Question

A certain flight arrives on time 81 percent of the time. Suppose 149 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly107 flights are on time.​

(b) at least 107 flights are on time.​

(c) fewer than 122 flights are on time.​

(d) between 122 and134 inclusive are on time.

Answer 1

Mean = n * P = ( 149 * 0.81 ) = 120.69
Variance = n * P * Q = ( 149 * 0.81 * 0.19 ) = 22.9311
Standard deviation = √(variance) = √(22.9311) = 4.7886

Part a)

P ( X = 107 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 107 - 0.5 < X < 107 + 0.5 ) = P ( 106.5 < X < 107.5 )

X ~ N ( µ = 120.69 , σ = 4.7886 )
P ( 106.5 < X < 107.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 106.5 - 120.69 ) / 4.7886
Z = -2.96
Z = ( 107.5 - 120.69 ) / 4.7886
Z = -2.75
P ( -2.96 < Z < -2.75 )
P ( 106.5 < X < 107.5 ) = P ( Z < -2.75 ) - P ( Z < -2.96 )
P ( 106.5 < X < 107.5 ) = 0.0029 - 0.0015
P ( 106.5 < X < 107.5 ) = 0.0014

Part b)

P ( X >= 107 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 107 - 0.5 ) =P ( X > 106.5 )

X ~ N ( µ = 120.69 , σ = 4.7886 )
P ( X > 106.5 ) = 1 - P ( X < 106.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 106.5 - 120.69 ) / 4.7886
Z = -2.96
P ( ( X - µ ) / σ ) > ( 106.5 - 120.69 ) / 4.7886 )
P ( Z > -2.96 )
P ( X > 106.5 ) = 1 - P ( Z < -2.96 )
P ( X > 106.5 ) = 1 - 0.0015
P ( X > 106.5 ) = 0.9985

Part c)

P ( X < 122 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 122 - 0.5 ) = P ( X < 121.5 )

X ~ N ( µ = 120.69 , σ = 4.7886 )
P ( X < 121.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 121.5 - 120.69 ) / 4.7886
Z = 0.17
P ( ( X - µ ) / σ ) < ( 121.5 - 120.69 ) / 4.7886 )
P ( X < 121.5 ) = P ( Z < 0.17 )
P ( X < 121.5 ) = 0.5675

Part d)

P ( 122 <= X <= 134 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 122 - 0.5 < X < 134 + 0.5 ) = P ( 121.5 < X < 134.5 )

X ~ N ( µ = 120.69 , σ = 4.7886 )
P ( 121.5 < X < 134.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 121.5 - 120.69 ) / 4.7886
Z = 0.17
Z = ( 134.5 - 120.69 ) / 4.7886
Z = 2.88
P ( 0.17 < Z < 2.88 )
P ( 121.5 < X < 134.5 ) = P ( Z < 2.88 ) - P ( Z < 0.17 )
P ( 121.5 < X < 134.5 ) = 0.998 - 0.5672
P ( 121.5 < X < 134.5 ) = 0.4309