Unformatted text preview: bus to arrive. P ( T 5 > 0.5 ) = P ( X 0.5 â‰¤ 4 ) = P ( Poisson ( 2 ) â‰¤ 4 ) = 0.947 . OR P ( T 5 > 0.5 ) = ( ) âˆ« âˆž Î“5 . 4 1 5 5 5 4 dt t t e = âˆ« âˆž5 . 4 4 5 4 4 ! dt t t e = â€¦ b) (6) Find the probability that the fifth bus arrives during the second hour. P ( 1 < T 5 < 2 ) = P ( T 5 > 1 ) â€“ P ( T 5 > 2 ) = P ( X 1 â‰¤ 4 ) â€“ P ( X 2 â‰¤ 4 ) = P ( Poisson ( 4 ) â‰¤ 4 ) â€“ P ( Poisson ( 8 ) â‰¤ 4 ) = 0.629 â€“ 0.100 = 0.529 . OR P ( 1 < T 5 < 2 ) = ( ) âˆ«Î“ 2 1 4 1 5 5 5 4 dt t t e = âˆ«2 1 4 4 5 4 4 ! dt t t e = â€¦...
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 Spring '08
 Kim
 Statistics, Probability, Alex, 15 minutes, 30 Minutes, Bus stop

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