Answer:
a) [tex]P(X \ge 490) = 0.628[/tex]
b) [tex]P(X \leq 190) = 0.165[/tex]
Step-by-step explanation:
a) This is an exponential distribution question since it is a poisson type problem
Mean rate is 0.95 per 1000 passengers
Mean rate, [tex]\lambda = 0.95/1000[/tex]
[tex]\lambda = 0.00095[/tex] per passenger
Probability that at least 490 passengers will have their baggage handled properly before the next mishandling occurs :
[tex]P(X \ge x) = e^{- \lambda x}[/tex]
[tex]P(X \ge 490) = e^{-490 * 0.00095}[/tex]
[tex]P(X \ge 490) = e^{-0.4655}\\P(X \ge 490) = 0.628[/tex]
b) Probability that the number will be fewer than 200 passengers
[tex]P(X \leq 190) = 1 - P(X \geq 190)\\P(X \leq 190) = 1 - e^{-190 \lambda} \\P(X \leq 190) = 1 - e^{-190 *0.00095}[/tex]
[tex]P(X \leq 190) = 1 - 0.835\\P(X \leq 190) = 0.165[/tex]
