2. Suppose Swift Jet Airlines estimates that it’s no-show rate is approximately 9%. What is the probability that no passenger will be bumped if the airline booked 230 passengers on a 200-seat plane?
2. Suppose Swift Jet Airlines estimates that it’s no-show rate is approximately 9%. What is the probability that no passenger will be bumped if the airline booked 230 passengers on a 200-seat plane?
Homework Answers
| n= | 230 | p= | 0.9100 | |
| here mean of distribution=μ=np= | 209.3 | |||
| and standard deviation σ=sqrt(np(1-p))= | 4.3402 | |||
| for normal distribution z score =(X-μ)/σx | ||||
| therefore from normal approximation of binomial distribution and continuity correction: | ||||
probability that no passenger will be bumped if the airline booked 230 passengers on a 200-seat plane :
| probability = | P(X<200.5) | = | P(Z<-2.03)= | 0.0212 |
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2. Suppose Swift Jet Airlines estimates that it’s no-show rate is approximately 9%. What is the probability that no passenger will be bumped if the airline booked 230 passengers on a 200-seat plane?
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