# A4 - Assignment 4 Weihan Ni A20252394 MBA 507 Decision...

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Assignment 4 Weihan Ni A20252394 MBA 507 Decision Making 9.6 a) (91.80-85.60)/6.20=1, from table z, we find that area z>=1=0.1587=15.87% b) (98.00-85.60)/6.20=2, from table z, we find that area z<=2=0.9772=97.72% c) (73.20-85.60)/6.20=-2, from table z, we find that area z<=-2=0.0228=2.28% (98.00-85.60)/6.20=2, from table z, we find that area z<=2=0.9772=97.72% Area -2<=Z>=2 = 97.72%-2.28%=95.44% d) (93.00-85.60)/6.20=1.2, from table z, we find that area z>=1.2=0.1151=11.51% (70.00-85.60)/6.20=-2.5, from table z, we find that area z<=-2.5=0.0062=0.62% So, Price below \$70 is more unusual than above \$93. 9.8 a) From table z, we find that lowest 16% is below z=-1 85.60+6.20*(-1)=\$79.4, below \$79.4 b) From table z, we find that highest 0.15% is above z=3 85.60+6.20*3=\$104.2, above \$104.2 c) (1-68%)/2=16% From table z, we find that lowest 16% is below z=-1, highest 16% is above z=1. 85.60+6.20*(-1)=\$79.4 85.60+6.20*1=\$91.8 Between \$79.4 and \$91.8 d) Normal Model, so the highest half is above mean \$85.60 9.30 a) (40000-32000)/2500=3.2, from table z, we find that area z>=3.2 = 0.0007=0.07% So the possibility of the hope is 0.07%, it’s not reasonable. b) (30000-32000)/2500=-0.8, from table z, we find that area z<=-0.8 =0.2119=21.19% c) (30000-32000)/2500=-0.8, from table z, we find that area z>=-0.8 =0.7881=78.81% (35000-32000)/2500=1.2, from table z, we find that area z>=1.2 =0.1151=11.51% 78.81%-11.51%=67.30% d) From table z, we find that lowest 25% is below z=-0.67, highest 25% is above z=0.67. IQR=middle half coverage=2*0.67*2500=3350 miles

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## This note was uploaded on 11/07/2011 for the course ECON 101 taught by Professor Johnson during the Spring '11 term at Sciences Po.

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A4 - Assignment 4 Weihan Ni A20252394 MBA 507 Decision...

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