This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
This** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**Unformatted text preview: **238 ? =0)=(5 (± ^ ² ^( −5 ) )/0!= .0067 ? = (± 1 )=(5^ 1 ^(−5))/ ² 1 != ° (°= ( 2 )=(5^ 2 °^(−5))/ 2 != ° (°=3)=(5^ ( 3 °^(−5))/3!=125(.0067)/6= .1396 ° (°= ( 4 )=(5^ 4 °^(−5))/ 4 != ° (°= ( 5 )=(5^ 5 °^(−5))/ 5 != ° (°= ( 6 )=(5^ 6 °^(−5))/ 6 != Answer average tme per order = 2 minuTes = 2/60(hr) = 1 / mean service tme =60/2 = 30 / hr Q: WhaT percenTage of The orders will Take less Than one minuTe To process A P(±<=1) = P(±<=1/60 hr) Q: whaT percenTage of The orders will be processed in exacTly 3 minuTes? A: since The exponental disTributon is a contnuous disTributon The probabiliTy a service tme exacTly equals any Q: whaT percenTage of The orders will require more Than 3 minuTes To process? (slide 24) A: ° ? ≤1)=1 − ( (± ² ° ^(−30/(1/60)) = 1 - .6065 = .3935 or 39.35% °° (°≥3/60)=° ( = ° ^(−30/(3/60)) = = .3935 or 39.35% specifc value is 0...

View
Full Document

- Spring '16
- Dr. Chao
- Normal Distribution, Probability theory, orders, average arrival rate