# 9 009 8 003 7 001 6 000 013 students solutions manual

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Student’s Solutions Manual and Study Guide: Chapter 2Page 795.15n= 15p= .20a)P(x= 5)=15C5(.20)5(.80)10=3003(.00032)(.1073742) =.1032b)P(x> 9):Using Table A.2P(x= 10) +P(x= 11) + . . . +P(x= 15) =.000 + .000 + . . . + .000 =.000c)P(x= 0)=15C0(.20)0(.80)15=(1)(1)(.035184) =.0352d)P(4 <x<7):Using Table A.2P(x= 4) +P(x= 5) +P(x= 6) +P(x= 7) =.188 + .103 + .043 + .014 =.348e)Graph
Student’s Solutions Manual and Study Guide: Chapter 2Page 805.17a)P(x=5= 2.3) =2356436343100259120523.!(.) ( .).e=.0538b)P(x=2= 3.9) =39215210202422239.!(.) ( .).e=.1539c)P(x<3= 4.1) =P(x=3) +P(x=2) +P(x=1) +P(x=0) =413689210165736341.!(.) ( .).e=.190441216810165732241.!(.) ( .).e=.1393411410165731141.!(.) ( .).e=.0679410!10165731041.() ( .).e=.0166.1904 + .1393 + .0679 + .0166 =.4142d)P(x=0= 2.7) =270!1067211027.() ( .).e=.0672e)P(x=1= 5.4)=5415400451661154.!(.) ( .).e=.0244f)P(4 <x< 8= 4.4):P(x=5= 4.4) +P(x=6= 4.4) +P(x=7= 4.4)=445544.!.e+446!644..e+447744.!.e=
Student’s Solutions Manual and Study Guide: Chapter 2Page 81(.) ( .)1649162201227734120+(.) ( .)7256313901227734720+(,.) ( .)31927781012277345040= .1687 + .1237 + .0778 =.37025.19a)= 6.3mean =6.3Standard deviation =6.3=2.51xProb0.00181.01162.03643.07654.12055.15196.15957.14358.11309.079110.049811.028512.015013.007314.003315.001416.000517.000218.000119.0000
Student’s Solutions Manual and Study Guide: Chapter 2Page 82
Student’s Solutions Manual and Study Guide: Chapter 2Page 83b)= 1.3mean =1.3standard deviation =1.3=1.14xProb0.27251.35422.23033.09984.03245.00846.00187.00038.00019.0000
Student’s Solutions Manual and Study Guide: Chapter 2Page 84c)=8.9mean =8.9standard deviation=8.9=2.98xProb0.00011.00122.00543.01604.03575.06356.09417.11978.13329.131710.117211.094812.070313.048114.030615.018216.010117.005318.002619.001220.000521.000222.0001
Student’s Solutions Manual and Study Guide: Chapter 2Page 85d)=0.6mean =0.6standard deviation=0.6=.775xProb0.54881.32932.09883.01984.00305.00046.0000
Student’s Solutions Manual and Study Guide: Chapter 2Page 865.21=x/n= 126/36 =3.5Using Table A.3a)P(x= 0) =.0302b)P(x>6) =P(x= 6) +P(x= 7) + . . . =.0771 + .0385 + .0169 + .0066 + .0023 +.0007 + .0002 + .0001 =.1424c)P(x< 410 minutes)Double Lambda to= 7.010 minutesP(x< 4) =P(x= 0) +P(x= 1) +P(x= 2) +P(x= 3) =.0009 + .0064 + .0223 + .0521 = .0817d)P(3 <x<610 minutes)= 7.010 minutesP(3 <x<6) =P(x= 3) +P(x= 4) +P(x= 5) +P(x= 6)= .0521 + .0912 + .1277 + .1490 =.42e)P(x= 815 minutes)Change Lambda for a 15 minute interval by multiplying the original Lambdaby 3.= 10.515 minutesP(x= 815 minutes)=λxeλx!=(10.58)(e10.5)8!=.1009
Student’s Solutions Manual and Study Guide: Chapter 2Page 875.23= 1.2 collisions4 monthsa)P(x=0= 1.2):from Table A.3 =.3012b)P(x=22 months):The interval has been decreased (by ½)New Lambda == 0.6 collisions2 monthsP(x=2= 0.6):from Table A.3 =.0988c)P(x<1 collision6 months):The interval length has been increased (by 1.5)New Lambda == 1.8 collisions6 monthsP(x<1= 1.8):from Table A.3xProb.0.16531.2975.4628The result is likely to happen almost half the time (46.26%). Ship channel andweather conditions are about normal for this period.Safety awareness isabout normal for this period.There is no compelling reason to reject thelambda value of 0.6 collisions per 4 months based on an outcome of 0 or 1collisions per 6 months.
Student’s Solutions Manual and Study Guide: Chapter 2Page 885.25n= 100,000p= .00004a)P(x>7n= 100,000p= .00004):=µ=np= 100,000(.00004) = 4.0Sincen> 20 andnp<7, the Poisson approximation to this binomial problem isclose enough.

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