# assignment 5 answers - 2 Mean= 600 Standard Deviation= 200...

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2. Mean= 600 Standard Deviation= 200 10000 customers in 16 hours Therefore, (10000/16) customers per hour= 625 Z= (625-600)/200= 0.125 P(z>0.125) 1-P(z<0.125) 1-0.549738 0.450262 Ans. 0.4503 3. Mean= 275 Standard Deviation= 75 1500 faxes in 5 days Therefore, (1500/5) faxes per day= 300 Z= (300-275)/75= 0.33 P(z>0.33) 1-P(z<0.33) 1-0.6293 0.3707 Ans. 0.3707 4. n= 1000 p= 30% Mean= np= 1000x.30= 300 Standard Deviation= = = 0.01449 P(>0.32) P(z>(0.32-0.30)/0.01449) P(z>1.38) 1-P(z<1.38) 1-0.9162 0.0838 Ans. 0.0838

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5. Worker 1: Mean= 75 Standard Deviation= 20 Worker 2: Mean= 65 Standard Deviation= 21 Mean= 75-65= 10 Standard Deviation= = 12.97 P(z> ) P(z> -0.77) 1-P(z<-0.77) 1-0.2206 0.7794 6. Waiters/Waitresses who introduce themselves: Mean= 18 Standard Deviation= 3 Waiters/Waitresses who don’t introduce themselves: Mean= 15 Standard Deviation= 3 Mean= 18-15= 3 Standard Deviation= = 1.34 P(z> ) P(z> -2.24) 1-P(z<-2.24) 1-0.0125 0.9875 8. a) E(x)= 1x 0.10+2x0.30+3x0.25+4x0.20+5x0.15= 3 Variance= (x0.10+x0.30+x0.25+x0.20+x0.15)- = 1.5 Standard Deviation= = 1.22 b) E(x)= 150x3= 450 Variance= x1.5= 33750 Standard Deviation=

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• Fall '12
• HilaCohen;OlgaKaminer;AlanMarshall;AlexShoumarov,DoritNevo
• Environmental standard, standard deviation=

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