Question

An airline has determined that 20% of its international flights are not on time. If the airline randomly selects the next 80 flights, using a suitable approximation, what is the probability that 17 or more flights will not be on time?

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 80 * 0.2 ) = 16
Variance = n * P * Q = ( 80 * 0.2 * 0.8 ) = 12.8
Standard deviation = √(variance) = √(12.8) = 3.5777

P ( X >= 17 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 17 - 0.5 ) =P ( X > 16.5 )

X ~ N ( µ = 16 , σ = 3.5777 )
P ( X > 16.5 ) = 1 - P ( X < 16.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 16.5 - 16 ) / 3.5777
Z = 0.14
P ( ( X - µ ) / σ ) > ( 16.5 - 16 ) / 3.5777 )
P ( Z > 0.14 )
P ( X > 16.5 ) = 1 - P ( Z < 0.14 )
P ( X > 16.5 ) = 1 - 0.5557
P ( X > 16.5 ) = 0.4443