Question

certain flight arrives on time 85 percent of the time. Suppose 127 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that ​(a) exactly 120 flights are on time. ​(b) at least 120 flights are on time. ​(c) fewer than 100 flights are on time. ​(d) between 100 and 120​, inclusive are on time.

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 127 * 0.85 ) = 107.95
Variance = n * P * Q = ( 127 * 0.85 * 0.15 ) = 16.1925
Standard deviation = √(variance) = √(16.1925) = 4.024

a)

P ( X = 120 ) = ?
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 120 - 0.5 < X < 120 + 0.5 ) = P ( 119.5 < X < 120.5 )

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 119.5 - 107.95 ) / 4.024
Z = 2.87
Z = ( 120.5 - 107.95 ) / 4.024
Z = 3.12
P ( 2.87 < Z < 3.12 )
P ( 119.5 < X < 120.5 ) = P ( Z < 3.12 ) - P ( Z < 2.87 )
P ( 119.5 < X < 120.5 ) = 0.9991 - 0.9979
P ( 119.5 < X < 120.5 ) = 0.0011

b)

P ( X >= 120 ) = ?
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 120 - 0.5 ) =P ( X > 119.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 119.5 - 107.95 ) / 4.024
Z = 2.87
P ( ( X - µ ) / σ ) > ( 119.5 - 107.95 ) / 4.024 )
P ( Z > 2.87 )
P ( X > 119.5 ) = 1 - P ( Z < 2.87 )
P ( X > 119.5 ) = 1 - 0.9979
P ( X > 119.5 ) = 0.0021

c)

P ( X < 100 ) = ?
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 100 - 0.5 ) = P ( X < 99.5 )

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 99.5 - 107.95 ) / 4.024
Z = -2.1
P ( ( X - µ ) / σ ) < ( 99.5 - 107.95 ) / 4.024 )
P ( X < 99.5 ) = P ( Z < -2.1 )
P ( X < 99.5 ) = 0.0179

d)

P ( 100 <= X <= 120 ) = ?
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 100 - 0.5 < X < 120 + 0.5 ) = P ( 99.5 < X < 120.5 )

Standardizing the value
Z = ( X - µ ) / σ
Z = ( 99.5 - 107.95 ) / 4.024
Z = -2.1
Z = ( 120.5 - 107.95 ) / 4.024
Z = 3.12
P ( -2.1 < Z < 3.12 ) =?
P ( 99.5 < X < 120.5 ) = P ( Z < 3.12 ) - P ( Z < -2.1 )
P ( 99.5 < X < 120.5 ) = 0.9991 - 0.0179
P ( 99.5 < X < 120.5 ) = 0.9812