Question

A certain flight arrives on time 81 percent of the time. Suppose 175 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 152 flights are on time.

​(b) at least 152 flights are on time.

​(c) fewer than 141flights are on time.

​(d) between 141 and 148 inclusive are on time.

Answer 1

Mean = n * P = ( 175 * 0.81 ) = 141.75
Variance = n * P * Q = ( 175 * 0.81 * 0.19 ) = 26.9325
Standard deviation = = 5.1897

Part a)

P ( X = 152 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 152 - 0.5 < X < 152 + 0.5 ) = P ( 151.5 < X < 152.5 )

P ( 151.5 < X < 152.5 )
Standardizing the value

Z = ( 151.5 - 141.75 ) / 5.1897
Z = 1.88
Z = ( 152.5 - 141.75 ) / 5.1897
Z = 2.07
P ( 1.88 < Z < 2.07 )
P ( 151.5 < X < 152.5 ) = P ( Z < 2.07 ) - P ( Z < 1.88 )
P ( 151.5 < X < 152.5 ) = 0.9808 - 0.9699
P ( 151.5 < X < 152.5 ) = 0.011

Part b)

P ( X >= 152 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 152 - 0.5 ) =P ( X > 151.5 )

P ( X > 151.5 ) = 1 - P ( X < 151.5 )
Standardizing the value

Z = ( 151.5 - 141.75 ) / 5.1897
Z = 1.88

P ( Z > 1.88 )
P ( X > 151.5 ) = 1 - P ( Z < 1.88 )
P ( X > 151.5 ) = 1 - 0.9699
P ( X > 151.5 ) = 0.0301

Part c)

P ( X < 141 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 141 - 0.5 ) = P ( X < 140.5 )

P ( X < 140.5 )
Standardizing the value

Z = ( 140.5 - 141.75 ) / 5.1897
Z = -0.24

P ( X < 140.5 ) = P ( Z < -0.24 )
P ( X < 140.5 ) = 0.4052

Part d)

P ( 141 <= X <= 148 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 141 - 0.5 < X < 148 + 0.5 ) = P ( 140.5 < X < 148.5 )

P ( 140.5 < X < 148.5 )
Standardizing the value

Z = ( 140.5 - 141.75 ) / 5.1897
Z = -0.24
Z = ( 148.5 - 141.75 ) / 5.1897
Z = 1.3
P ( -0.24 < Z < 1.3 )
P ( 140.5 < X < 148.5 ) = P ( Z < 1.3 ) - P ( Z < -0.24 )
P ( 140.5 < X < 148.5 ) = 0.9033 - 0.4048
P ( 140.5 < X < 148.5 ) = 0.4985