ch05 SOLUTION.doc

9 009 8 003 7 001 6 000 013 512 the highest

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9 .009 8 .003 7 .001 < 6 .000 .013 5.12 The highest probability values are for x = 15, 16, 14, 17, 13, 18, and 12. The expected value is 25(.60) = 15. The standard deviation is 2.45. 15 + 2(2.45) = 15 + 4.90 gives a range that goes from 10.10 to 19.90. From Subscribe to view the full document.

Chapter 5: Discrete Distributions 7 table A.2, the sum of the probabilities of the values in this range (11 through 19) is .936 or 93.6% of the values which compares quite favorably with the 95% suggested by the empirical rule. 5.13 n = 15 p = .20 a) P(x = 5) = 15 C 5 (.20) 5 (.80) 10 = 3003(.00032)(.1073742) = .1032 b) P(x > 9): Using Table A.2 P(x = 10) + P(x = 11) + . . . + P(x = 15) = .000 + .000 + . . . + .000 = .000 c) P(x = 0) = 15 C 0 (.20) 0 (.80) 15 = (1)(1)(.035184) = .0352 d) P(4 < x < 7): Using Table A.2 P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) = .188 + .103 + .043 + .014 = .348 e) 5.14 n = 18 a) p =.30 µ = 18(.30) = 5.4 p = .34 µ = 18(.34) = 6.12 Chapter 5: Discrete Distributions 8 b) P(x > 8) n = 18 p = .30 from Table A.2 x Prob 8 .081 9 .039 10 .015 11 .005 12 .001 .141 c) n = 18 p = .34 P(2 < x < 4) = P(x = 2) + P(x = 3) + P(x = 4) = 18 C 2 (.34) 2 (.66) 16 + 18 C 3 (.34) 3 (.66) 15 + 18 C 4 (.34) 4 (.66) 14 = .0229 + .0630 + .1217 = .2076 d) n = 18 p = .30 x = 0 18 C 0 (.30) 0 (.70) 18 = .00163 n = 18 p = .34 x = 0 18 C 0 (.34) 0 (.66) 18 = .00056 The probability that none are in the \$500,000 to \$1,000,000 is higher because there is a smaller percentage in that category which is closer to zero. 5.15 a) Prob(x=5  = 2.3)= (2.3 5 )(e -2.3 ) = (64.36343)(.1002588) = .0538 5! (120) b) Prob(x=2  = 3.9) = (3.9 2 )(e -3.9 ) = (15.21)(.02024) = .1539 2! (2) c) Prob(x < 3  = 4.1) = Prob(x=3) + Prob(x=2) + Prob(x=1) + Prob(x=0) = (4.1 3 )(e -4.1 ) = (68.921)(.016574) = .1904 3! 6 + (4.1 2 )(e -4.1 ) = (16.81)(.016573) = .1393 2! 2 + (4.1 1 )(e -4.1 ) = (4.1)(.016573) = .0679 1! 1 Subscribe to view the full document.

Chapter 5: Discrete Distributions 9 + (4.1 0 )(e -4.1 ) = (1)(.016573) = .0166 0! 1 .1904 + .1393 + .0679 + .0166 = .4142 d) Prob(x=0  = 2.7) = (2.7 0 )(e -2.7 ) = (1)(.0672) = .0672 0! 1 e) Prob(x=1 = 5.4)= (5.4 1 )(e -5.4 ) = (5.4)(.0045) = .0244 1! 1 f) Prob(4 < x < 8 = 4.4) = Prob(x=5 = 4.4) + Prob(x=6 = 4.4) + Prob(x=7 = 4.4)= (4.4 5 )(e -4.4 ) + (4.4 6 )(e -4.4 ) + (4.4 7 )(e -4.4 ) = 5! 6! 7! (1649.162)(.012277) + (7256.314)(.012277) + (31927.781)(.012277) 120 720 5040 = .1687 + .1237 + .0778 = .3702 5.16 a) Prob(x=6 = 3.8) = .0936 b) Prob(x>7 = 2.9): x Prob 8 .0068 9 .0022 10 .0006 11 .0002 12 .0000 x > 7 .0098 c) Prob(3 < x < 9 = 4.2)= x Prob 3 .1852 4 .1944 5 .1633 6 .1143 7 .0686 8 .0360 9 .0168 3 < x < 9 .7786 d) Prob(x=0 = 1.9) = .1496 Chapter 5: Discrete Distributions 10 e) Prob(x < 6 = 2.9)= x Prob 0 .0050 1 .1596 2 .2314 3 .2237 4 .1622 5 .0940 6 .0455 x < 6 .9214 f) Prob(5 < x < 8 = 5.7) = x Prob 6 .1594 7 .1298 8 .0925 5 < x < 8 .3817 5.17 a) = 6.3 mean = 6.3 Standard deviation = 3 . 6 = 2.51 x Prob 0 .0018 1 .0116 2 .0364 3 .0765 4 .1205 5 .1519 6 .1595 7 .1435 8 .1130 9 .0791 10 .0498 11 .0285 12 .0150 13 .0073 14 .0033 15 .0014 16 .0005 17 .0002 18 .0001 19 .0000 Subscribe to view the full document.

Chapter 5: Discrete Distributions 11 b) = 1.3 mean = 1.3 standard deviation = 3 . 1 = 1.14 x Prob 0 .2725 1 .3542 2 .2303 3 .0998 4 .0324 5 .0084 6 .0018 7 .0003 8 .0001 9 .0000 c) = 8.9 mean = 8.9 standard deviation = 9 . 8 = 2.98 x Prob 0 .0001 1 .0012 2 .0054 3 .0160 4 .0357 Chapter 5: Discrete Distributions 12 5 .0635 6 .0941 7 .1197 8 .1332 9 .1317 10 .1172 11 .0948 12 .0703 13 .0481 14 .0306 15 .0182 16 .0101 17 .0053 18 .0026 19 .0012 20 .0005 21 .0002 22 .0001 d) = 0.6 mean = 0.6 standard deviation = 6 . 0 = .775 x Prob 0 .5488 1 .3293 2 .0988 3 .0198 4 .0030 5 .0004 6 .0000 Subscribe to view the full document.

Chapter 5: Discrete Distributions 13 5.18 = 2.8 4 minutes a) Prob(x=6 = 2.8) from Table A.3 .0407 b) Prob(x=0 = 2.8) = from Table A.3 .0608 c) Unable to meet demand if x > 4 4 minutes: x Prob.  • Summer '18
• Keyur Popat
• Poisson Distribution, Discrete probability distribution, C5-convertase, = P, + p, 4C3

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