. E(X) = -$1.35 c. SD(X) = $15.06 d. E(X + X + X) = -1.35 + -1.35 + -1.35 = -$4.05 SD(X + X + X) = 15.062 15.062 15.062 $26.08 3. SAT Scores M – score for a randomly selected male F – score for a randomly selected female E(M) = 1532 E(F) = 1506 SD(M) = 312 SD(F) = 304 a. M + F; E(M + F) = 1532 + 1506 = 3038 points SD(M + F) = 3122 3042 435.61 points b. M – F E(M – F) = 1532 – 1506 = 26 points SD(M – F) = 3122 3042 435.61 points c. Assume a normal model with M – F ~ N(26, 435.61). Female scored higher than male P(F > M) P(M -- F < 0) P(M –F < 0) = normalcdf(-E99, 0, 26, 435.61) = 0.4762 There is a 47.62% chance that a randomly selected female scored higher than a randomly selected male on the SAT. 4. Random Variables a. -2X E(-2X) = -2(12) = -24 SD(-2X) = 2(5) = 10 b. 4Y – 7 E(4Y – 7) = 4(18) – 7 = 65 SD(4Y – 7) = 4(8) = 32 c. X + Y E(X + Y) = 12 + 18 = 30 d. X – Y E(X – Y) = 12 – 18 = -6 SD(X + Y) = 52 82 9.43 e. X1 + X2 E(X1 + X2) = 12 + 12 = 24 SD(X1 + X2) = SD(X – Y) = X Y Mean 12 18 SD 5 8 52 82 9.43 f. 2X – 4Y E(2X – 4Y) = 2(12) – 4(18) = -48 52 52 7.07 E(2X – 4Y) = 22 (52 ) 42 (82 ) 33.53 5. The probability distribution below represents (X) the length of long distance calls in minutes. X P(X) 5 .1 10 .2 15 .3 20 .3 25 .1 a. What is the expected length of a long distance call? What is the standard deviation? E(X) = 15.5 mins SD(X) = 5.679 mins b. Suppose that there is an initial connection charge of $.60 and an additional charge of $.015 for each minute. What is the expected cost and standard deviation of the cost of a long distance call? E(0.60 + 0.015X) = 0.60 + 0.15(15.5) = $0.8325 SD(0.60 + 0.015X) = SD(0.015X) = $0.0852 c. The expected cost of sending a text message is $.20 with a standard deviation of $.10. What is the mean and standard deviation for the cost of making a long distance call and sending a text message? E(C) = $0.8325 E(T) = $0.20 E(C + T) = $1.0325 SD(C) = $0.0852 SD(T) = $0.10 SD(E + T) = $0.1314 d. Which costs more, the call or the text? How much more? With what standard deviation? E(C – T) = $0.6325 Calls cost more, on average $0.6325 with std. dev of $0.1314 SD(E – T) = $0.1314 6. Check: Bernoulli? 1. 2 Outcomes – has jumper cables or doesn’t have jumper cables 2. p = 0.40 and p is constant 3. 10% Condition – we can assume independence as long as the number of drivers asked is less than 10% of all drivers. a. G(0.40) b. G(0.40) c. Geom(0.40) E(X) = P(X = 7) = geometpdf(0.40, 7) = 0.0187 P(X < 5) = geometcdf(0.40, 5) = 0.92224 1 1 2.5 drivers p 0.40 d. B(8,0.40) P(X = 3) = binompdf(8, 0.4, 3) = 0.2787 e. B(6,0.40) P X 4 1 P( X 3 ) 1 binomcdf ( 6,0.4,3 ) 0.1792 f. B(10,0.40) P X 6 1 P( X 5 ) 1 binomcdf (10,0.4,5 ) 0.1662 g. B(12,0.40) E(X) = np = 12(0.40) = 4.8 drivers h. B(80, 0.40) i. Check? np & nq > 10 (80)(0.40) & (80)(0.60) > 10 ii. N(32, 4.382) iii. P(X < 30) = normalcdf(-E99, 30, 32, 4.382) = 0.324 iv. P(42 < X < 61) = normalcdf(42, 61, 32, 4.382) = 0.01124 YES! 7. p = 0.15 a. B(6, 0.15) P(X = 2) = 0.1762 b. B(8, 0.15) P(X > 3) = 1 – P(X < 2) = 0.1052 c. G(0.15) P(X = 4) = 0.09212 d. G(0.15) P(X = 5 U 6) = P(X = 5) + P(X = 6) = 0.0783 + 0.0666 = 0.1449 e. B(10, 0.15 P(X < 2) = P(X < 1) = 0.5443 f. B(12, 0.15) E(X) = 1.8 passes SD(X) = 1.237 passes g. G(0.15) E(X) = 6.667 passes h. B(25, 0.15) P(9 < X < 14) = P(10 < X < 13) = P(X < 13) – P(X < 9) = 0.00214 i. B(70, 0.15) i. Check? np & nq > 10 (70)(0.15) & (70)(0.150) > 10 YES! ii. N(10.5, 2.987) iii. P(X = 20) = binompdf(70, 0.15, 20) = 0.0016 Note: with the normal model you cannot calculate exact probabilities (ones with “equal to”). It will give you 0 as an answer. You must go back to binompdf to calculate exact probabilities of binomials. iv. P(11 < X < 29) = normalcdf(11, 29, 10.5, 2.987) = 0.4335