Nagy_BUS207.920_Ch7&8

# Nagy_BUS207.920_Ch7&8 - Chris Nagy BUS207.920 Chapter 7...

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Chris Nagy BUS207.920 10/16/2007 Chapter 7, pages 247-250 #37-65 EOO

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37 a mean= (11.96+12.05)/2=12.005 b c P(X<12) = 1/(12.05-11.96)*[(12.00-11.96)/1]=.04/.09= .44 d P(X>11.98) = 1/(12.05-11.96) * [(12.05-11.98)/1] = .07/.09=.78 e 100% since all cans have more than 11.00 ounces 41 a (170-180)/25= -0.4 for net sales; (1850-1500)/120= 2.92 for employees b Net sales are 0.4 standard deviations below mean. Employees is 2.92 standard deviations above me c 65.54% of aluminum fabricators have greater net sales than Clarion: 0.1554+0.5000. 0.18% have mo 45 a 0.5000-0.1985= 0.3015 b 0.4564-0.1985= 0.2579 c 0.5000-0.4989= 0.0011 d 1280+1.28(420)= 1818 49 1.65 = (X-4000)/60 X=4099 53 a mean=60(0.64)=38.4 standard deviation= √13.824=3.72 (31.5-38.4)/3.72= -1.85, area is .4678 0.5000+.04678=0.9678 b (43.5-38.4)/3.72=1.37, area is 0.4147; 0.5000-0.4147=0.0853 c 0.4441+0.3643=0.8084 d 0.4495-0.4147=0.0348 57 a 1.65=(45- mean)/5, mean=36.75 b 1.65=(45- mean)/10, mean=28.5 c z= (30-28.5)/10 = 0.15: 0.5000+0.0596=0.5596
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Nagy_BUS207.920_Ch7&8 - Chris Nagy BUS207.920 Chapter 7...

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